## Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion

We can state whether an object is in a state of rest or motion only with reference to another object. The object which is taken as reference is the reference body. Reference body is the object with respect to which the state of rest or motion of an object is described.

Distance and Displacement

If an object travels from initial position to final position through different routes the distance will not be equal. But if it travels through straight line the distance is called displacement. The distance has only magnitude. But displacement has both magnitude and direction.

While stating the displacement, it is necessary to indicate the direction along with the magnitude of the distance travelled. Such physical quantities having both magnitude and direction are referred to as vector quantities. Physical quantities of which the direction is not to be indicated, are scalar quantities When a body travels along a straight line in the same direction, the magnitude of its distance and displacement will be equal.

Kerala Syllabus 8th Standard Physics Notes Speed and Velocity

Speed is the distance travelled in unit time. Velocity is the displacement in unit time.

The unit of speed and velocity is metre/second (m/s)

Uniform speed and Non uniform speed

If a body in motion covers equal distance in equal intervals of time the body is said to have uniform speed. That is if a car travels 20 km in first 2 seconds, again travel 20 km in successive 2 seconds, the body is said to have uniform speed. But the body travels 20 km in first 2 seconds, 40 km in next 2 seconds and then travels 50 km in next 2 seconds the body is said to have non uniform speed.

The speed of a car in non uniform speed is different. So to find out the speed calculate the average speed

Uniform velocity and Non uniform velocity

A body has uniform velocity if it covers equal displacements in the same direction in equal intervals of time. But if it is not in equal velocity in equal interval it is in non uniform velocity.

Hss Live Guru 8th Physics Kerala Syllabus Acceleration

A body with a velocity of o m/s takes 5 seconds to rise the velocity to 5 m/s the change of velocity is 5 – o = 5 m/ s. The change is taken place in 5 sec. $$\frac{change of velocity}{time}$$ the rate of change of velocity which is called acceleration.

Class 8 Physics Notes Kerala Syllabus Retardation

When the velocity of a body is decreasing the acceleration is negative. This type of acceleration is called retardation.

Carelessness of pedestrians is also one of the reasons for road accidents.
To reduce the accidents following precuations are to be taken.

• walk along the footpath only
• walk only along the right side of the road
• cross roads only at the zebra crossing.
• while walking along the road late in the evening or night avoid wearing black or dark coloured clothes.
• ATUL Pivot Point Calculator

### Motion Textbook Questions and Answers

Kerala Syllabus 8th Standard Biology Notes Question 1.
Which of the following does not belong to the group?
(velocity, acceleration, speed, displacement)
speed

Hsslive Guru 8th Class Physics Kerala Syllabus Question 2.
The statement of a child is as follows: “My displacement is zero though I ran 250 m”. What is meant by this?
The child reaches the point which he started. So the displacement is o.

Class 8 Physics Kerala Syllabus  Question 3.
All objects having uniform speed need not have uniform velocity. Describe with the help of examples.
If a body is in uniform velocity the direction and quantity will be equal.

In figure (1) from A to B and B to C the direction is different. Therefore it is not in uniform velocity. In figure (2) it is in uniform velocity.

Kerala Syllabus 8th Standard Physics Question 4.
Bus A covered 75 m in 5 s. Bus B covered 169 m in 13 s
a. Which bus covered a greater distance?
b. Which bus has a higher speed?
The bus B covered more distance.
b. Speed = $$\frac{distance}{time}$$
Speed of A= $$\frac{75}{5}$$ = 15 m/s
Speed of B = $$\frac{169}{13}$$ = 13 m/s
So the bus A has higher speed

Kerala Syllabus 8th Standard Physics Notes Pdf Question 5.
What is the acceleration of a car which started from rest and acquired a velocity of 40 m/s in 8 s?

Physics Notes Class 8 Kerala Syllabus Question 6.
A car covered the first 400 m distance with a speed of 8 m/s, the next 1200 111 with a speed 12 m/s and the last 360 m with a speed of 12 m/s. Calculate the average speed of the car.
Average speed = $$\frac{Total distance travelled}{time taken to trvel the distance}$$
Time taken to travel the distance 400m = $$\frac{400}{8}$$ = 50s
Time taken to travel the distance 1200m = $$\frac{1200}{10}$$ = 120s
Time taken to travel the distance 360m = $$\frac{360}{12}$$ = 30s
Total distance travelled = 400 + 1200 + 360 = 1960m
Total time to travel = 50 + 120 + 30 = 200s
Average speed=$$\frac{1960}{200}$$= 9.8 m/s

Physics Notes For Class 8 Kerala Syllabus Question 7.
Does a body have acceleration in the following situations? Why?

• body travelling along a straight line with uniform velocity.
• body travelling along a straight line with non-uniform velocity.
• body travelling along a circular path with uniform speed.
• body travelling along a circular path with non-uniform speed.

• body travelling along a straight line with non-uniform velocity.
• body travelling along a circular path with non-uniform speed.

If an object should have acceleration it should be travelled in non uniform speed or in non uniform velocity.
Change in velocity time
Acceleration = $$\frac{change in velocity}{time}$$

Basic Science Class 8 Chapter 9 Kerala Syllabus Question 8.
A lorry travelling with a velocity of 30 m/s came to rest in 5 s. What is its acceleration?
Acceleration = $$\frac{0 – 30}{5}$$ = 6m/s²
∴ Retardation = 6m/s²

Hsslive Guru 8th Basic Science Kerala Syllabus Question 9.
What is the displacement of a car in 30 s if it is travelling with a velocity of 15 m/s?
Velocity = $$\frac{dispalcement}{time}$$
Displacement = velocity × time = 15 × 30 = 450m

Hss Live Guru 8th Basic Science Kerala Syllabus Question 10.
Observe the figure showing the path of a body which started from
A and moved to C through B.

a. Calculate the speed of the body.
b. What is the velocity of the body?
c. What is the velocity of the body if it has taken 5 s to reach A hack from C?
d. Compare the velocity of the body when it reached C from A and also when it reached A from C.
a. Distance = 60 + 40 = 100 m
Time = 6 + 4 = 10 sec
∴ Speed = $$\frac{100}{10}$$ = 10m/s

b. Velocity = $$\frac{dispalcement}{time}$$ = $$\frac{70}{10}$$ = 7m/s
c. displacement = 0, Velocity = 0
d. Velocity of the body when it reached C from A = 7 m/s
Velocity of the boby when it reached A from C velocity = 0 m/s

Hsslive Guru Physics 8th Standard Kerala Syllabus Question 1.
Observe the figure in which a child travelled from A to B and returned from B to A.

a. Are the distance from A to B and B to A through C be equal?
b. What is the difference between the distance between A to B and return to B through C?
c. Which is the displacement ?
a. No
b. The path from A to B is straight path. But the path from B to A is curved.
c. A to B

Basic Science For Class 8 Chapter 9 Kerala Syllabus Question 2.
If a stone is thrown up 8 m perpendicularly and it falls in hand itself,
a. What is the distance travelled by the stone when it is thrown up and what is the displacement?
b. What is the distance when it fell in hand and what is the displacement?
a. Distance = 8m, displacement = 8m
b. Distance = 16m, displacements = 0

Basic Science Class 8 Chapter 9 Solutions Kerala Syllabus Question 3.

A man start from A travelled through B, C and D reaches at A. Complete the table below.

Basic Science Class 8 Pdf Notes Kerala Syllabus Question 4.
Tabulate the difference between distance and displacement

 Distance displacement Scalar quantity Vector quantity Motion in different direction Straight line motion

Question 5.
A car travels a distance of 15m from A to B in 2 s, a distance of 25 m from B to C in 3 s, a distance of 8 m from C to D in 1 s.
a. What is the total distance travelled?
b. What is the total time taken to travel this distance?
c. What is the average speed of the car?
a. 48 m
b. 6 Sec

Question 6.
Write examples of uniform velocity and non uniform velocity.
mniform velocity : Travelling 9# light non uniform velocity : Travelling of bus

Question 7.
What is the acceleration when a car started from rest raised a velocity 80 m/s in 10 s?

Question 8.
Complete the table using the figure below.

Travelled position acceleration

Question 9.
Write 3 occasions of occurrence of acceleration.
1. Falling of coconut from coconut tree.
2. Rolling of wheel on a slope.
3. A train starting from the station.

Question 10.
write three methods to reduce road accidents.
1. walk along the footpath only
2. Avoid over speed, Avoid drink and drive

Question 11.
Brakes are applied suddenly to a racing car travelling at 50m/s. If the car stops after 20 seconds, calculate the retardation of the car.
Initial velocity of the car (u) = 50 m/s
Final velocity of the car (v) = o (The car stops)
Time = 20s

When it is written as retardation, no negative sign is required.
∴ Retardation of the car = 2.5 m/s²

Question 12.
Classify the following quantities into Scalar and Vector quantities.
Distance, Displacement, Time, Speed, Velocity, Acceleration.

 Scalar Vector Distance Displacement Time Velocity Speed Acceleration

Question 13.
Velocity is a vector quantity. What do you mean by vector quantities? Write examples.
Quantities having both magnitude and direction are called vector quatities. eg: Displacement, Acceleration.

Question 14.
Compare distance and displacement.
The length of the path covered by a body is known as distance. This is a scalar quantity. This is measured in a standard unit metre. The straight line distance between initial and final positions of a moving body is called displacement. This is a vector quantity. This is measured in metre.

Question 15.
A cheetah can start from rest and attain the velocity 72 km/h in 2 seconds. Calculate the acceleration of cheetah.

Question 16.
Given figure shows the two ways in which a person reached at C

Reach at C from A through B ,
then
a. From fig.1 what is the total di-stance travelled? What is the total displacement covered?
b. From fig.2 what is the total distance travelled by the person? What is the total displacement covered by him?
c. find out the situation in which distance and displacement are equal by comparing given measures?
a. Distance = 14 m
Displacement = 10 m
b. Distance = 14 m,
Displacement = 14 m
c. Only when a moving object travelling in a straight line on same direction.

## Kerala Syllabus 8th Standard Basic Science Solutions Chapter 12 Why Classification? in Malayalam

Students can Download Basic Science Chapter 12 Why Classification? Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Basic Science Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 3 मेरे बच्चे को सिखाएँ

You can Download मेरे बच्चे को सिखाएँ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 3 मेरे बच्चे को सिखाएँ (पत्र)

### मेरे बच्चे को सिखाएँ पाठ्यपुस्तक के प्रश्न और उत्तर

Mere Bache Ko Sikhaye Notes 8th Kerala Syllabus प्रश्ना 1.
“मेहनत से कमाया एक पैसा भी, हराम में मिली नोटों की गड्डी से कहीं अधिक मूल्यवान होता है।” अपना दृष्टिकोण प्रकट करें।

उत्तर:
यह कथन बिलकुल ठीक है। हराम की चीजें हमारे हक का नहीं है। मेहनत से कमाया पैसा ही मूल्यवान है। मानव को ईमानदारी के साथ जीना है। मेहनत ईमानदारी में चार चाँद लगाता है।

Kerala Syllabus 8th Standard Hindi Notes प्रश्ना 2.
‘बदमाशों को आसानी से काबू में किया जा सकता है। ऐसा क्यों कहा होगा?

उत्तर:
बदमाशों के अंदर भी कुछ सच्चाई होती है। उपदेश और सत्संग के द्वारा उन सच्चाइयों को बाहर ला सकते हैं। इसलिए ऐसा कहा गया है।

Hindi State Syllabus 8th Standard प्रश्ना 3.
‘नकल करके पास होने से फेल होना बेहतर है’ इस प्रस्ताव से क्या आप सहमत है? क्यों?

उत्तर:
मैं इससे शतप्रतिशत सहमत हूँ। नकल से मिली जीत में ज्ञान की गहराई नहीं होती। यह तत्काल लाभ दे सकता है। लेकिन भविष्य में इससे कोई मुनाफ़ा नहीं होता। इसलिए फेल से सीख लेना ही बेहतर है।

8th Standard Hindi Guide Kerala Syllabus प्रश्ना 4.
‘भीड़ से अलग होकर अपना रास्ता बनाना’ का मतलब क्या है?

उत्तर:
भीड़ एक ही मानसिकता के आधार पर चलती है। भीड़ की मानसिकता से अलग होकर सोचने से ही नई दृष्टि और नए विचार मिलते हैं। इस नए दृष्टिकोण से ही सामाजिक प्रगति संभव होती है। संसार के सभी महत् व्यक्ति इस प्रकार सोचनेवाले थे। इसलिए उन्हें संसार में बदलाव ला सका।

### मेरे बच्चे को सिखाएँ Textbook Activities

8th Standard Hindi Notes State Syllabus प्रश्ना 1.
लघु-लेख लिखें।
‘सफल जीवन’ विषय पर लघु-लेख लिखें।

उत्तर:
सफल जीवन
जीवन को सफल बनाने के लिए मनुष्य को आत्मविश्वास, दृढ़संकल्प, अदम्य उत्साह और लगन चाहिए। केवल पढ़ने से ही नहीं, अच्छे चरित्र के निर्माण में भी मानव को ध्यान देना चाहिए। उसे परिश्रमी होना चाहिए। उसे यह समझना चाहिए कि नुशासन जीवन को सफल और उज्ज्वल बनाने के लिए आवश्यक है। व्यक्ति को सादा जीवन और उच्च विचार का आदर्श ग्रहण करना चाहिए। उसे बुरी आदतों और बुरे सहवास से बचकर रहना भी होगा। अपने पाठों को लगन से पढ़ना, बड़ों के सदुपदेशों का पालन करना, बड़ों से आदर और छोटों से प्यार करना आदि की आवश्यकता है। उसे पथभ्रष्ट करनेवाली बातों से बचकर रहना भी चाहिए। जो व्यक्ति इस प्रकार का जीवन बिताएगा, वह अपने जीवन में सफल बनेगा।

मेरे बच्चे को सिखाएँ मेरी रचना में

उचित चौकार में ✓ लगाएं।

विषय का विश्लेषण किया है।

प्रस्तुतीकरण में क्रमबद्धता है।

उचित भाषा का प्रयोग किया है।

अपना दृष्टिकोण प्रस्तुत किया है।

उचित शीर्षक दिया है।

### मेरे बच्चे को सिखाएँ Summary in Malayalam and Translation

मेरे बच्चे को सिखाएँ शब्दार्थ Word meanings

## Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 2 पिता का प्रायश्चित

You can Download पिता का प्रायश्चित Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 2 पिता का प्रायश्चित (संस्मरण)

पिता का प्रायश्चित पाठ्यपुस्तक के प्रश्न और उत्तर

Pitha Ka Prayaschit Questions And Answers Kerala Syllabus 8th प्रश्ना 1.
वह झूठ बोला, “कार तैयार नहीं थी, इसलिए देर हो गई।” इस तरह झूठ बोलना क्या सही है? क्यों?

उत्तर:
झूठ बोलना कभी भी सही नहीं है। क्योंकि यह एक बुरी आदत है। झूठ बोलने से तत्काल फ़ायदा हो सकता है। लेकिन इससे भविष्य में नुकसान ही होगा।

Pitha Ka Prayaschit Notes Kerala Syllabus 8th प्रश्ना 2.
“घर तक की अठारह मील की दूरी पैदल चलकर ही तय करूँगा।” मनीलाल गाँधी के इस निर्णय से आप सहमत हैं? क्यों?

उत्तर:
मनीलाल गाँधी का यह निर्णय बिलकुल सही है। क्योंकि बेटे की गलती का कारण वे अपने को मानते हैं। इसके द्वारा उन्होंने अपने बेटे को अपनी गलती पर सोचविचार करने मौका दिया।

### पिता का प्रायश्चित Textbook Activities

Pitha Ka Prayaschit In Malayalam Kerala Syllabus 8th प्रश्ना 1.
सही मिलान करें।

उत्तर:
वर्ष — बरस
सुदूर — दूरदराज
प्रदेश — इलाका
अवसर — मौका
प्रतीक्षा — इंतज़ार
ढूँढ़ — तलाश

Prayashchit Hindi Lesson Question Answer Kerala Syllabus 8th प्रश्ना 2.
अर्थभेद समझें।

डरबन से 18 मील दूर एक आश्रम में रहता है।

डरबन से 18 मील दूर एक आश्रम में रहता था।

मैं और मेरी दो बहिनें हमेशा शहर जाने की इंतज़ार में रहते हैं।

मैं और मेरी दो बहिनें हमेशा शहर जाने की इंतज़ार में रहते थे।

वहाँ दूर तक गन्ने के खेत हैं।

वहाँ दूर तक गन्ने के खेत थे।

Prayaschit Questions And Answers Kerala Syllabus 8th प्रश्ना 3.
पत्र लिखें।

संस्मरण कैसा लगा? पुत्र की गलती पर पिता ने अपने आप को सज़ा दी। इसी दर्द के एहसास से अरुण गाँधी ने यह निर्णय लिया- मैं कभी झूठ नहीं बोलूँगा। अपना दर्द वह दोस्त से बाँटे बिना नहीं रह सका। उसने मित्र को पत्र लिखा। वह पत्र कल्पना करके लिखें।

उत्तर:

डरबन
20 अगस्त 1950

प्रिय मित्र,
नमस्कार।
तुम कैसे हो? सोचता हूँ कुशल से हो। हम यहाँ डरबन में खुशी से जी रहे हैं। अपने जीवन के एक विशिष्ट बात बताने के लिए मैं यह चिट्ठी लिख रहा हूँ। कल पिताजी को मेरी गलती पर प्रायश्चित करना पड़ा। हुआ यह कि पिताजी को शहर में कल एक मीटिंग थी। उन्हें मैंने कार से शहर छोड़ा। शाम पाँच बजे उन्हें लेने जाना था। लेकिन बेन जॉन का सिनेमा देखकर मैं समय भूल गया। देरी के कारण पूछने पर झूठ बोला कि कार ठीक करके गैरेज से नहीं मिला। लेकिन पिताजी बात पहले ही समझ गए थे।

पिताजी ने मेरे झूठ को अपनी गलती माना। वे प्रायश्चित करते हुए घर तक का रास्ता पैदल चले। यह देखकर मुझे बहुत दुख हुआ। मैं यह निश्चय किया हूँ कि आइंदा झूठ नहीं बोलूंगा। अगर पिताजी मुझे कोई सज़ा दी होती तो मैं ऐसा कोई निर्णय नहीं लेता। मैं यह घटना कभी नहीं भूलूँगा। उसकी याद ज़िंदगी में मुझे सही रास्ते पर ज़रूर ले जाएगी।

अपना दोस्त
अरुण गाँधी।

सेवामें

अरविंद
वर्धा आश्रम
पोरबंदर
गुजरात
भारत

पत्र लिखते समय ध्यान दें…
स्थान और तारीख है।
उचित संबोधन है।
विषय का सही संप्रेषण है।
स्वनिर्देश है।
पता है।

पिता का प्रायश्चित मेरी रचना में

उचित चौकोर में ✓ लगाएँ।

स्थान और तारीख हैं।

उचित संबोधन है।

विषय का सही संप्रेषण है।

स्वनिर्देश है।

पता है।

### पिता का प्रायश्चित Summary in Malayalam and Translation

पिता का प्रायश्चित शब्दार्थ Word meanings

## Kerala State Syllabus 8th Standard Maths Solutions Chapter 6 Construction of Quadrilaterals

Textbook Page No. 108

Construction Of Quadrilaterals Class 8 State Syllabus Question 1.
Can you draw these patterns of squares in your notebook?

Solution:
1. Draw a square of side 6 cm. Draw lines horizontally and vertically 2cm apart. Rub off unwanted parts. We get the required pattern.

2. Draw a square of side 7 cm. In it draw horizontal and vertical lines at intervals of 4 cm, 2 cm and 1 cm. Erase the unwanted part. We get the required pattern.

3. Draw a square of diameter 6 cm. (Draw a circle of diameter 6 cm and draw two perpendicular diameters. Join their ends.) Mark the points on diagonal 2 cm a part. Join the points as in the following figure. Rub off the unwanted part. We get the required pattern.

Text Book Page No. 111

Class 8 Mathematics Construction Of Quadrilaterals Kerala Scert Solutions Question 2.
Draw the figures below in your notebook.

Solution:
1. Draw a rectangle of diagonal 9 cm and angle between the diagonal and one side is 30°. Mark a point on the diagonal at a distance 6 cm from one end of the diagonal. Draw lines perpendicular to the sides of the rectangle through this point. Erase unwanted parts, we get the required pattern.

2. Construct an equilateral triangle with side 3 cm.

The other two sides of a triangle are made by equal diagonals of the rectangle. Then the diagonal of the rectangle is 6 cm. Draw a rectangle in a horizontal position with diagonals 6 cm and the angle between the diagonals 60°. Draw another rectangle of the same measure in the vertical position at the middle of the first rectangle. Draw the wanted part in bold lines. Erase unwanted part.

Construction Of Quadrilaterals Class 8 Kerala Syllabus Question 3.
Draw a rectangle CBD, which di agonal (AB) is 6 cm and AC making an angle 30° with AB. The arc with centre at B and BC as radius and the arc with centre at A and AC as radi us meet at E. Complete ∆ AEB. Draw an arc with centre A and EB as radi us. Also draw arc with B as centre and AE as radius meet at F. Complete the rectangle AEBF.

Text Book Page No. 114

8th Class Maths Construction Of Quadrilaterals Kerala Syllabus Question 3.
Draw a rhombus of diagonals 5.5 cm and 3 cm in your notebook.
Solution:
Draw a line of length 5.5 cm, and find its midpoint by drawing the perpendicular bisector. Mark the points on the upper and lower part of the bisector.
Mark the points on the upper and lower part of the bisector line at a distance 1.5 cm from the intersecting point of the first line and perpendicular bisector. Join these points to the end of the first line.

Construction Of Quadrilaterals 8th Class Kerala Syllabus Question 4.
Draw also a rhombus of diagonals 5.5 cm and 3.5 cm.
Solution.
It is difficult to measure 1.75 cm (half of 3.5 cm) using scale, so draw a rectangle of 5.5 cm and breadth of 3.5 cm. By drawing the perpendicular bisectors find the midpoints of the sides. By joining the midpoints of the sides, we get a rhombus.

Textbook Page No. 117

Class 8 Construction Of Quadrilaterals Kerala Syllabus Question 5.
Draw these figures
Solution:
1. Two equal rhombuses :

Solution:
1. From the figure two sides of A BCD are equal, angles opposite these sides are also equal. We can calculate them as 50° each. In the same way find other angles in the figure.

Draw a line BD vertically, 3 cm long. At D draw angles of 50° on both sides. At B also draw angles of 50° on both sides. Then we get a rhombus ABCD. Extend BC to G such that BC = CG and extend DC to E such that DC = CE. Draw GE. Draw angle of 50° at G and E to find F.

2. Draw a circle of radius 2 cm. Divide the centre of the circle into angles of 60° each. These lines meet the circle at the points A, B, C, D, E and F.

Draw arcs of 2 cm from A and B to get G. Similarly find H and I. Draw the required parts and rub off unwanted parts.

Construction of Quadrilaterals Class 8 Question 3. Draw a circle of radius 2 cm. Mark the points A, B, C, D, E, F, G, and H on the circle by making 45° angles at the centre. Draw arc of 2 cm A and B to get I. Similarly find J, K and L. We get the required figure.

4. Draw AC, 4 cm long and mark its midpoint B. Since all are rhombuses, ABI is a equilateral triangle. Its angle are 60° each.

In the rhombus BCDJ, ∠ JBC = ∠JDC = 60°, ∠BCD = ∠BJD = 120°
In the rhombus BJFI, ∠IBJ =∠IFJ = 60°, ∠BJF = ∠FIB = 120°. Draw each rhombus and complete the pattern.

5. Draw a line AB, 4 cm long and mark its midpoint C. CJFI is a square, all its angle are 90° each. Also calculate ∠ICA = 45° and ∠CAH = 135°. Taking measures of the sides and angles draw each rhombus.

6. Draw a square of side 3 cm. And draw two parallelograms with sides 3 cm, 2 cm and angle between them 45°, on the sides of the square.

Text Book Page No. 124

Construction Of Quadrilaterals Class 8 Solutions Kerala Syllabus  Question 6.
Draw the figures below :
1. Three equal isosceles trapeziums:

Solution:
1. Draw DE, 5 cm long and mark G on DE such that DG = 2 cm (length of AB) ∆ ABC and ∆ DEF are equilateral triangles. Their each angle is 60°. Find B by drawing ∠EGB = ∠BEG = 30°. Draw ∠ADG = 30°and BA = 2 cm to get A. Now we got one trapezium. Draw the other two trapezium in the same way.

2. Draw two circle of radius 1 cm and 3 cm with the same centre. Divide the circumferance of the both circle into 6 equal parts. And join them to obtain six equal isosceles trapeziums.

3. Draw two parallel sides AB = 8 cm, CD = 4 cm. AE= FB = 2 cm, EF = 4 cm. Also GD = HC = 2 cm. ∠B + ∠B CD = 180°. The angles at C are equal. They are equal to B. So ∠B = $$\frac{180}{3}$$ = 60°. ∠A = 60°. Now draw the pattern.

4. Draw a rectangle with length 8 cm and breadth 4 cm. Divide this rectangle into two squares with side 4 cm. Consider one square and half of second. Mark the midpoint of the sides. Join as in the figure.

5. Draw a square of side 8 cm. And draw lines horizontally and vertically 2cm apart. Complete the figure.

Text Book Page No. 128

8th Class Maths Notes Kerala Syllabus Question 7.

Solution:
1.Draw AB = 5 cm. Draw AD such that ∠A = 80° and AD =3 cm. Mark C such that ∠D = 120° and DC = 4 cm. Join BC.

2. Draw AB = 5 cm. Draw AD such that ∠A = 60° and AD = 3 cm. Mark point C such that ∠B = 80° and ∠D = 100°. Join DC and BC and complete the quadrilateral ABCD.

3. Draw a ABD with AB = 7 cm, BD = 8 cm and AD = 4 cm. Mark the point C at a distance 6 cm from A and 5 cm from D. Join BC and CD.

Class 8 Maths Construction Of Quadrilaterals Kerala Syllabus Question 1.
In the figure, ABCD is a square whose diagonals intersect at O. If AD = 10 cm, find the length of BD and CD ?

Solution:
Tn a square diagonal are equal and perpendicular bisectors of each other.
OD = 5 cm and OC = 5 cm.
CD2 = OD2 + OC2 = 52 +52 = 50 cm
CD = $$\sqrt{50}$$ = BD = $$5\sqrt{2}$$ cm.

Kerala Syllabus 8th Standard Maths Notes Question 2.
Draw quadrilateral PQRS, PQ = 7 cm, QR = 5 cm, RS= 4cm, ∠Q = 60° and ∠R = 140°.
Solution:
Draw PQ = 7 cm and draw QR of length 5 cm which makes an angle 60° with PQ. Draw RS such that RS = 4 cm and ∠R = 140°.

8th Standard Maths Notes State Syllabus Question 3.
(a) Write any two peculiarities of the diagonals of a square.
(b) The length of a diagonal of a square is 7 cm. Draw the square.
Solution:
(a) The diagonals of a square are equal.
The diagonals bisect each other perpendicularly.
(b)

Question 4.
In rhombus PQRS , PR = 7 cm and Question = 5 cm. Construct rhombus PQRS.
Solution:

Question 5.
In the figure, ABCD is a parallelogram. ∠D = 80°. Find all other angles?

Solution:
ABCD is a parallelogram
Opposite angles are equal ∠B = 80°.
Sum of the angles on the same side is 180°.
∠A + ∠B = 180°.
∠A = 180° – 80° = 100°
And ∠C = 100° (opposite angles are equal).

Question 6.
In the figure, ABCD is a parallelogram. Find x, y, z.

Solution:
∠Y = 112° (opposite angles are equal)
In ADC, ∠x + ∠y + 40 = 180° (sum of angles in a triangle)
∠x + 112° + 40° = 180°
∠x = 180° – 152° = 28°
∠z = 28°(transversal alternate interior angles are equal).

Question 7.
Construct a quadrilateral ABCD, AB = 6 cm, BC = 3 cm, CD = 2 cm, AD = 4 cm and AC = 5 cm.
Solution:
Draw AB = 6 cm. Then find C by drawing arcs of radius 5 cm and 3 cm from A and B. Find D by drawing arcs of radius 4 cm and 2 cm from A and C. Join BC, CD and AD to get quadrilateral ABCD.

Question 8.
Construct a quadrilateral ABCD, AB = 8 cm, BC = 6 cm, CD = 5.5 cm, DA = 3 cm and ∠B = 50°.
Solution:
Draw AB = 6 cm. Draw BC such that BC = 6 cm and ∠B = 50°. Draw an arc of radius 5.5 cm with C as centre and another arc of radius 3 cm with A as centre. Mark the point of intersection as D. Join CD and AD.

Question 9.
In parallelogram ABCD, the diagonals AC and BD intersect at O. AC = 6.5 cm, BD = 7 cm and ∠AOB = 100°. Construct the parallelogram.
Solution:

Question 10.
The diagonals of a rhombus are of lengths 16 cm and 12 cm. What is its perimeter?
Solution:

In right angled AOB,
AO = 8 cm, BO = 6 cm, ∠AOB = 90°
AB2 = AO2 + BO2 = 82 + 62
= 64 + 36 = 100
Side, AB = $$\sqrt{100}$$ = 10 cm
Perimeter = 4 × 10 = 40 cm.

Question 11.
Draw the following patterns, a. 6 equal rhombuses :
(a) 6 equal rhombuses:

(b) 3 equal rhombuses

Solution:
(a) Draw a circle of radius 2 cm with centre O. Divide the centre of circle into angles of 60° each. These lines meet the circle at the points A, B, C, D, E and F. Draw arc of 2 cm from A and B to get G. In the same way find H, I, J, K, L. Draw needed part.

(b) Draw square of side 4 cm. Draw rhombuses with side 4 cm and an gle 30° on top and bottom side of the square. Complete the figure.

(c) Angles around the point at which three rhombuses joined together is 120° each. Since one angle of the rhombus is 120°, another angle 60°. Draw three rhombuses with side 4 cm and angle 60°. Complete the figure.

(d) Draw a semicircle of radius 2 cm with centre O. Divide the centre of circle into angles of 45° each .These lines meet the circle at the point A, B, C, D and E. Draw arc of 2 cm from A and B to get F. In the same way find G, H and I. Complete the figure.

(e) Draw a rectangle of length 6 cm and breadth 3 cm. Draw two rhombuses on both side of the rectangle which makes angle 45° and 135° with length and breadth respectively.

Question 12.
In a parallelogram ABCD, find x, y, z from the adjoining figure.

Solution:
ABCD is a parallelogram,
∠C = 45° (Opposite angles are equal)
∠C + Z = 180° (linear pair)
Z = 180° – 45° = 135°
45° + Y = 180° (sum of the angles on the same side is 180°)
Y = 180° – 45° = 135°
sinceY = 135°
X = 135°(Opposite angles are equal)

## Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter

You can Download Properties of Matter Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter

### Properties of Matter Questions and Answers

Properties of matter

The matter commonly seen in three states, solid, liquid and gas. Matter has volume and mass. Matter is anything which occupies space and has volume. The solid has a definite shape. But liquid occupy the shape of the container. Gas has no definite shape.

Properties Of Matter Class 8 Questions And Answers Question 1.
explain the peculiarities of materials.
Tiny particles of matter

The substances are made up of many tiny particles which cannot be seen by naked eye. Even though we cannot see the tiny particles of sugar in sugar solution, we can understand that tiny particles of sugar is dissolved in the solution while we taste it.

All the substances are made up of tiny particles which cannot be seen by naked eye and these tiny particles bear all the properties of the substances.

Properties Of Matter Class 8 Notes Pdf Kerala Syllabus Question 2.
identify and picturise the arrangement of particles in different states of matter.
Arrangements of particles in different states of matter

The particles of solids are very close to each other. The freedom of movement of particles is limited. The attractive force between particles is very high.

The particles of liquid are relatively further apart and have more freedom for movement than in the solid state. The attractive force is less than that of solid state. The particles in gaseous state remains far away from one another and the attractive force is very low.

Change of State

When ice is heated, it changes into water and when water is further heated it boils and changes into steam.

Properties Of Matter Questions And Answers Kerala Syllabus Question 3.
Which form of energy is responsible for the change of state here?
Here the form of energy is responsible for the change of state is heat.

When heat is absorbed energy of the particles, distance between the particles, movement of the particles are increased. The attractive force between particles is decreased.

Some substances, when heated, change directly into gas without forming liquid. This phenomenon is sublimation.

Diffusion of substances in different states

The fragrance of incense stick fills the room when the stick is lit. The fragrance spreads because its particles are spread in air. If a drop of ink is added carefully into the water in a beaker, the ink spread in the water.

The spontaneous mixing of different particles having freedom of movement is Diffusion.

Hsslive Guru 8th Class Chemistry Kerala Syllabus Question 4.
Distinguish pure substances and mixtures and tabulate them?
Pure Substances and Mixtures
Depending on their nature, substances can be classified into two.
1. Pure Substances
2. Mixtures Materials made of particles of identical nature are called pure substances
Eg: Water, common salt, sugar

The substances made of particles of different nature are called mixtures.
Eg: Salt solution, sugar solution

Separating the Components of a Mixture.

Most of the substances found in nature are mixtures. We use different methods of separation of components in rice mixed with stones, tea mixed with tea dreg, mixture of methanol and ethanol.

Distillation

Distillation is used for separation of components in salt solution.

When one component of the mixture is volatile and the others do not vapourise under the same condition, they can be separated by distillation. More over components of a mixture possess a large difference in their boiling points, they can be separated by distillation. Eg:Mixture of water (boiling point ioo°C)and acetone(56°C)

Fractional Distillation

If the boiling points of components have very small differences, fractional distillation is to be used to separate them.

Eg: Mixture of Ethanol (boiling point 78°C) and methanol (boiling point 65°C).

When vapours of the mixture pass through the fractionating column, repeated. liquefaction and vapourisation take place. Subsequently, the vapours of low boiling methanol enter the condenser from the fractionating column, condense to liquid and get collected in the round bottomed flask first. Similarly, ethanol with higher boiling point can be collected later in another round bottomed flask.

Separating Funnel

Separating funnel is an apparatus used for separating immiscible liquids from their mixture.

We can use separating funnel to separate the mixture of water and kerosene which are having difference in density.

Sublimation

This method can be used to separate the components which have the property of sublimation from the mixture. This method can be used for separating the components of a mixture of ammonium chloride and sand.

Centrifugation

This is a method for separating components from a mixture, based on the difference in the mass of particles. This method is used in clinical laboratories to separate blood cells from blood samples and also for separation of butter from curd.

In order to separate insoluble particles in a liquid mixture on the basis of their mass difference, centrifuge is used. The process is known as centrifugation.

Chromatography

Chromatography is the method used to separate more than one solute dissolved in the same solvent. This method was first employed for separating coloured substances and hence this process came to be known as Chromatography. Chromatography is employed to separate components from dyes and to separate poisonous substances mixed with blood.

### Properties of Matter Textbook Questions and Answers

Basic Science Class 8 Solutions Chapter 4 Kerala Syllabus Question 1.
A few mixtures are given below. Tabulate the methods to separate their components and give the reasons for selecting the method.

 Mixture Method Reason Common salt and ammoni­um chloride Sugar solution Petrol and ker­osene Camphor and glass powder Iron powder and sand

 Mixture Method Reason Common salt and ammonium Chlo- Sublimation Components having the prope- Sugar solution Distillation Components having the property of evaporation Petrol and ker­osene Fractional dist­illation Components have small differanee in boiling po­int Camphor and glass powder Sublimation Components having the property of sublimation Iron powder and sand Magnetic separation Components ha­ving magnetic properties

8th Class Biology Notes Pdf Kerala Syllabus  Question 2.
Given below are certain changes taking place to the particles during change of state. From this, find out and tabulate the changes in the particles when water boils to form steam and also when steam condenses to form water.

• distance increases
• attractive force decreases
• energy increases
• rate of movement increases
• distance decreases
• energy decreases
• attractive force increases
• rate of movement decreases

 Water changes to Steam steam to water Distance increases Distance decreases Energy increases Energy decreases Attractive force decreases Attractive force increases Movement of particles increases Movement of particles decreases

Kerala Syllabus 8th Standard Chemistry Notes Question 3.
Spirit kept open in a.watch glass disappears after some time. Which among the following phenomena are responsible for this?
a) sublimation
b) distillation
c) evaporation
d) diffusion
a) sublimation

Basic Science For Class 8 Chapter 4 Kerala Syllabus Question 4.
Which are the methods that can be used to separate the components of a mixture made of common salt, ammonium chloride and sand? Write the methods in the order in which they are applied.
sublimation → filtration → distillation

Kerala Syllabus 8th Standard Physics Notes  Question 5.
Many minerals are present in ordinary water.
a. Which is the method that can used to remove the mineals and obtain pure water?
h.In which type of mixtures is this method employed?
c. Water purified by this method is distilled water .Write two instances of its use.
a. distillation
b. When one component of the mixture is volatile and the others do not vapourise under the same condition, they canbe separated by distillation
c. For vaccination, In storage batteries

Basic Science Class 8 Pdf Notes Kerala Syllabus Question 6.
From the following statements, tick (V) those which apply to solid substances alone.
□ Particles have little freedom of movement.
□ Distance between particles is very high
□ Particles remain very close to each other
□ Energy of particles is very high
✓ Particles have little freedom of movement.
✓ Particles remain very close to each other

Question 7.
Take a small wooden rectangular block and find its volume (volume = length x breadth x height).
Take a big measuring jar and fill three-fourth of it with water and mark the‘water level. Then dip the block in water in the jar. (To prevent floating, nails can be inserted in the block). Mark the difference in the water level.
a. Is there any relation between the difference in the water level and the volume of the block?
b. Which property of matter is revealed by this experiment?
a.Yes. the volume of water rises is the volume of wooden block.
b. The substance that occupies a space.

Question 8.
Electronic balances are very popular now. On an electronic balance, find the weight of an empty balloon. Again, find its weight after filling air. Now, can you find the weight of the air in the balloon?
Repeat the experiment using balloons of different size by filling them with varying quantity of air.
Reduce the weight of balloon from the weight of balloon with air

Question 9.
Take water mixed with chalk powder in a bottle. Tie a string to the bottle and swirl it at high speed along a circular path. Observe after sometime.
Repeat the activity using other mixtures which are suspensions. To which method of separation of components of a mixture can this be connected? Are there instances where this principle is made use of. Prepare a note.
Centrifugation is the method used. This is a method for separating components from a mixture, based on the difference in the mass of particles. This method is used in clinical laboratories to separate blood cells from blood samples and also for separation of butter from curd.

Question 10.
Take a long white chalk piece and put a mark with black ink slightly above the bottom. Keep the chalk piece dipped perpendicularly in water in a watch glass. After sometime observe the changes. Repeat the experiment using different chalk pieces marked with sketch pens of different colours. To which of the methods of separation you have studied is this related?
Chromatography

Question 1.
Find the odd one out
a. Evaporation, Fermentation, Filtration, Chromatography.
b. Sea water, Muddy water, Table salt, Curd, Air
c. Distillation, Separating funnel, Filtration, Evaporation
a. Fermentation
b. Table salt
c. Separating funnel

Question 2.
What is the use of chromatography?
Solutes dissolved in the same solvent can be separated using chromatography.

Question 3.
What is the importance of evaporation? Where is this technique used on a large scale?
Evaporation is used to separate solids dissolved in liquids, mainly water. It is used on a large scale to obtain common salt from sea water. Sea water is left in the open in large containers on sunny days. After a few days salt remains as residue and water evaporates. Common salt is then further purified.

Question 4. What is the advantage of distillation over evaporation?
Distillation is more advantageous than evaporation, because liquid can be recover in distillation while in evaporation liquid is lost.

Question 5.
Which process is used to dry clothes in a washing machine?
Centrifugation.

Question 6.
Method for separating drugs (medicine) which are miscible in blood Centrifugation, Evaporation, Filtration, Chromatography.
Chromatography

Question 7.
From the below statements select correct one?
a. Every substance has a definite mass
b. Every substance has a definite shape
c. Every substance has a definite volume
a. Every substance has a definite mass

Question 8.
Melting: Converting solid into liquid Sublimation:
a. Observe the first pair and fill up. the blanks in the second pair suitable.
b. Write two examples of sublimation
a. Sublimation. A solid directly changes into gas
b. Camphor, iodine, naphthalene.

Question 9.
From Gopu’s hand the bottle containing coconut oil fall into a bucket containing water. Is it possible to separate oil and water?
Using a separating funnel, they can be separated. The denser water will be at the bottom and the lighter oil will be at the top. Water is removed through the tap.

Question 10.
Classify the following substances according to its state and tabulate.
Water, pencil, book, air, kerosene, oxygen

 solid liquid gas Pencil Book Water kerosene Air Oxygen

Question 11.
Will water fill in tumbler when a glass tumbler is immersed perpendicularly into the water in a turf ? Give reason.
No because there is air in the tum-bler. Air need space to occupy so water cannot enter into the tumbler.

Question 12.
A balloon filled with air and an empty balloon weigled in a digital balance found difference in weight. What are the properties of matter understood by this experiment?
Matter needs space to occupy and has mass.

Question 13.
Complete the table.

Question 14.
Which energy causes the change of state when ice is heated to water and water to water vapour? The three figures shows the arrangements of particles. Recoga- nise and write the state of matter.

Heat energy figure
(1) solid, figure
(2) liquid, figure
(3) gas

Question 15.
Mention three examples from daily life for diffusion.
Spreads the smell of perfumes, smell of fruits and food item. Ink spreads in water.

Question 16.
Complete the table

Question 17.
When naphthalene balls are kept somewhere after some days its size diminishes and disappears. Name the phenomenon caused for this process.
Sublimation

Question 18.
Mention the reason behind the smell spreads everywhere in the room while in scent stick is lighted. Sublimation, Evaporation, Diffusion
Diffusion

Question 19.
Tabulate following as pure substances and mixtures.
Gold, soda water, steam, sugar, soil, water, common salt, sugar solution, carbon dioxide, salt solution.

 Pure substances Mixtures Gold Soda water Sugar Sugarsolution Salt Soil Water Salt solution Steam Carbon dioxide

Question 20.
Find the method of separation of components following mixt¬ures.
b. butter from curd
c. tea dreg from tea
d. salt from sea water
e. blood cells from blood sample
f. mixture of petrol and kerosene
g. salt and ammonia
h. poison from blood
i. water and kerosene
j. water and acetone
a. by picking
b. centrifugation
c. filtration
d. distillation
e. centrifugation
f. fractional distillation
g. sublimation
h. chromatography
i. using separating funnel
j. distillation

Question 21.
Write two examples of sublimation from daily life.
1.camphor
2. naphthalene balls becomes smaller

Question 22.
Salt, rava and mustard are mixed while Suresh has brought it from shop. Say methods to separate these.
Salt dissolves in water. Then filter it. Rava and mustard are separating by picking, salt is collected by distillation.

Question 23.
What will be seen when a filter paper is marked by sketch pen and one end of the paper is dipped in water? what is this process called?
different colours are separated. Chromatography

Question 24.
Using given methods write one example for separating components of mixture.
Distillation, magnetic separation, centrifugation, fractional distillation, filtration, sublimation.
• Separating common salt from sea water – Distillation
• Mixture if iron powder and sulphur – Magnetic separation
• Butter from curd – Centrifugation
• Separation of kerosene, petrol, die sel – Fractional distillation
• Separating tea dreg from tea – Fil-tration
• Ammonium chloride and common salt – Sublimation

Question 25.
Solid, Liquid, Gas are the three states of matter
a. Which of this having definite volume hut not definite shape?
b. Write any two changes occurs when we change a substance from liquid state to gaseous state?
c. How to convert a gasecfas substances into liquid?
a. Liquid
b. Increase the distance between molecules.
• Decrease the attractive force between molecules.
• Increase the speed of molecules.
• Increase the energy of molecules.
c. to be cooled (or) Reduce temperature

Question 26.
Name of the mixtures in column A, properties of components are given in column B, method of separating of component are given in the column C also. Find out the match items from B & C to A.

 Mixture Character of components Method of Separation 1. Alcohol and water difference in boiling point Centrifugation 2. Soil and mentor difference in eva­porating Character Distillation 3. Curd and butter difference in weight of inso­luble precipitate Chromatography Sublimation

 Mixture Character of components Method of Separation 1. Alcohol and water difference in boiling point Distillation 2. Soil and mentor difference in eva­porating character Sublimation 3. Curd and butter difference in weight of inso­luble precipitate Centrifugation

Question 27.
a. Which device given in the picture?
b. Why should this device doesn’t use for separating petrol and kerosene mixture?
c. Which method is used to separate the mixture containing petrol and kerosene?

a. Separating funnel
b. Petrol and kerosene are miscible liquids.
c. Fractional distillation

Question 28.
Diffusion is a phenomenon which
a. What is the relation between the rate of diffusion and states of substance?
b. What is the relation between diffusion and temperature?
c. Given an example from everyday life which shows relation between diffusion and temperature.
a.Rate of diffusion is more in gaseous state, less in solid state.
b. Rate of diffusion increases with increase in temperature.
c. The smell of hot food material spread quickly.

Question 29.
Find out the relation and fill in the blanks.
a. Separate salt from saline water :: Evaporation; Separation of components from black coloured ink :: ………..
b. Distillation :: Condenser; Fractional distillation :: ………..
c. Write down some situations for which the fractional distillation used?
d. When we use fractional distillation for separating components of mixture?
a. Chromatography
b. Fractionating column
c. • Separating components to petroleum
• Separating components of air.
d. Used when slight differences in their boiling points.

## Kerala State Syllabus 8th Standard Maths Solutions Chapter 5 Money Maths

### Money Maths Text Book Questions and Answers

Textbook Page No. 90

Money Maths Class 8 Kerala Syllabus Chapter 5 Question 1.
Sandeep deposited Rs 25000 in a bank which pays 8% interest compounded annually. How much would he get back after two years?
Solution:
Interest in the first year
= 25000 × $$\frac{8}{100}$$
=Rs 2000
Principal in the second year
= 25000 + 2000
= 27000
Interest in the second year
= 27000 × $$\frac{8}{100}$$
= 2160
Total amount gets back at the end of 2 year
= 27000 + 2160
= 29160

Money Math Class 8 Kerala Syllabus Chapter 5 Question 2.
Thomas took out loan of 15000 rupees from a bank which charges 12% interest, compounded annually. After 2 years, he paid back 10000 rupees. To settle the loan, howmuch should he pay at end of three years?
Solution:
Amount borrowed = 15000
Interest for the first year
= 15000 × $$\frac{12}{100}$$ = 1800
Loan for the second year
= 15000 + 1800 = Rs.16800
Interest for the second year
= 16800 × $$\frac{12}{100}$$ = 2016
Loan for the end of second year
= 16800 + 2016 = 18816
Amount payback at the end of second year = 10000
Loan for the third year
= 18816 – 10000 = 8816
Interest for the 3rd year
= 8816 × $$\frac{12}{100}$$ = 1057.92
Amount to be repair at the end of third year
= 8816 + 1057.92 = Rs.9873.92

Money Maths Class 8 Solutions Kerala Syllabus Chapter 5 Question 3.
Rs 200 was got as interest from a bank for 2 years at the rate 5%. Compute the compound interest for the same principal for 2 years at the same rate of interest.
Solution:
Interest for 2 years = Rs 200
Interest for one year = Rs 100
Excess amount in the 2nd year as the compound interest
= 100 × $$\frac{5}{100}$$ = 5
Compound interest for two years = Rs. 205

Textbook Page No. 92

Class 8 Money Maths Kerala Syllabus Chapter 5 Question 1.
Anas deposited Rs 20000 in a bank where compound interest in computed 6% annually. How much amount he will get at the end of 3rd year?
Solution:
Amount Anas gets at the end of third year

Money Maths Class 8 Questions And Answers Kerala Syllabus Question 2.
Diya Deposited Rs. 8000 in a bank where compound interest is computed annually at 10%. Rs. 5000 was withdrawn by her at the end of 2 years. How much amount will he there in the account of Diya after 1 more year?
Solution:
Amount deposited = 8000
Rate of interest = 10%
Amount in the account of Diya at the end of 2 years
= 8000 × (1 + $$\frac{10}{100}$$)2
= 8000 × $$\frac{110}{100}$$ × $$\frac{110}{100}$$ = Rs.9680
Amount withdrawn =Rs. 5000
Principal for the 3rd year
= 9680 – 5000 = Rs.4680
Amount she get at the end of 3 rd year
= 4680 (1 + $$\frac{10}{100}$$)
= 4680 × $$\frac{110}{100}$$
= Rs. 5148

Money Maths Class 8 Pdf Kerala Syllabus Chapter 5 Question 3.
Varun borrowed rupees 25000 from a bank where compound interest in computed at 11% annually. He repaid Rs 10000 at the end of 2 years. How much amount he has to repay after one more year?
Solution:
Amount borrowed = 25000
Rate of interest = 11%
Total liability after 2 year

Amount he repaired = Rs.10000
Principal for the third year
= 30802.50 – 10000
= Rs.20802.50
Amount he has to repay at the end of third year
= 20802.50 × (1 + $$\frac{11}{100}$$)
= 20802.50 × $$\frac{111}{100}$$
= Rs. 23090.78

Textbook Page No. 93

Hss Live Guru 8 Maths Kerala Syllabus Chapter 5 Question 1.
Arun deposited Rs 5000 in a bank where compound of interest is computed half yearly. Mohan deposited RS 5000 in a bank where compound interest is computed quarterly she rate of interest at both the banks is 6%. Money was withdrawn by both of them after an year. How much amount Mohan got more than Arun.
Solution:
Amount deposited by Arun = 5000.
Rate of interest = 6%
Amount Arun gets after an year
= 5000 × (1 + $$\frac{8}{100}$$)2
= 5000 × $$\frac{103}{100}$$ × $$\frac{103}{100}$$
= 5304.50
Amount deposited by Mohan
= Rs 5000
Rate of interest = 6%
Amount Mohan gets after 1 year

= Rs 5306.82
Amount Mohan gets more than
Arun = 5306.82 – 5304.50
= Rs 2.32

Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 5 Question 2.
Rs. 16000 was borrowed by a man from a bank where compound interest is computed quarterly. Annual interest rate is 10%. How much he has to pay after 9 months to clear his liabilities?
Solution:
Amount borrowed = Rs 16000
Interest rate = 10 %
9 Months = 3 quarterly years
Amount to be repaired after 9 months
= 16000 × (1 + $$\frac{2.5}{100}$$)3
= 16000 × ($$\frac{102.5}{100}$$)3
= 16000 × $$\frac{102.5}{100}$$ × $$\frac{102.5}{100}$$ × $$\frac{102.5}{100}$$
= 17230.25

Hss Live Guru 8th Maths Kerala Syllabus Chapter 5 Question 3.
Manu deposited Rs 15000 in a financial institution. Interest is calculated in every 3 months and added to the amount. Rate of interest in 8 %. How much he gets after 1 year.
Solution:
Amount deposited = Rs 15000
Rate of interest = 8%
Amount he gets after 1 year

Kerala Syllabus 8th Standard Notes Maths Chapter 5 Question 4.
John deposited Rs 2500 on 1st January in a co-operative bank. Bank computes compound interest half yearly. Annual interest rate is 6%. Again he deposited Rs 2500 on 1st July? How much amount he will have in his account at the end of the year?
Solution:
Amount John deposits on 1st January = Rs.2500
Principal after the half year up to July
1 st = 2500 (1 + $$\frac{3}{100}$$)
= 2500 × $$\frac{103}{100}$$
= Rs.2575
Amount deposited in July 1st
= Rs 2500
Principal after July 1st= 2575 + 2500 = 5075
Amount he gets at the end of the year 3
= 5075 × (1 + $$\frac{3}{100}$$)
= 5075 × $$\frac{103}{100}$$
= Rs 5227.25

Hss Live Guru Class 8 Maths Kerala Syllabus Chapter 5 Question 5.
Ramlath deposited Rs 30,000 in a financial institution where compound interest is computed in every four months. The ann¬ual interest rate in 9%. How much amount Ramlath gets after 1 year?
Solution:
Amount deposited =Rs 30,000
Rate of interest = 9 %
Amount gets after 1 year
= 30000 (1 × $$\frac{3}{100}$$)3
= 30000 × $$\frac{103}{100}$$ × $$\frac{103}{100}$$ × $$\frac{103}{100}$$
= Rs.32781.81

Textbook Page No. 95

Hss Live Guru 8 Kerala Syllabus Chapter 5 Question 1.
The e-waste increases by 15 % every year, according to the study report. There was around 9 crore tonnes of e-waste in 2014. Then how many tonnes of e-waste will be there in 2020.
Solution:
E-waste in 2014 – 9 crore tonnes
Rate of increase = 15 %
The E-waste in 2020 = 9 × (1 + $$\frac{15}{100}$$)6
= 9 × ($$\frac{115}{100}$$)6
= 20. 82 crore tonnes

Hss Live 8 Maths Kerala Syllabus Chapter 5 Question 2.
A T.V manufacturer reduces the price of a particular model by 5% every year. The current price of this model is Rs. 8000. What would be the price after 3 years?
Solution:
The present price of the T.V = Rs. 8000
rate of reduction = 5% ,
Price after 2 years = 8000 (1 – $$\frac{5}{100}$$)2
= 8000 × ($$\frac{95}{100}$$)2
= 8000 × $$\frac{95}{100}$$ × $$\frac{95}{100}$$
= Rs.7220

Class 8 Maths Hsslive Kerala Syllabus Chapter 5 Question 3.
Tiger is our national animal. Ac-cording to a statistics the number of tigers is reduced annually by 3%. There are 1700 tigers in 2011 according to the senses or tiger protection authority. Then how many tigers will be there in 2016?
Solution:
The number of tigers in 2011 = 1700
Reduction rate = 3%
No. of tigers in 2016 after 5 years of the senses

= 1459.84
= 1460

Kerala Syllabus 8th Standard Maths Notes Chapter 5 Question 1.
Nirmala deposited Rs. 20,000 in a bank where compound interest of rate 7% is computed annually Rs. 5000 was with drawn after one year. How much amount she will get after 2 years.
Solution:
Amount deposited in the bank = Rs. 2000
Rate of interest = 7%
principal after one year
= 20000 + [20000 × $$\frac{7}{100}$$] = 20000 + 1400 = 21400
Amount withdrawn after one year = Rs. 5000 .
Principal for the 2-nd year
= 21400 – 5000 = Rs. 16400
Amount she gets after 2 years
= 16400 + [16400 × $$\frac{7}{100}$$] = 16400 + 1148
= Rs. 17548

Question 2.
Aswathy deposited Rs. 10,000 in a bank where compound interest is calculated at the rate of 10%. Rs.5000 was deposited in the beginning of the 2-nd year and Rs.5000 in the beginning of the 3-rd year. How much amount she gets after 3 years?
Solution:
Amount deposited in the first year = Rs. 10,000
rate of interest = 10%
Amount at the end of first year
= 10000 + [10000 × $$\frac{10}{100}$$] = 10000 + 1000 = Rs. 11000
Principal’for the 2-nd year
= 11000 + 5000 = Rs. 16000
Amount at the end of the 2-nd year
= 16000 + [16000 × $$\frac{10}{100}$$] = 16000 + 1600 = Rs. 17600
Principal for the 3-rd year
= 17600 + 5000 = Rs.22600
Amount at the end of 3-rd year
= 22600 + [22600 × $$\frac{10}{100}$$] = 22600 + 2260 = Rs. 24860
Amount Aswathy will get at the end of 3-rd year = Rs. 24860

Question 3.
Raj an borrowed Rs. 15000 from a co-operative bank for business purpose. The bank computes 9% interest. How much money she has to repay after 5 months.
Solution:
Amount borrowed = Rs. 15000
Rate of interest = 9%
Time duration = 5 month months
Interest .
= 15000 × $$\frac{9}{100}$$ × $$\frac{5}{12}$$ = 562.50
Amount he has to repay
= 15000 + 562.50
= Rs 15562.50

Question 4.
Rasiya deposited Rs. 20,000 in a bank where compound interested is computed half yearly. If the rate of interest is 11%, How much amount she will get after an year?
Solution:
Capital for first year = Rs. 20,000
Rate of interest = 11 %
Interest of first half year
= 20000 × $$\frac{11}{100}$$ × $$\frac{1}{2}$$ = Rs. 1100
Principal of 2-nd half year
= 20000 + 1100
= 21100
Interest of the 2-nd half year
= 21100 × $$\frac{11}{100}$$ × $$\frac{1}{2}$$ = Rs. 1160.5
Amount she gets after one year
= 21000 + 1160.5
= Rs. 22160.5

Question 5.
The price of a car is rupees 5 lakh and it depreciates by 6% every year. What would be the price after 2 year ?
Solution:
Here the price every year is 6% less than the previous years price.
First year’s price = Rs 500000
First year’s depreciation
= 500000 × $$\frac{6}{100}$$ = Rs. 300000
Second year’s price = Rs 4,70000
Second year’s depreciation
= 470000 × $$\frac{6}{100}$$ = 28200
The price of the car after 2 years
= Rs 4,70,000 – Rs 28200
= Rs 441800

Question 6.
The simple interest of an amount is Rs 50 for two years and the compound interest is Rs 55. The rate of interest is same in both the cases. Find the rate? Find the amount?
Solution
The simple interest of the amount for two years = Rs 50
The simple interest of the amount for 1 year = Rs 25
The compound interest of the amount for 2 years = Rs 55
The interest in the 2 nd year
= 55 – 25 = Rs 30
Interest for Rs 25 = Rs 5

Question 7.
A company which manufactures computers increases its production by 10% every year. In 2009 the company produced 80,000 computers. How many computers would it produced in 2011?
Solution:
Here the number of computers produced every year is 10% more than the number produced the year before. So starting from 80,000. We have to find the number of computers produced every year after that for two years.
Number of computers produced in 2009 = 80,000
Number of computers produced in 2010
= 80000 + 80000 × $$\frac{10}{100}$$
= 80000 + 8000 = 88000
Number of computers produced in 2011
= 88000 + 88000 × $$\frac{10}{100}$$
= 88000 + 8800 = 96800

Question 8.
Ramu borrowed Rs 50,000 from a bank at the interest rate of 10% for agricultural purpose. The interest rate will be reduced by 5% if he repays the amount properly within 2 years. If he fails to repay in time there will be fine as 1%. Ramu could not repay the amount in time. How much amount Ramu repair?
Solution:
Amount borrowed =Rs 50,000
Rate of interest = 10%
Interest rate including fine = 10 + 1=11%
Amount to be repaid
= 50000 + [50000 × $$\frac{11}{100}$$ × 2]
= 50000 + 11000 = 61000
= Rs 61000

Question 9.
Raju has Rs 800 with him. He spent 25 % of it. How much amount left with him?
Solution:
Total amount = Rs 800
Amount spent = 800 × $$\frac{25}{100}$$ = 200
Amount left with Raju = 800 – 200
= Rs 600

Question 10.
Balu and Ramu decided to borrow Rs 15000 each for a joint business. Balu borrowed Rs 15000 from a financier Who imputes Rs 5 per month for Rs 100 as interest. Ramu borrowed Rs 15000 from a bank where compound interest of 12 % is computed. How much money both of them have to repay after 2 years?
Solution:
Interest Balu has to pay after 2 years
= 15000 × $$\frac{60}{100}$$ × 2 = Rs. 18000
(Interest for Rs 100 per month in Rs 5. Interest for Rs 100 in an year = 5 × 12
= Rs. 60.
The rate of interest = 60%)
Amount Balu has to repay after 2 years
= 15000 + 18000
= Rs 33000
In the case of Ramu interest is computed as compound interest.
Principal for first year = Rs 15000
Interest for 1 st year
= 15000 × $$\frac{12}{100}$$ × 1 = Rs 1800
Principal for 2 nd year
= 15000 + 1800
= Rs 16800
Interest for the second year = 16800 × $$\frac{12}{100}$$ × 1 = Rs. 2018
Amount Ramu has to repay
= 16800 + 2018
= Rs 18816

Question 11.
The population of Kerala increases by 3% every year. The current population is 5 crore. What would he the population after 2 years.
Solution:
Current population = 50000000
Percentages increase in every year = 3%
The population in Kerala after 3 year
= 50000000 [1 + $$\frac{3}{10}$$]2
= 50000000 × $$\frac{(103)^{2}}{10000}$$
= 530,45000

Question 12.
A financial company claims that it charges only 20% interest on loans. But if a person takes out a loan of Rs 100 he would get only Rs 80, after subtracting the annual interest of rupees 20 at the outset. And he has to pay back Its 100 after 1 year. How much is their real interest?
Solution:
A borrower gets only Rs 80 when he borrows Rs 100
He has to repay Rs 100.
Real interest
$$\frac{20}{80}$$ × 100 = 25%

## Kerala Syllabus 8th Standard English Solutions Unit 2 Chapter 3 Marvellous Travel

You can Download From Marvellous Travel Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard English Solutions Unit 2 Chapter 3  helps you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 3 Marvellous Travel (Joshua Fernandez)

### Marvellous Travel Textbook Questions And Answers

Marvellous Travel Kerala Syllabus 8th Question 1.
Why does the poet say that he travels with his ‘eyes’ and ‘thoughts’?

The poet keenly observes and explores everything that comes along and enjoys them.

Marvellous Travel Question Answers Kerala Syllabus 8th Question 2.
What does the poet mean by ‘I travel with my pen’?

He records his travel experience to write travelogues.

Marvellous Travel Summary Kerala Syllabus 8th Question 3.
That something new, would spring into my horoscope…’ What does this line mean?

This line means that the poet hopes to have new experiences that would change his life.

Marvellous Travel Poem Summary Kerala Syllabus 8th Question 4.
Who are the different types of people the poet meets during his journey?

Children, women, men, tribesmen and little girls.

Marvellous Travel Poem Questions And Answers Question 5.
‘Every experience is worth life’s journey.’ What does the poet mean by this?

Life is a journey from birth till death. The journey of life offers a mixture of good and bad experience which is true about travelling too. Such experiences may help the individual in facing the challenges that life throws to him.

Marvellous Travel Poem Kerala Syllabus 8th Question 1.
Read the lines from the poem ‘ Marvellous Travel’ and answer the following questions.
I travel with my eyes
Watching those silently cry
Someone left them without saying goodbye
I travel with my thoughts
I travel with my pen
To write about children, women and men
I travel with my voice,
I travel with my hope,
That something new would spring into my horoscope,
Whether in Asia, America or Europe,
There’ll always be something interesting to scope;
a. What does the poet mean by saying 1 travel with my eyes’?
b. Why does the poet say that he travels with his pen?
c. Pick out an instance of alliteration from the above stanzas.
d. What does the poet hope for?
e. Pick out a pair of rhyming words.
a. The poet keenly observes everything
b. He records his travel experiences to write travelogues.
c. something, scope/ something, spring
d. The poet hopes that something new would spring into his horoscope.
e. pen-men / cry-why

Marvelous Travel Poem Summary Kerala Syllabus 8th Question 2.
Prepare a short profile of Joshua Fernandez using the hints given below.
Name: Joshua Fernandez
Born : 19 April 1974, Kuala Lumpur, Malaysia
Famous as: Popular Malaysian Film director and designer
Major Works: Clock on the wall, Where Am I, Choices, Granted All That has Happen.
Joshua Fernandez:
Joshua Fernandez was born on 19 April 1974 in Kaula Lumpur in Malaysia. He was a popular Malaysian film director and designer. His major works are Clock on the wall, Where Am I, Choices, Granted All that has happened.

### Marvellous Travel Summary in English

[Travelling is a method to know the world around us. By travelling we can find new people and their lives.]
I watch people who are crying. I see people who went away without saying good-bye. As I travel I meet women, men and children whose stories are written with my pen. While travelling I hope to see new things around. Irrespective of the places everywhere I find interesting things. When I am travelling I find girls, tribesmen and. different types of people. I do not take money with me during my journey. Life as a journey is with full of good, bad and ugly people. Each and every experience in life is valuable.

### Marvellous Travel Summary in Malayalam

Marvellous Travel Glossary

## Kerala Syllabus 8th Standard Basic Science Solutions Chapter 14 For the Continuity of Generations in Malayalam

Students can Download Basic Science Chapter 14 For the Continuity of Generations Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Basic Science Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Syllabus 8th Standard Maths Solutions Chapter 8 Area of Quadrilaterals in Malayalam

Students can Download Maths Chapter 8 Area of Quadrilaterals Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units

You can Download Measurements and Units Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units

In olden days different scales were used for measuring length in different countries. Vaara, One forearm (muzham) etc are some of such units. These units were not accurate. For solving this problem a standard scale of a definite length has been recognised world widw for measuring length. This standard scale is the unut of length. The basic unit of length is metre (m).

Smaller units of length

Examine the metre scale. You can see small measurements on it. These are centimetre (cm) and millimetre(mm)
1m = 1oocm 1cm=1omm
The smallest length that can be accurately measured using a metre scale is its least count.

Thickness of paper

We cannot measure the thickness of the paper even with these small units. Take too sheets of paper and pile it up to make a bundle. Measure the height of this bundle using a scale. If it falls in between two markings, make suitable changes in the number of sheets. Then measure its height. By dividing this height by the number of sheets, we will get the thickness of a sheet of paper.

Similarly to find out the thickness of thin wire roll it on a pencil and mea-sure the length of ten rolls Then divide it by 10.

Length of a curved line

To measure the length of the curved line put a thread on it and measure its length using a meter scale

Distance of a sphere

A ball kept in between two wooden blocks, whose diameter is to be measured. Then we will get its diameter.

Various measurements of length

The basic unit of length is metre. Smaller units like centimetre (cm), millimetre (mm), micrometre (micron – ™ nano-metre (nm) are also used in certain situa-tions for convenience. We use a bigger unit like kilometre (km) to measure the distance between two places. Units like astronomical unit (AU), light year (ly) are also used to measuring distance to planets or stars.

Measurement And Units In Physics Class 8 Questions And Answers Mass

Mass is the measure of quantity of matter contained in it. The basic unit of mass is kilogram(kg).

Except kilogram, milligram(mg), gram(g), quintal, tonne are also the units of mass.

Measurement And Units In Physics Class 8 Kerala Syllabus  Time
The symbol is s.
60s = iminute; 6ominute = 1 hour One second is 1/ 86400 part of an average solar day.

Fundamental units

There are some quantities which are not related to one another and cannot be expressed using other quantities. Such quantities are fundamental quantities. The units of the fundamental quantities are the Fundamental Units. The system based on these fundamental units is the | International System of Units. Its short form is SI Units.

Measurements And Units Class 8 Notes Kerala Syllabus Derived unit

Units which are expressed in terms of fundamental units or those units which are dependent on fundamental units are derived units.
Ex. Area = length x breadth

Unit of area = unit of length x unit of breadth = m xm = m2.

Salient features of SI units

1. Unified units
2.  Internationally accepted
3.  Adequate to express all physical quantities.

Measurement And Units In Physics Class 8 Notes Density and volume

The space occupied by a body is referred to as its volume.
The unit of volume is cubic metre (m3)

Mass per unit volume of substance is its density. Density = mass/volume Unit of density = unit of mass/unit of volume = kg/m3

The rules to be followed while writing units

We have to observe certain rules according to international standards while using these units and their symbols. They are given below:

1. The symbols of units are normally written using small letters in the English alphabet, eg. m (metre), s (second), kg (kilogram)
2. But there are certain occasions on which capital letters of the English alphabet are used as symbols. The units named after persons are written like this.
3. While writing the names of units never use capital letters, eg. kelvin (correct) Kelvin (wrong) newton (correct) Newton (wrong)
4. Never use the plural form for symbols, eg. 10 kg (correct) 10 kgs (wrong) 75 cm (correct) 75 cms (wrong)
5. Never use full stop or comma after a
symbol except at the end of a sentence. eg. 75 cm is the length of a table, (correct) 75 cm. is the length of a table, (wrong),
6. While writing derived units a slash (/) is used to denote division. But never use more than one slash in one derived unit. eg. m/s1 (correct) m/s/ s (wrong)
7. When a derived unit is expressed as the product of other units use a dot or a space between them. eg. N.m or Nm
8. 8. Do not mix the name of a unit with the symbol.
9. While writing units along with a value, there must be single space between them.
10. Never use more than one unit to express a physical quantity.

### Measurements and Units Textbook Questions and Answers

Units And Measurements Class 8 Kerala Syllabus Question 1.
Write down the following units in the ascending order of their values
a) mm nm cm µ m.
b) 1 m 1 cm 1 km 1 mm
a. nm → µ m → mm → cm
b. 1 mm → 1 cm → 1 m → 1 km

Basic Science Question Answer Chapter Wise Class 8 Question 2.
Which of the following units does not belong to the group?
a) kg mg g mm
b) m mm km mg
a. mm
b. mg

Chemistry Textbook For Class 8 Kerala Syllabus Question 3.
Imagine that the distance from school to your friend’s house is 2250 m. State this in kilometres.
2$$\frac{1}{4}$$ km or 2.25 km

8th Class Physics Question And Answers Kerala Syllabus Question 4.
Convert the following units, into SI units without changing their values.
a. 3500 g
b. 2.5 km
c. 2 h
a. 3.5 kg
b. 2500 m
c. 7200 s

Kerala Syllabus 8th Standard Notes Physics Question 1.
Take a small quantity of solution A and solution B in a jar. Density of A is 2 kg/m3 and density of B is 5 kg/m3. Which of these flot above the jar ? Give reasion.
Solution A
Density of A is less than the density of B. So solution with less density floats above the solution that has high density.

Class 8 Chemistry Chapter 1 Kerala Syllabus Question 2.
Find out the given word pair relation; and fill in the blanks.
a. Length : m ; Volume :………
b. Displacement : m ; Acceleration:…………….
a. m3
b. m/s²

Basic Science Class 8 Solutions Kerala Syllabus Question 3.
a. What is the basic unit of time?
b. What do you mean by one Solar day?
c. How much time earth will take to revolve once around the sun?
a. Second
b. A day or a solar day is the time period from one noon to the next noon.
c. One year (365 days)

Class 8 Science Notes Pdf State Syllabus Question 4.
a. meter, kilogram, kilogram/ meter3, second
b. Length, volume, mass, time
c. Volume, mass, density accelaration.
Ans:
a.kilogram/meter3 – meter, kilogram, second are basic units, kilogram/meter3 is unit of volume.
b. Volume – Volume is a derived quantity and others are basic quatities.
c. Mass – Mass is a basic quantity and others are derived quantity.

Class 8 Physics Kerala Syllabus Question 5.
What is meant by least count of a metre scale?
The smallest length that can be accurately measured using a metre scale is its least count. The least count is 1mm when metre scale is used.

Basic Science Class 8 Pdf Notes Kerala Syllabus Question 6.
Mention a method to measure the thickness of a paper sheet.
Take 100 sheets of paper and pile it up to make a bundle. Measure the height of this bundle using a scale. If it falls in between two markings, make suitable changes in the number of sheets. Then measure its height. By dividing this height by the number of sheets, we will get the thickness of a sheet of paper.

Hsslive Guru 8th Class Physics Kerala Syllabus Question 7.
Using following objects how can you measure the diameter of a ball?
hall, two wooden blocks, metre scale
The ball is kept in between two wooden blocks as shown in the figure whose diameter is to be measured.

Basic Science Class 8 Kerala Syllabus Question 8.
Fill in the blanks
a. 1 nm = ….. m
b. 1 km = …….cm
c. 1000g = …….kg
d. one light year = …… km
e. 60 sec = …… minute
a. 1 nm = 10-9 m
b. 1 km = 105
c. 1000g = 1kg
d. one light year = 365 × 24 × 60 × 60 × 105 km
e. 6o sec = 1 minute

Chemistry Class 8 Kerala Syllabus Question 9.
Calculate how many seconds in a solar day.
24 ×60 × 60s

Std 8 Biology Notes Malayalam Medium Kerala Syllabus Question 10.
Explain the derived units with examples.
Units which are expressed in terms of fundamental units or those units which are dependent on fundamen¬tal units are derived units.
Ex . Area = length × breadth=m × m = m2

Question 11.
Tabulate the fundamental quantities and their units.

Question 12.
What are the salient features of SI units.?

• Unified units
• Internationally accepted
• Adequate to express all physical quantities.

Question 13.
Find the volume of a box having length 2m,breadth lm and height 0.5m
volume = length × breadth × height =2 × 1 × 0.5 = 1 m3

Question 14.
Will the mass be same if you take one thermocol and a wooden block of same volume?
No. Mass/Volume is density. The density of thermocol is lesser than that of wooden block. The mass will not be the same.

Question 15.
Take one jar with salt water and into both jars. In which jar the egg floated over the liquid. Justify your answer.
In the jar having salt water because it has more density.

Question 16.
Tabulate following units and symbols
a. Length
b.Mass
c. Potential difference
d. Force
e.Pressure
f. Time
g.Density

Question 17.
Choose the derived quantities from following.
a. Length
b. volume
c. time
d. temperature
e. density
f. velocity
g. intensity of light
volume, density, velocity

Question 18.
Tabulate two units connected to the name of personals

Question 19.
Find the density of an object of mass 60 kg a having length 5m breadth 3m and height 2m.
density =mass ÷ volume
$$=\frac{60}{5 \times 3 \times 2}=\frac{60}{30}=2 \mathrm{kg} / \mathrm{m}^{3}$$

Question 20.
Find the volume of a brick which has a length, breadth and height of 3m, 2m and 4m respectively.
Volume = Length × breadth × height = 3 × 2 × 4 = 24m3

Question 21.
How to find the area of irregular surface? Explain with example.
The area of irregular surface can be calculated by using graph paper.

Question 22.
How do you find out the diameter of a torch cell
Place the cell between two wooden blocks tightly. Find out the distance between the two faces of the blocks, touching the cell using a metre scale. This reading gives the diameter of the cell.

Question 23.
The length of a room is 8m and the breadth ia 5m. Calculate the area of the room.
Aiea = length × breadth = 8m × 5m = 40m2

Question 24.
What is meant by density of a substance?
Density is the mass of a substance having unit volume. If the mass of the substance is M and V is its volume,
then $$\frac{M}{V}$$

Question 25.
Find out the fundamental units from the following table and mark them. Find out which unit is used to measure each of the physical quantity.

• Ampere
• Metre/second
• Kilogram
• Kilogram/metre
• Mole
• Kelvin
• Metre3
• Kilogram metre/second

Fundamental units

• Ampere
• Kilogram
• Mole
• Kelvin

Physical quantity

• Ampere – Electric current
• Kilogram – Mass
• Mole – Amount of substance.
• Kelvin – Temperature

## Kerala Syllabus 8th Standard Maths Solutions Chapter 6 Construction of Quadrilaterals in Malayalam

Students can Download Maths Chapter 6 Construction of Quadrilaterals Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 16 Water

Water is a precious natural resource. Water is essential not only for sustaining life but also for agriculture, industry, energy production and transportation. Water is formed by the combination of hydrogen and oxygen. Hydrogen and oxygen are produced by decomposition of water. Decomposition of water is done by electrolysis of water.

Basic Science Class 8 Chapter 16  Characteristics of water

Water is a substance found in nature in all the three states of matter namely solid, liquid and gas. The temperature at which a liquid boils at normal atmospheric pressure is its boiling point. The boiling point of water is 100°C. Once water starts boiling, the temperature will not change because all the heat supplied is utilised for the change of state. Water has the ability to hold more heat than other liquids. The ability to hold heat by a liquid is its heat capacity. Because of its high heat capacity water is utilised in radiators to regulate heat, to cool hot objects, to control the temperature of earth etc.

Water becomes ice in 0°C. The temperature of a liquid at which it freezes to solid at normal atmospheric pressure is called its freezing point. When water is converted to ice its volume increases and the density decreases. The unbalanced attractive force on the surface of a liquid causes the liquid, surface behaves like a stretched membrane. This is surface tension. The surface tension decreases the surface area of a liquid. The liquid drops remain like a ball to decrease the surface area. Adding soap to water is a way to decrease surface tension of water.

Class 8 Chemistry Notes Kerala Syllabus The components of water

When electricity is passed through water it decomposes to hydrogen and oxygen. For this, we can make water voltmeter shown in fig.

Iron nail and plastic bottle can be used for this. Adding a few drops of acid into water and battery are also used. When electricity is passed through it oxygen and hydrogen are filled in the test tubes. This process is called electrolysis. The apparatus used for the electrolysis of water is called Hoffmann Water Voltameter.

Kerala Syllabus 8th Standard Physics Notes Reaction of water with metals

Coldwater reacts with sodium, potassium etc. to liberate hydrogen. Magnesium reacts with hot water and iron reacts with steam to liberate hydrogen gas.

Hss Live Guru 8 Chemistry Kerala Syllabus Water the Universal solvent

Several substances dissolve in water. Since water can dissolve various substances and widely used for preparing solutions it is a universal solvent.

8th Standard Chemistry Textbook Soft water and hard water

The water in which soap does not lather easily is called hard water. Water in which soap gives lather readily is called soft water. The hardness of water containing calcium or magnesium bicarbonate is removed during boiling. This type of hardness is called temporary hardness. But the hardness of water containing the chlorides and sulphates of calcium and magnesium is not removed by boiling. Such hardness is called permanent hardness. Pure water has neither the properties of acid nor those of alkali, hence it is a neutral solvent.

Std 8 Chemistry Notes Kerala Syllabus Water pollution

The following activities cause the water pollution

• Dumping of waste in water resources.
• Rampant use of fertilisers
• Excessive use of detergents
• Insecticides getting mixed with water

### Water Textbook Questions and Answers

Kerala Syllabus 8th Standard Basic Science Notes Questions 1.
When water is heated at its boiling point or melting point, its temperature does not change.
a. What is meant by boiling point and melting point?
b. What are the boiling and freezing points of water?
c. Why is there no change in temperature?
a. The temperature at which a liquid boils at normal atmospheric pressure is its boiling point. The temperature of a liquid at which it freezes to solid at normal atmospheric pressure is called its freezing point.
b. 100°C, 0°C.
c. Once water starts boiling, the temperature will not change because all the heat supplied is utilized for the change of state. So the temperature does not vary.

8th Class Chemistry Notes Kerala Syllabus Question 2.
A definite quantity of water and coconut oil are heated in separate test tubes using the same amount of heat.
a. In which case does the temperature increase slowly?
b. What is the reason for this?
c. Write any one practical application of this property.
a. water
b. The high heat capacity of water
c. 1. To regulates the heat of radiator in vehicles
2. To regulate the heat of earth

8th Standard Water Lesson Kerala Syllabus Question 3.
100 ml. each of coconut oil and water are taken in two beakers and kept in the freezer.
a. What difference can be observed in their volumes during freezing?
b. What do you infer from the observation?
c. When water is frozen in glass bottles, it is advised not to fill the bottles completely. Explain the reason.
a. The volume of the beaker containing water is greater.
b. Due to the anomalous expansion of water.
c. The volume of water in the bottle filled with water is increased and the bottle is broken.

Hss Live Guru 8 Physics Kerala Syllabus Question 4.
Soap decrease the surface tension of water.
a. What is surface tension?
b. How does the decrease in surface tension benefit washing of clothes?
a. The unbalanced attractive force on the surface of a liquid causes the liquid surface behaves like a stretched membrane. This is surface tension
b. When we add soap to water it reduces the surface tension of water and water can enter into the fine threads of the fabric.

Basic Science For Class 8 Chapter 16 Question 5.
Surface tension tends to minimise the surface area of a liquid. Suggest an experiment to prove this. (Follow the pattern: Required materials, procedure, expected observation).
’Materials required: metallic loop, thread, soap water, pin Method of experiment: Tie a thread to a metallic loop, immerse it in soap water and create a soap film as shown in fig. a, b, c.

Prick a portion of film using a pin. The shape of remaining portion of soap film will be as such the surface area is reduced.

Basic Science Class 8 Ch 16 Kerala Syllabus Question 6.
Providing excess food for fish in an aquarium is a threat to its survival. Justify.
When we add more food materials into water, the dissolved oxygen is utilized to decompose the food materials. So the amount of oxygen is reduced and the water is polluted.

Hsslive Guru 8th Class Chemistry Question 7.
Some substances when dissolved in water cause hardness of water.
a. Which of the following substances cause hardness of water? Sodium chloride, Calcium bicarbonate, Calcium carbonate, Calcium sulfate, Magnesium sulfate, Calcium chloride, Magnesium carbonate
a. The hardness due to which of the above salts cannot be removed by boiling?
a.calcium bicarbonate, calcium sulfate, calcium chloride.
b. The hardness of water containing the chlorides and sulphates of calcium and magnesium is not removed by boiling.

Kerala Syllabus 8th Standard Chemistry Notes Question 1.
a. What are the gases liberated in the water voltameter arranged for decomposition of water?
b. Which gas is liberated at the negative pole of the battery?
a. Hydrogen and oxygen
b. Hydrogen
c. Ratio of hydrogen and oxygen – 2: 1

8th Class Biology Notes Pdf Question 2.
Heat equal quantity of water and coconut oil in two vessels.
a. The temperature of which liquid will be raised rapidly?
b. Which has more heat capacity?
a.Coconut oil
b.water

Hss Live Guru 8th Chemistry Kerala Syllabus Question 3.
Take water in a vessel and put a needle on the surface of water carefully.
b. Why do the water drops assume spherical shape?
a. No. Because of the surface tension of water.
b. The surface tension reduces the surface area.

Hsslive Guru 8th Basic Science Question 4.
Name two metals which react with cold water.
Potassium, sodium

Hsslive Guru Chemistry 8 Kerala Syllabus Question 5.
Which is the metal reacts with hot water? Which is the gas produced? Write the chemical equation.
Magnesium; Hydrogen.
Mg + H2O → MgO + H2

Hss Live Guru 8th Basic Science Question 6.
Why does water is known as universal solvent?
Since water can dissolve various substances and widely used for preparing solutions it is a universal solvent.

8th Physics Guide Kerala Syllabus Question 7.
What are hard water and soft water?
The water in which soap does not lather easily is called hard water. Water in which soap gives lather readily is called soft water.

Class 8 Physics Notes Kerala Syllabus Question 8.
Give the reason for temporary hardness of water.
The hardness of water is due to the presence of calcium or magnesium bicarbonate. This type of hardness is called temporary hardness.

8th Standard Chemistry Textbook Kerala Syllabus Question 9.
Hardness of water is not removed even after it is boiled. What may be the reason for this?
This is due to chloride and sulphate of magnesium and calcium are dissolved in water.

Class 8 Basic Science Chapter 16 Question 10.
Why is air continuously introduced into the water in an aquarium?
Aquatic plants and animals make use of oxygen dissolved in water. Water gets polluted as the amount of oxygen in it decreases. Air is introduced into water to make up the quantity of oxygen.

Question 11.
Write two remedies to prevent water pollution.

• Stop dumping of waste in water resources.
• Reduce the use of fertilisers and pesticides.

Question 12.
Write two methods to reduce the surface tension of water

• Boil the water.

## Kerala Syllabus 8th Standard English Solutions Unit 4 Chapter 1 Song of the Flower

You can Download From Song of the Flower Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard English Solutions Unit 4 Chapter 1  helps you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Hindi Solutions Unit 4 Chapter 1 Song of the Flower (Khalil Gibran)

### Song of the Flower Textbook Questions And Answers

Song of The Flower Questions and Answers Question 1.
Who is the speaker in the poem?
The Flower

Song of The Flower Notes Question 2.
What do the expression ‘blue tent’ and ‘green carpet’ refer to?
‘Blue tent’ refers to the sky and the ‘Green carpet’ refers to the earth full of vegetation.

Song of The Flower Appreciation Question 3.
How are the seasons related to the life of the flower?
The life of the flower is related to the season. Winter conceived it, spring gave birth to it, summer reared it and in autumn it passed away.

Song of the flower Poem Question 4.
What does the flower do at dawn and at dusk?
The flower joins the breeze and announces the arrival of the light in the morning. In the evening it joins the

Song Of The Flower Question Answer Kerala Syllabus 8th Standard Question 5.
How does the flower make the world beautiful?
lt decorate the plain and fills the air with its fragrance.

Song Of The Flower Kerala Syllabus 8th Standard Question 6.
What is described as ‘the eye of the day’?
The sun

Song Of The Flower Summary Kerala Syllabus 8th Standard Question 7.
How does the flower become a part of joy and sorrow?
The flower is used to make wreaths on both joyous and sorrowful occasions like wedding and death respectively. Thus it becomes the part of joy and sorrow.

Song Of The Flower Questions And Answers Kerala Syllabus 8th Standard Question 8.
Why does the flower consider itself as ‘the last gift of the living to the dead’?
Because it is used as a wreath to be placed on a dead body.

Song Of The Flower Activities Kerala Syllabus 8th Standard Question 9.
What, according to the flower, is the wisdom that man should learn?
The flower always looks up to see only the light. It never looks down to see the shadow. Similarly, man should see the brighter side of life and never brood over its darker side.

### Song of the Flower Textbook Activities And Answers

Let’s revisit

Song Of The Flower Lesson Plan Kerala Syllabus 8th Standard Question 1.
What does the flower symbolize? Explain.
The flower symbolizes kindness, beauty, fragrance, love – all the bright things in nature. It also symbolizes the experiences we have in our life – joy, sorrow, failure, victory, etc.

Song Of The Flower Poem Questions And Answers Kerala Syllabus 8th Standard Question 2.
Quote some instances in which the flower becomes part of joy and sorrow.
Wedding (Joy), Funeral (Sorrow)

Song Of The Flower Appreciation Kerala Syllabus 8th Standard Question 3.
Read the last stanza of the poem once again. What are the two contrasting ideas presented in it? Do you think the poet is optimistic? Why?
The light is contrasted with the shadow. Yes, the poet is optimistic. The attitude of the flower represents the poet’s optimism. The flower likes to see only the light and not the shadow.

Appreciation Of The Poem Song Of The Flower Kerala Syllabus 8th Standard Question 4.
Pick out the lines and expressions that impressed you the most. Give reasons.
But I look up high to see only the light And never look down to see my shadow I like these lines the most because here the poet says that man should see the brighter side of life and never brood over their darker side. I appreciate the poet’s optimistic view of life.

Song Of The Flower Poem Kerala Syllabus 8th Standard Question 5.
What do you think is the mood of the poem – happy, sad, humorous or philosophical?
The mood of the poem is happy and philosophical. The poem reflects upon the themes of life and death and imparts the wisdom that man must look always for light.

Activity 1.

Read the poem again and complete the following notes. One hint is given for you.

Song Of The Flower Stanza Wise Summary Kerala Syllabus 8th Standard Question 1.
What does the flower say about itself?
1. a kind word uttered and repeated by the voice of nature.
2. ………………………………
3. ………………………………
1. ‘a star fallen from the sky’
2. ‘daughter of the elements’
3. ‘was conceived by winter

Song Of The Flower Notes Kerala Syllabus 8th Standard Question 2.
What are the everyday activities that the flower is engaged with?
1. announce the coming of light
2. …………………..
3. …………………..
1. ‘announce the coming of the light’
2. ‘bid the light farewell’
3. ‘decorate the plains with beautiful colors’

Summary Of The Poem Song Of The Flower Kerala Syllabus 8th Standard Question 3.
Who watches over the flower?
……………………
Eyes of night’

Song Of The Flower Theme Kerala Syllabus 8th Standard Question 4.
How does the flower enjoy nature?
1. I drink dew
2. ………………..
3. ………………..
1. ‘listen to the voice of the birds’
2. ‘dances to the rhythmic swaying of the grass’

Song Of The Flower By Khalil Gibran Kerala Syllabus 8th Standard Question 5.
What is the flower used as
2. ………………….
3. ………………….
4. ………………….
2. wedding wreath
3. wreath for the dead body
4. memory of a moment of happiness

Appreciation of the Poem Song of the flower Question 6. What is the flower’s vision?
1. see only the light
2. ………………..
2. ‘never look down to see my shadow’

Activity 2.

I am a kind word uttered and repeated By the voice of Nature Here the flower is considered as a word uttered by Nature. This is an example of a metaphor. A metaphor is a figure of speech that develops a hidden comparison between two objects that share something in common. It is different from a ‘simile’ where the comparison is drawn using ‘like’ or ‘as’.
Pick out few more examples of metaphors from the poem.
1. ………………………………….
2. ………………………………….
3. ………………………………….
1. I am a star.
2. I am the daughter of the elements.
3. The eyes of the night
4. The eyes of the day

Write what the following metaphors imply. One is done for you.

Attempt writing your own examples of metaphors
1. …………………..
2. ………………….
3. ………………….
4. …………………
1. Her eyes were fireballs.
2. I was lost in a sea of nameless faces.
3. He is a rolling stone
4. The wheels of justice turn slowly

Summary Of Song Of The Flower Kerala Syllabus 8th Standard Activity 3.

Read the poem again and complete the following spider diagram.

Now, can you prepare an appreciation of the poem ‘Song of the Flower’ using the spider diagram?
……………………………………….
……………………………………….
……………………………………….

Appreciation of the poem.
Song of the flower is a beautiful poem about nature and its relationship with living beings. This poem was written by Khalil Gibran, the famous Lebanese poet. This poem speaks about the life cycle of a flower and its experience. The flower in the poem is a representative of every aspect of nature. The flower claims that it is a kind word uttered and repeated by nature’s voice. It is a star fallen from the sky. The sky is referred as ‘the blue tent’. All the four seasons play vital roles in the life of the flower. The flower is conceived by winter, given birth by spring, reached by summer and eternal sleep by autumn. The flower heralds the coming of the light in the dawn. It bids farewell to the light in the evening.

The plains are decorated with beautiful flowers. The air is filled with the lovely fragrance of the flower. Days and nights shower love, care and affection on the flower lavishly. The flower enjoys nature as much as it can. To the flower dew is wine. It sings with- the birds and dances according to the rhythm of the swaying grass. The poet is impressed by the beautiful image of nature. The flower becomes a part of joy as well as sorrow because it is used on occasions of wedding and funeral. The poem teaches us how to deal with life’s struggles, to learn from our past and face the future with confidence and hope. The poet uses appropriate similes and metaphors in the poem, that makes the poem more beautiful.

Questions l to 4: Read the lines from the poem ‘ Song of the Flower’ and answer the questions that follow:
I am a star fallen from the
Blue tent upon the green carpet
I am the daughter of the elements
With whom Winter conceived:
To whom Spring gave birth: I was
Reared in the lap of Summer and I
Slept in the bed of Autumn
At dawn, I unite with the breeze
To announce the coming of light:
1. I am a star fallen from the Blue tent upon the green carpet …………… What do the expressions ‘ blue tent and ‘green carpet’ imply?
2. What does the speaker do at dawn?
3. ‘I am the daughter of the elements …………… Identify the figure of speech used in this line.
4. Pick out an example of a visual image from the above lines.
1. Blue tent – Sky
Green carpet – Earth
2. Unite with the breeze to announce the coming of light.
3. Metaphor
4. A star fallen from the blue tent.

8th Standard English Question 2. Prepare a profile of the famous writer KHALIL GIBRAN using the hints given below.
Birth: 1883
Nationality: Lebanese
Famous as: Artist, poet, writer, philosopher, Considered as third most popular poet in history.
Works: ‘ The Prophet’ (1923)
Death: 1931
Khalil Gibran:
Khalil Gibran the famous Lebanese poet was born in 18 83. He was famous as a Artist, poet, philosopher. He was considered as the third most popular poet in history. His major work is The Prophet in the year 1923. He passed away in the year 1931.

### Song of the Flower Summary in English

This beautiful poem portrays the character of a flower. According to the poet, the flower is a word spoken and repeated by nature. It is a star that has fallen from the blue sky upon the green grass on the Earth. Then the poet describes how the different seasons show their role in the life of the flower. It is conceived (created as embryo)in the winter, born in the spring, summer looks after it like its mother and it bids farewell(dies) to the world in autumn. When morning arrives, it joins the breeze to announce the coming of the sun and in the evening along with the birds, it bids farewell to the sun. The plains are all decorated by the flowers and the air is filled by its sweet aromatic smell.

In the night, when it sleeps, the night becomes a watchman for it and in the morning it wakes to see the sun without which even our eyes wouldn’t be helpful. It drinks dew like wine, listens to the birds singing and the dances when the grasses sway in the wand. When lovers meet, flower is exchanged and so is it when they get married. Also, it is the last gift that we give to our beloved ones when they pass away. But the flower wants us to be optimistic and look high and dream. It doesn’t want us to look back on the sorrows of our lives and regret. It teaches us the lesson of self-worth and wants us to spread joy just as it does.

### Song of the Flower Summary in Malayalam

Song of the Flower Glossary

## Kerala Syllabus 8th Standard Maths Solutions Chapter 8 Area of Quadrilaterals

Textbook Page No 149

Area Of Quadrilateral Class 8 Scert Chapter 8 Question 1.
Draw a parallelogram of sides 5 centimetres, 6 centimetres and area 25 square centimetres.
Solution:

• Area of parallelogram = one side × distance to the opposite side.
• Since 5x distance to the opposite side is 25 sq.cm. Distance to the opposite side is 5 cm.
• Draw a square ABCD of side 5cm.
• Draw an arc of radius 6cm with centre A to cut DC at E.
• Length of radius 6cm with centre B and arc of radius 5 cm with centre E meet at F.

Class 8 Mathematics Area Of Quadrilaterals Kerala Scert Solutions Question 2.
Draw a parallelogram of area 25 square centimetres and perimeter 24 centimetres.
Solution:
Area of the parallelogram to be 25 cm2. The product of one side and distance to the opposite side to be 25 cm2.
One side and distance to the opposite side can be 5 cm each.
The other side is = 12 – 5 = 7 cm.
Draw a square with side cm and draw an arc of radius 7 cm with A as centre such that it intersects the side CD. Mark the point E and join AE. Draw an arc of radius 7 cm with B as centre and draw another arc of radius 5 cm with E as centre, Two arcs meet at the point
P. JoinBF and CF.

Area Of Quadrilaterals Class 8 Kerala Syllabus Question 3.
In the figure, the two bottom corners of a parallelogram are joined to a point on the top side.

Solution:
Area of dark triangle = $$\frac{1}{2} b h$$ cm2
Area of parallelogram = bh
= 2 × area of triangle 10 cm2

Class 8 Mathematics Construction Of Quadrilaterals Kerala Scert Solutions Question 4.
The picture below shows the parallelogram formed by the intersection of two pairs of parallel lines?

What is the area of this parallelogram? And the perimeter?
Solution:
One side of the parallelogram = 4 cm
Distance to the opposite side = 3 cm
Area of the parallelogram = 4 × 3 = 12 sq.cm
BC × DE = 12
BC × 2 = 12
BC = 6, AD = 6

Perimeter = 4 + 6 + 4 + 6 = 20 cm.

Area Of Quadrilateral Questions For Class 8 Kerala Syllabus Question 5.
Compute the area of the parallelogram below:

Solution:

Textbook Page No 153

Hss Live Guru 8 Maths Kerala Syllabus Chapter 8 Question 6.
Draw a square of area 4½ square centimetres.
Solution:

Draw a circle of radius 3 cm. Draw the diameter AC and construct the perpendicular bisector of AC which meets the circle at B and D. Join AC, AD, CD and BD to get the square ABCD.

Area Of Quadrilateral Formula Class 8 Kerala Syllabus  Question 7.
Draw a non-square rhombus of area 9 square centimetres.
Solution:

• Draw AC = 1 cm
• Find midpoint of AC by drawing per perpendicular bisector
• Find B and D, such that OB = OD = 9 cm

Similarly we can draw the other two rhombus
AC = 9 cm,
OB = OD = 1 cm
AC = 3 cm, OD = OB = 3 cm

8th Standard Maths Notes Kerala Syllabus Chapter 8 Question 8.
The area of the dark triangle in the figure is 5 square centimetres. What is the area of the parallelogram?
Solution:

iii. Perimeter = 4 × 15 = 60 m
iv. Rhombus is a parallelogram. So,
Area of parallelogram = one side × distance to the opposite side
15 × h = 216
h = $$\frac{216}{15}=14.4 \mathrm{cm}$$

8th Standard Maths Notes PdfKerala Syllabus Chapter 8 Question 9.
A 68 centimetre long rope is used to make a rhombus on the ground. The distance between a pair of opposite corners is 16 metres.
i. What is the distance between the other two corners?
ii. What is the area of the ground bounded by the rope?
Solution:

Length of one side of a parallelogram

Diagonals are perpendicular bisector

Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 8 Question 10.
In the figure, the midpoints of the diagonals of a rhombus are joined to form a small quadrilateral:

i. Prove that this quadrilateral is a rhombus.
ii.The area of the small rhombus is 3 square centimetres. What is the area of the large rhombus?
Solution:

The diagonals of the rhombus intersect at O and they bisect each other at right angles.
OA = OC
$$\frac{\mathrm{OA}}{2}=\frac{\mathrm{OC}}{2}$$
OQ = OS
Similarly, since OB = OD
ie., OR = OP
The diagonals of PQRS bisect each other. (Since OQ = OS and OR = OP)
Since the diagonal AC is perpendicular to BD, the diagonals of PQRS are also mutually perpendicular bisectors.
therefore PQRS is a rhombus

ii. Area of quadrilateral PQRS = 3 sq. cm.

8th Area Of Quadrilateral Formula Class 8 Kerala Syllabus Question 11.
What is the area of the largest rhombus that can be drawn inside a rectangle of sides 6 centimetres and 4 centimetres?
Solution:
If we join the midpoints of the opposite sides of rectangle, we get the longest diagonal of the rhombus.

Textbook Page No 156

Hsslive 8th Maths Kerala Syllabus Chapter 8 Question 12.
Draw a rectangle of sides 7 centimetres and 4 centimetres. Draw isosceles trapeziums of the same area, with the following specifications.
i. Lengths of parallel sides 9 centimetres, 5 centimetres.
ii. Lengths of non-parallel sides 5 centimetres.
Solution:
i. Draw a rectangle of length 7 cm and breadth 4 cm.

Extent AB either side by 1 cm and mark E and F. Also mark the points G and H on CD at a distance 1 cm from C and D respectively. Draw the rhombus EFGH.

ii. Draw the rectangle ABCD.The arc of radius 5cm drawn with A as centre intersects DC at P. The arc of 5cm drawn with C as the centre intersect extented AB at Q.

Std 8 Maths Kerala Syllabus Kerala Syllabus Chapter 8 Question 13.
Calculate the area of the isosceles trapezium drawn below:

Solution:

8th Std Maths Notes Kerala Syllabus Chapter 8 Question 14.
The parallel sides of an isosceles trapezium are 8 centimetres and 4 centimetres long; and non – parallel sides are 5 centimetre long. What is its area?
Solution:
Divide the isosceles trapezium into a rectangle and two triangles.

Textbook Page No 158

Area Of Quadrilateral Class 8 Kerala Syllabus Chapter 8 Question 15.
The lengths of the parallel sides of a trapezium are 30 centimetres, 10 centimetres and the distance between them is 20 centimetres. What is its area?
Solution:

Question 16.
Compute the area of the trapezium shown below:

Solution:
Consider the ∆ PQS, it is a right angled triangle

Textbook Page No 159

Question 17.
Compute the area of the hexagon below.

Solution:

Total area = 108 + 108 + 168 = 384 cm2

Question 18.
This is a picture drawn in the lesson Construction of Quadrilaterals.

What is the area of the large trapezium made up of the four smaller ones?
Solution:

Question 19.
What is the area of the quadrilateral shown below?

Solution:

Question 20.
Prove that for any quadrilateral with diagonals perpendicular, the area is half the product of the diagonals.
Solution:

Question 21.
Compute the area of the quadrilateral shown below:

Solution:

Question 22.
Compute the area of the parallelogram shown below:

Solution:
The diagonal divides the parallelogram into two equal triangles.

Area of the parallelogram = 2 × area of triangle = 2 × 96 = 192 cm2

Question 23.
The three blue lines in the picture below are parallel:

Prove that the areas of the quadrilaterals ABCD and PQRS are in the ratio of the lengths of the diagonals AC and PR.
i. How should the diagonals be related for the quadrilaterals to have equal area?
ii. Draw two quadrilaterals, neither parallelograms nor trapeziums, of area 15 square centimetres
Solution:

i. For the quadrilaterals to have equal area, the diagonals AC and PR should be equal.
ii. Draw three parallel lines, distance between them is 5 cm each. Also draw two quadrilaterals with diagonals PQ and RS 6 cm each.

Question 1.
Which of the following has greater area; a parallelogram with one side 12 cm and distance between parallel sides is 6 cm or a square having diagonals 12 cm each?
Solution:

Question 2.
Area of a trapezium is 128 sq. cm and the distance between its parallel sides is 8 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Solution:

Length of the other Parallel side = 4 cm

Question 3.
Find the area of a rhombus of one side 6 cm and one diagonal 6 cm.
Solution:

Question 4.
It was decided to lay tiles in the room of Hari’s house with tiles in the shape of trapezium. The parallel sides of a tile is 40 cm and 20 cm and the distance between them is 10 cm. If 200 tiles are needed to lay tiles in the hall, find the area of the hall in square metres.
Solution:

Number of tiles required = 200
Area of the hall = 200 × 300
= 60000 cm2
= 6 m2

Question 5.
In the pictures given below, which is has more area?

Solution:
Rectangle has the maximum area among the parallelograms with same sides.

Question 6.
The ratio of the two adjacent sides of a parallelogram is 3 : 2. Distance between longer sides is 10 cm. If area is 900 cm2, find the sides of the parallelogram?
Solution:
Let the sides be 3x and 2x

Question 7.
If one side of a parallelogram is ‘a’ and height of that side is h, prove that area = ah.
Solution:
In the figure, ABCD is a parallelogram. AB = a

The perpendicular distance from D to AB = h
By drawing the diagonal BD, we can divide the parallelogram into two equal triangles. ∆ ABD and ∆ BCD are equal triangles. ∴ their areas are equal, ie; area of the parallelogram is two times of the area of ∆ ABD.

Question 8.
PQRS is a rhombus. If the diagonals are 8 cm and 9 cm each, compute the area.

Solution:
Let the diagonals of the rhombus are d1 and d2

Question 9.
The perimeter of a rhombus is 40 cm. If the length of one diagonal is 16 cm. What is the length of the other diagonal. Find the area?
Solution:
Perimeter = 40 cm
One side = 10 cm .
∆ POQ is a right angled triangle Since d1 = 16 cm, OP = 8 cm
∴ 82 + OQ2 = 102

Question 10.
Draw a square of area 12.5 cm2 and write the geometric principles of the construction.
Solution:

Draw a line of length 5 cm and its perpendicular bisector.
Draw a circle with the midpoint of this line as centre and the distance to one end as radius. This line will be the diameter of the circle. Mark the point of intersection of the perpendicular bisector as the circle and complete the square.

Question 11.
Compute the area of the trapezium shown below.

Solution:

Question 12.
Compute the area of the quadrilateral ABCD

Solution:

Question 13.
In the figure AB || CD. AD = BC = 13 cm
Distance between the parallel side is 12 cm. If CD = 20 cm. Find the area of ABCD.

Solution:
To compute AB

Question 14.
Draw a square and mark the mid-points. How is the area of the first square with the second square.
Solution:
Draw the square ABCD.
The perpendicular bisector of the side AB bisects CD also at right angles.

Similarly draw the perpendicular bisectors of AD and BC. Which intersect the sides. Join the points of intersection to complete the square.

We see, 8 triangles in the figure. The square PQRS is formed by joining 4 among these. The area of the square PQRS is half of the area of the square ABCD.

Question 15.
What is the maximum area of parallelogram of sides 8 cm and 5 cm? What is the speciality of the parallelogram of maximum area?
Solution:
The area will be maximum for a rect¬angle and the maximum area is 40 cm2.

Question 16.
See the quadrilateral Ravi drew. In it AC = 12 cm and perpendiculars to AC are 6 cm and 12 cm.

a. Find the area of the quadrilateral.
b. Raju has to draw a parallelogram of this area. If its base is 12 cm what should be its height?
Solution:

b. If the height of the parallelogram is h its area is bh = 12 × h
This is 54 sq. cm

Question 17.
The area of a rhombus is 112 sq. cm and one of its diagonal is 16 cm long. Find the length of the other diagonal.
Solution:
Area of the rhombus $$=\frac{1}{2} \mathrm{d}_{1} \times \mathrm{d}_{2}$$
= 112 sq. cm

Question 18.
Draw a parallelogram of sides 6 cm, 4 cm and area 18 cm2.
Solution:
Draw a line AB of length 6 cm. Draw a circle with centre A and radius 4 cm

In the circle (as shown in the picture) mark a point D and join AD. Draw an arc of radius 4 cm with centre B as centre and radius 4 cm and draw another arc with D as centre and radius 6 cm. Two arcs meet at C. Join CD and BC.

ABCD will be a parallelogram.

Question 19.
In the figure ABCD, AB parallel to CD and the distance between them is 8 cm. ?

AB = 12 cm, CD = 10 cm. Compute the area of quadrilateral (trapezium) ABCD?
Solution:

Question 20.
Compute the area of the quadrilateral ABCD in the figure.

Solution: