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You can Download Similar Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles

Kerala Syllabus 9th Standard Maths Similar Triangles Text Book Questions and Answers

Textbook Page No. 100

Similar Triangles Class 9 Kerala Syllabus Question 1.
The perpendicular from the square corner of a right triangle cuts the opposite side into two parts of 2 and 3 centimetres length.

i. Prove that the two small right triangles cut by the perpendicular have the same angles.
ii. Taking the length of the perpendicular as h, prove that \(\frac{h}{2} = \frac{3}{h}\)
iii. Calculate the perpendicular sides of the large triangle.
iv. Prove that if the perpendicular from the square corner of a right triangle divides the opposite side into parts of lengths a and b and if the length of the perpendicular is h, then h² = ab.
Answer:
i. ∠ADB = 90°
Let ∠BAD = x°
∠DAC = ( 90 – x)°

∴ ∠B = 180 – (x + 90) = 90 – x
and ∠C = 180 – (90 – x + 90) = x
Angles of ΔABC = x°, (90 – x)°, 90°
Angles of ΔACD = x°, (90 – x)°, 90°
The two small triangles have the same angles.

ii. Triangle with the same angles, sides opposite equal angles are scaled by the factor.
2 : h = h : 3 ⇒ \(\frac{2}{h} = \frac{h}{3}\)
i.e \(\frac{h}{2} = \frac{3}{h}\)

iii. \(\frac{h}{2} = \frac{3}{h}\)
h² = 6 ⇒, h = \(\sqrt 6\)
In ΔABD
AB² = BD² + AD² = 2² + (\(\sqrt 6\))²
= 4 + 6 = 10
AB = \(\sqrt 10\)
In ΔADC
AC² = DC² + AD² = 3² + (\(\sqrt 6\))²
= 9 + 6 = 15
AC = \(\sqrt 15\)
The perpendicular sides are \(\sqrt 10\) and \(\sqrt 15\)

iv. Triangle with the same angles, sides opposite equal angles are scaled by the same factor.
BD : AD = AD : DC

\(\frac{h}{a} = \frac{b}{h}\)
h² = ab

Similar Triangles Class 9 Question 2.
At two ends of a horizontal line, angles of equal size are drawn and two points on the slanted lines are

i. Prove that the parts of the horizontal line and parts of the slanted line are in the same ratio.
ii. Prove that the two slanted lines at the ends of the horizontal line are also in the same ratio.
iii. Explain how a line of length 6 cm can be divided in the ratio 3 : 4
Answer:

i. ∠A = ∠B
∠AMC = ∠BMD
∴ ∠C = ∠D
(The triangles AMC and DMB are similar).
∴ \(\frac{MC}{MD} = \frac{AC}{BD} = \frac{AM}{MB}\)
\(\frac{AM}{MB} = \frac{MC}{MD}\)

ii. \(\frac{MC}{MD} = \frac{AC}{BD} = \frac{AM}{MB}\)
(From the similar triangles AMC and DMB)

iii. Draw a line segment of length 6 cm. At one end of this line AB, 3cm long and at the other end draw CD, 4cm long in the opposite direction.

∠CAB = ∠ACD
Draw BD to cut A at O.
Since AB : CD = OA : OC
∴ OA : OC = 3 : 4

Scert Class 9 Maths Similar Triangles  Question 3.
The mid point of the bottom side of a square is joined to the ends of the top side and extended by the same length. The ends of these lines are joined and perpendiculars are drawn from these points to the bottom side of the square extended.

i. Prove that the quadrilateral obtained thus is also a square.
ii. Explain how we can draw a square with two corners on a semicircle and the other two corners on its diameter as in the figure

Answer:

i. ΔDCM and ΔSRM are similar.
The angles of the ΔMBC and ΔMAD are equal.
ΔPMS, ΔQMR are also similar triangles.
∠D = ∠S; ∠C = ∠R; ∠A = ∠B; ∠B = ∠Q
Let the ratio of the equal sides are be k.
\(\frac{QR}{BC}\) = k ⇒ \(\frac{QR}{2a}\) = k
∴ QR = 2ak
Similarly, PS = PQ = SR = 2ka
∴ PQRS is a square.

ii.

Draw a line of AB, 8cm long. Find midpoint O on the line. Mark P and Q such that OP = OQ = 1cm. Draw the square PQRS whose sides are 2cm long.

Draw a semicircle with O as centre and OA as radius. Extend OS and OR to meet the semicircle at G and F. From G and F draw GD and FE perpendicular AB. Draw GF. Quadrilateral DEFG is the required square.

Kerala Syllabus 9th Standard Maths Chapter 7 Question 4.
The picture shows a square drawn sharing one corner with a right triangle and the other three corners on the sides of this triangle.

i. Calculate the length of a side of the square.
ii. What is the length of a side of the square drawn like this within a triangle of sides 3, 4 and 5 centmetres?
Answer:
i.

ΔABC and ΔAPQ are similar.
\(\frac{x + 2}{2} = \frac{x + 1}{x}\) = x(x + 2) = 2(x +1)
x² + 2x = 2x + 2
x² = 2; x = \(\sqrt 2\)
Side of square = \(\sqrt 2\)cm.

ii. AB = 4, BC = 3, AC = 5
ΔAPQ and ΔABC are similar triangles
∴ \(\frac{AP}{PB} = \frac{PQ}{BC} ⇒ \frac{4 – x}{4} = \frac{x}{3}\)
3(4 – x) = 4x
12 – 3 = 4x
12 – 7x; x = \(\frac {12}{7}\)

The length of the side of the square = \(1\frac {5}{7}\)

Chapter 7 Similar Triangles Kerala Syllabus Question 5.
Two poles of heights 3 m and 2 m are erected upright on the ground and ropes are stretched from the top of each to the foot of the other.

i. At what height above the ground do the ropes cross each other?
ii. Taking the heights of the poles as a and b and height above the ground of the point where the ropes cross each other as h, And the relation between a, b and h.
iii. Prove that this height would be the same, whatever be the distance between the poles.
Answer:
i. Consider ΔABC and ΔAFE. They are similar.
\(\frac{b}{h} = \frac{x + y}{x}\) ………..(1)

Consider ΔADB and ΔFEB. They are also similar
\( \frac{a}{h} = \frac{x + y}{y}\) ………..(1)
\( \frac{h}{a} = \frac{y}{x + y}, \frac{h}{b} = \frac{x}{x + y}\)
(1) + (2) ⇒ \( \frac{h}{a} + \frac{h}{b} = \frac{y}{x + y} + \frac{x}{x + y}\)
h\((\frac{1}{a} + \frac{1}{b}) = \frac{x + y}{x + y}\)
h\((\frac{1}{a} + \frac{1}{b})\) = 1
\(\frac{1}{a} + \frac{1}{b} = \frac{1}{h}\) ……..(2)
a = 3 and b = 2
\(\frac{1}{h} = \frac{1}{3} + \frac{1}{2} + \frac{5}{6}\)cm
h = \(\frac{6}{5}\) = 1.2

ii. The heights of the poles are a and b and the height above the ground of the point where the ropes cross each other is h, then the relation between a, b and h is \(\frac{h}{a} + \frac{h}{b}\) = 1 or \(\frac{a + b}{ab} = \frac{1}{h}\) from equation (3).

iii. Only change its height according to the height of the poles not the distance.

Textbook Page No. 107

9th Class Maths Notes Kerala Syllabus Question 1.
Draw a triangle of angles the same as those of the triangle shown and sides scaled by \(1\frac{1}{4}\)

Answer:
One side is 6 cm. Its \(1\frac{1}{4}\) part is 7.5 cm
Other sides are 4 × \(1\frac{1}{4}\) = 5 cm
8 × \(1\frac{1}{4}\) = 10 cm
Draw one side in 10 cm. Complete the triangle with given measures.

Kerala Syllabus Class 9 Maths Solutions Question 2.
See this picture of a quadrilateral.

i. Draw a quadrilateral with angles the same as those of this one and sides scaled by \(1\frac{1}{2}\)
ii. Draw a quadrilateral with angles different from those of this and sides scaled by \(1\frac{1}{2}\).
Answer:
i. Draw a quadrilateral by increasing all the sides including the diagonal by \(1\frac{1}{2}\)
4 × \(1\frac{1}{2}\) = 4 × 1.5 = 6
2 × \(1\frac{1}{2}\) = 2 × 1.5 = 3
3 × \(1\frac{1}{2}\) = 3 × 1.5 = 4.5
5 × \(1\frac{1}{2}\) = 5 × 1.5 = 7.5
6 × \(1\frac{1}{2}\) = 6 × 1.5 = 9

ii. Draw a quadrilateral by increasing all the sides increased by \(1\frac{1}{2}\) except the diagonal.

Textbook Page No. 111

Kerala Syllabus 9th Standard Maths Notes Question 1.
The picture shows two circles with the same centre and two triangles formed by joining the centre to the points of intersection of the circles with two radii of the larger circle. Prove that these triangles are similar

Answer:

AP = AQ (radius)
∠APQ = ∠AQP
(Base angles of an isosceles triangle)
∠A = ∠A (Common angle)
AC = AB (radius)
∴ ∠ACB = ∠ABC
∴ \(\frac{AP}{AC} = \frac{AQ}{AB}\)
That is two sides of these triangles are scaled by the same factor.
Since two sides of these triangles are scaled by the same factor and the angle between them the same ΔAPQ and ΔACB are similar. So the triangles are similar

Kerala Syllabus 9th Standard Maths Guide In Malayalam Question 2.
The lines joining the circumcentre of a triangle to the vertices are extended to meet another circle with the same centre and these points are joined to make another triangle.

i. Prove that the two triangles are similar.
ii. Prove that the scale factor of the sides of the triangle is the scale factor of the radii of the circles.
Answer:
∴ ΔAOB and ΔPOQ are similar triangles

∴ \(\frac{AB}{PQ} = \frac{OB}{OQ}\) ……..(1)
ΔBOC, ∠QOR are similar
∴ \(\frac{OB}{OQ} = \frac{BC}{QR} = \frac{OC}{OR}\) ……..(2)
Similarly
\(\frac{OC}{OR} = \frac{AC}{PR}\) ……..(3)
(1), (2), (3) consider
∴ \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}\)
ΔABC and ΔPQR are similar triangles.

ii. From equation (1), (2) and (3)
The scale factor of the sides of the triangle is the scale factor of the radii of the circles.

9th Class Maths Kerala Syllabus Question 3.
A point inside a quadrilateral is joined to its vertices and the lines are extended by the same scale factor. Their ends are joined to make another quadrilateral.

i. Prove that the sides of the two quadrilateral are scaled by the same factor.
ii. Prove that the angles of the two quadrilaterals are the same.
Answer:
i.

ΔAOB, ΔPQO are similar
\(\frac{OA}{OP} = \frac{AB}{PQ} = \frac{OB}{OQ}\) …….(1)
ΔOBC, ΔOQR are similar
\(\frac{OB}{OQ} = \frac{BC}{QR} = \frac{OC}{OR}\) …….(2)
ΔOCD, ΔORS are similar
\(\frac{OC}{OR} = \frac{CD}{RS} = \frac{OD}{OS}\) …….(3)
ΔODA, ΔOSP are similar
\(\frac{OD}{OS} = \frac{AD}{PS} = \frac{OA}{OP}\) …….(4)
Consider (1), (2), (3), (4)
\(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{DC}{SR} = \frac{AD}{PS}\)
Since the sides of quadrilateral EFGH and quadrilateral ABCD are scaled by the same factor.
∴ The angles of the quadrilateral are also same (Angles of similar triangles).

ii. Since the sides of quadrilateral PQRS and quadrilateral ABCD are scaled by the same factor, the angles of the large quadrilateral will be the same as the angles of the small quadrilateral.

Kerala Syllabus 9th Standard Maths Similar Triangles Exam Oriented Text Book Questions and Answers

Class 9 Maths Notes Kerala Syllabus Question 1.

ΔABC, ΔXYZ are similar triangles in the picture. Then
\(\frac{AB}{XY} = \frac{BC}{….} = \frac{….}{XZ}\)
Answer:
\(\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}\)

Hss Live Class 9 Maths Kerala Syllabus Question 2.
In ΔXYZ PQ is parallel to XY.

a. \(\frac{YP}{PZ} = \frac{XQ}{….}\)
b. \(\frac{x}{a} = \frac{…..}{b}\)
Answer:
a. \(\frac{YP}{PZ} = \frac{XQ}{QZ}\)
b. \(\frac{x}{a} = \frac{y}{b}\)

Kerala Syllabus 9th Standard Maths Solutions Question 3.


Fill in the blanks according to the picture.
∠K = ∠A; ∠B = …….
∠C = ……; KL = ……cm
KM = ….. cm
Answer:
∠B = ∠L
∠C = ∠M
KL = 8cm
KM = 10cm

9th Standard Maths Notes Kerala Syllabus Question 4.
In the diagram AP and BQ are perpendicular to AB. AP = 4cm, BQ = 2cm Then show that AC : CB = 2 : 1

Answer:
∠PAC = 90° ∠CBQ = 90°
∠ACP = ∠BCQ (opposite angles)
∴ ΔPAC ≅ ΔBCQ
\(\frac{PA}{BQ} = \frac{AC}{CB} = \frac{PC}{CQ}\)
\(\frac{AC}{CB} = \frac{PA}{BQ} = \frac{AC}{CB} = \frac{2}{1}\)
AC : CB = 2 : 1

Class 9 Maths Chapter 7 Kerala Syllabus Question 5.
In ΔABC and ΔPQR we have \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}\). In triangle ABC, the altitude through P meets BC at D, and in triangle PQR. The altitude through P meets QR at ‘S’. Prove that \(\frac{AB}{PQ} = \frac{AD}{PS}\)

Answer:
ΔABC, ΔPQR are similar.
∠A = ∠P; ∠B = ∠Q;
∠C = ∠R;
Consider ΔADB ΔPSQ
∠B = ∠Q, ∠ADB = ∠PSQ = 90°
(AD ⊥ BC & PS ⊥ QR)
[each equal to 90°]
∴ ΔADB, ΔPSQ are similar.
[Since the corresponding sides of similar triangles are proportional]
∴ \(\frac{AD}{PS} = \frac{DB}{SQ} = \frac{AB}{PQ} ⇒ \frac{AB}{PQ} = \frac{AD}{PS}\)

Chapter 7 Maths Class 9 Kerala Syllabus Question 6.
Prove that if the area of two similar triangles are proportional to the squares on their corresponding sides.

Answer:
ΔABC ~ ΔPQR.
AM, PN are perpendicular, \(\frac{AM}{PN} = \frac{BC}{QR}\)
Area of ΔABC = \(\frac{1}{2}\) BC × AM
Area of ΔPQR = \(\frac{1}{2}\) QR × PN
\(\frac{Area of ΔABC}{Area of ΔPQR}\)

Std 9 Maths Kerala Syllabus Kerala Syllabus Question 7.
In the given figure, PC ⊥ QR and QD ⊥ PR. Prove that ΔPCR and ΔQDR are similar.

Answer:
In ΔPCR, ∠PCR = 90°, and in ΔQDR = 90°
∠R is common to both the triangles
ΔPCR, ΔQDR. ∠PCR = ∠QDR = 90°
∠PRC = ∠QRD (common)
(Two triangles having two pairs of corresponding angles equal. The triangles are similar.)

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 8.
In ΔABC, P is a point on BC. Where D, E, F are the mid-points on BP, AP and CP respectively, then prove that ΔABC ~ ΔDEF.

Answer:
Consider ΔABP, ΔDPE.
ΔABP ~ ΔDPE (∠A is ∠BAP = ∠DEP and ∠ABP = ∠EDP)
∴ \(\frac{AB}{ED} = \frac{AP}{EP}\) …..(1)
similarly ΔAPC ≅ ΔEPF.
∴ \(\frac{AC}{EF} = \frac{AP}{EP}\) …..(2)
(1) = (2) \(\frac{AB}{ED} = \frac{AC}{EF} = \frac{AP}{EP}\)
ΔABC ~ ΔDEF

Kerala Syllabus 9th Standard Maths Guide Pdf Question 9.
In figure AD || BC, AB || DE prove that ΔABC – ΔEDA .

Answer:
∠DAE = ∠C
(AD || BC corresponding angles)
∠BAC = ∠DEA
(AB || DE corresponding angles)
ΔABC ~ ΔEDA

Question 10.
ΔABC is a right angled triangle. ∠B = 90° a perpendicular line from B to AC intersect AC at D. Prove that ΔABC, ΔADB and ΔBCD are similar to each other.
Answer:

ΔABC = ΔABD
[∠ABC = ∠ADB = 90°, ∠ACB = ∠ABD = 90 – X]
∠A (common)
similarly ΔABD ~ ΔBDC
In ΔABC, consider
∠A = X°
∠B = 90°
∠A + C = 90°, ∠C = 90 – X
In ΔABD, ∠ADB = 90°
Therefore ∠A = X°, ∠ABC = 90°
∠A + ∠ABD = 90°
∠ABD = 90 – ∠A = 90 – X;
In ΔBCD, ∠BDC = 90°
∠C = (90 – X); ∠C + ∠DBC = 90°;
∠DBC = 90 – (90 – X) = X°
Hence angles of three triangles are 90°, X° and 90° – X°.
Hence three triangles are similar to each other.

You can Download Parallel Lines Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines

Kerala Syllabus 9th Standard Maths Parallel Lines Text Book Questions and Answers

Textbook Page No. 88

Parallel Lines Class 9 Kerala Syllabus Question 1.
Draw an 8 cm long line and divide it in the ratio 2 : 3.
Answer:
Draw AB whose length is 8 cm. Draw AC with length 5 cm. Join BC. Let the point D divides AC in the ratio 2 : 3. Draw a line through D parallel to BC. This line divides AB in the ratio 2 : 3.

Kerala Syllabus 9th Standard Maths Chapter 6 Question 2.
Draw a rectangle of perimeter 15 cm and sides in the ratio 3 : 4
Answer:
Perimeter is 15 cm.
So length + breadth = 7.5 cm
Draw AB, a line segment of length 7.5 cm.
Draw AC from A, 4 + 3 = 7cm long.
Mark D on this line such that AD = 4 cm. Draw BC.
Draw a line parallel to BC and passing through D, which meets AB at E. Draw rectangle AEGF whose length as AE and breadth as EB.

Class 9 Maths Parallel Lines Kerala Syllabus Question 3.
Draw rectangles specified below each of perimeter 10 cm.
i. Equilateral triangle.
ii. Sides in the ratio 3 : 4 : 5
iii. Sides in the ratio 2 : 3 : 4
Answer:
i. Draw a line AB of length 10 cm.
Draw AC with length 9 cm. Join BC.
AD = DE = EC = 3
AF : FG : GB = 3 : 3 : 3 = 1 : 1 : 1
Divide it in the ratio 1 : 1 : 1 and draw a triangle.

Draw ΔFHG with FG as its side.

ii. Draw a line segment AB of length 10 cm. Draw AC = 12 cm, join BC. Divide it in the ratio 3 : 4 : 5.
AD : DE : EC = 3 : 4 : 5
AF : FG : GB = 3 : 4 : 5
Draw ΔGHB with sides AF, FG and GB.

iii. Draw a line segment AB of length 10 cm. Draw AC with length 9 cm, join BC. Divide it in the ratio 2 : 3 : 4.

AD : DE : EC = 2 : 3 : 4
AF : FG : GB = 2 : 3 : 4
Draw ΔGHB with sides AF, FG and GB.

Parallel Lines 9th Standard Kerala Syllabus Question 4.
In the picture below, the diagonals of the trapezium ABCD intersect at P. Prove that PA × PD = PB × PC.

Answer:

Sides AB, CD are parallel.
Draw EF parallel to AB and passing through P.
Now lines AB, EF and DC are parallel.
These lines cut the lines AC and BD in the same ratio.
So, PA : PC = PB : PD
i.e., \(\frac {PA}{PC} = \frac {PB}{PD}\)
From this by cross multiplication we get PA × PD = PC × PB

Textbook Page No. 93

Parallel Lines Class 9 Kerala Syllabus Question 1.
In the picture, the perpendicular is drawn from the midpoint of the hypotenuse of a right triangle to the base. Calculate the length of the third side of the large right triangle and the lengths of all three sides of small right triangle.

Answer:
AC = 10 cm
∴ AM = 5 cm

MN, CB are perpendicular to AB
MN, CB are parallel lines.
∴ MN divides AB and AC in the same ratio.
∴ AN = BN = 4cm
BC = \(\sqrt{10^{2} – 8^{2}} = \sqrt {100 – 64} = \sqrt {36}\) = 6 cm
M is the midpoint of the AC. N is the midpoint of AB.
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side. So MN is half of BC.
∴ MN = 3cm

Parallel Lines Chapter Class 9 Kerala Syllabus Question 2.
Draw a right triangle and the perpendicular from the midpoint of the hypotenuse to the base.

i. Prove that this perpendicular is half the perpendicular side of the large triangle.
ii. Prove that perpendicular bisects the bottom side of the larger triangle.
iii. Prove that in the large triangle the distances from the mid point of the hypotenuse to all the vertices are equal.
iv. Prove that the circumcentre of a right triangle is the mid point of its hypotenuse.
Answer:
MN is perpendicular to AB.
MN and CB are parallel lines. MN divides AC and AB in the same ratio.

i. AM = \(\frac{1}{2}\)AC, AN = \(\frac{1}{2}\)AB
∴ MN = \(\frac{1}{2}\)CB

ii. We get two right triangles of same base and perpendicular. So their hypotenuses are also equal.
∴ MA = MC = MB
AN = NB

iii. ∠ANM = ∠BNM = 90°
MN = MN
ΔANM ≅ ΔBNM
AM = MB …….(1)
M is the midpoint of AC
AM = MC …….(2)
(1) = (2) ⇒ AM = MB = MC

iv. The points A, B and C are at same distance from M. So a circle can be drawn with M as centre and pass through there three points. So this circle is the circum circle of ΔABC.

9th Standard Maths Notes Kerala Syllabus Question 3.
In the parallelogram ABCD, the line drawn through a point P on AB, parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R.

Prove that \(\frac {AP}{PB} = \frac {AR}{RD}\)
Answer:
In ΔABC, RQ the parallel line of BC, divides AB and BC in the same ratio When we consider the relation
\( \frac {AR}{RD} = \frac {AQ}{QC}, \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}\)

\( \frac {AR}{RD} = \frac {AQ}{QC}\)
\( \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}\)
∴ \( \frac {AP}{PB} = \frac {AR}{RD}\)

Hsslive Guru 9th Maths Kerala Syllabus Question 4.
In the picture below, two vertices of a parallelogram are joined to the mid points of two sides. Prove that these lines divide the diagonal in the picture into three equal parts.

Answer:

In the parallelogram ABCD, AB = CD and AB parallel CD.
P is the mid point of AB and Q is the midpoint of DC.
Since P and Q are the midpoints of AB and CD respectively.
PB = ½AB
DQ = ½CD = ½AB
∴ PB = DQ
i.e., PB = DQ and PB parallel DQ.
∴ PBQD is a parallelogram.
Since the parallel lines PD and BQ cut AB into equal parts, they cut AY also into equal parts.
AX = XY
The parallel lines PD and BQ cut CD into equal parts. So they cut CX also into equal parts.
YC = XY
∴ AX = XY = YC

9th Standard Maths Kerala Syllabus Question 5.
Prove that the quadrilateral formed by joining the mid points of a quadrilateral is a parallelogram. What if the original quadrilateral is a rectangle? What if it is a rhombus?
Answer:

P, Q, R and S are the midpoints of the sides of the quadrilateral ABCD. Draw the diagonals AC and BD.
The length of the line joining the mid-points of two sides of a trianglle is half the length of the third side.
In ΔABC, PQ = ½AC
In ΔADC, SR = ½AC
∴ PQ = SR
In ΔABD, PS = ½BD
In ΔBCD, QR = ½BD
∴ PS = QR
In quadrilateral PQRS, both pairs of opposite sides are equal. So it is a parallelogram.
PQ = ½AC, RS = ½AC
∴ RS = PQ = ½AC
PS = ½BD, QR = ½BD
∴ PS = QR = ½BD = ½AC
∴ PQ = RS = PS = QR
All sides of the quadrilateral PQRS are equal. So it is a rhombus.
Since ABCD is a rhombus. AC perpendicular to BD.
PQ and SP are perpendicular, i.e., ∠SPQ = 90°
One angle of the parallelogram PQRS is 90° and therefore it is a rectangle.

Kerala Syllabus 9th Standard Maths Parallel Lines Exam Oriented Text Book Questions and Answers

Class 9 Maths Chapter 6 Kerala Syllabus Question 1.
Draw ΔABC with AB = 5cm, BC = 4cm, AC = 6cm, ΔABC. In BC mark P such that BP = 1cm, PC = 3cm. What is the relation between the areas of ΔABP and ΔAPC?
Answer:
Area of ΔAPC is 3 times the area of ΔABP.

If two triangles have the common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.
⇒ \(\frac{Area of ΔABP}{Area of ΔAPC} = \frac{BP}{PC} = \frac{1}{3}\)

Chapter 6 Maths Class 9 Kerala Syllabus Question 2.
In ΔABC, the line parallel to BC meet AB and AC at D and E respectively. AE = 4.5cm, \(\frac{AD}{DB} = \frac{2}{5}\) = then find EC.
Answer:

\(\frac{AD}{DB} = \frac{AE}{EC}\)
\(\frac{2}{5} = \frac{4.5}{EC}\) = EC = \(\frac{4.5 \times 5}{2} = \frac{22.5}{2}\)
= 11.25 cm

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
In ΔPQR, PQ = 9cm, QR = 14 cm, PR = 12 cm The bisector of ∠PRS meets QR in S. Find QS and RS.
Answer:

\(\frac{PQ}{PR} = \frac{QS}{SR}\);
\(\frac{18}{24} = \frac{6}{RS}\)
= RS = \(\frac{24 \times 6}{18}\) = 8 cm

Hsslive Guru 9 Maths Kerala Syllabus Question 4.
In ΔABC, the point P is on AB such that the length of AP is double the length of PB. The line through P parallel to BC meets AC at Q and the length of AQ is 1 cm more than that of QC, what is the length of AC?
Answer:

AP = 2 × PB
AP : PB = 2 : 1
\(\frac{AP}{PB} = \frac{AQ}{QC} = \frac{2}{1}\) ⇒
AQ = 2QC
Consider QC = x cm
x + 1 = 2x ⇒ x = 1 ⇒ QC = 1
AQ = 2 × 1 = 2
AC = AQ + QC = 2 + 1 = 3 cm

Hss Live Guru Maths Kerala Syllabus Question 5.
In ΔABC, a line parallel to BC cuts a B and AC at X and Y

(a) If AB = 3.6cm, AC = 2.4cm, AX = 2.1 cm, what is the length of AY?
(b) If AB = 2cm, AC = 1.5cm, AY = 0.9cm
what is the length of BX?
Answer:
a. \(\frac{AB}{AX} = \frac{AC}{AY} ⇒ \frac{3.6}{2.1} = \frac{2.4}{AY}\);
⇒ 3.6 × AY = 2.4 × 2.1 ⇒ AY = 1.4 cm

b. Let length of AX = xcm
\(\frac{AX}{AB} = \frac{AY}{AC} ⇒ \frac{x}{2} = \frac{0.9}{1.5}\);
1.5 × x = 1.8 ⇒ x = 1.2;
BX = AB – AX; BX = 2 – 1.2 = 0.8 cm

Hss Live Guru Maths 9 Kerala Syllabus Question 6.
In the figure, ABCD is a trapezium, and AB || CD. P is the midpoint of AD. If PQ || AB, then prove that Q is the midpoint of BC.
Answer:

Given PQ || AB,
therefore
PQ || DC also. Line segments AD and BC are dividing 3 parallel lines EC, PQ and AB. Given P is the midpoint of AD
⇒ DP = PA ∴ \( \frac{DP}{PA}\) = 1, ∴ \( \frac{CQ}{QB}\) = 1
∴ CQ = QB ∴ Q is the midpoint of BC.

Ch 6 Maths Class 9 Kerala Syllabus Question 7.
In the figure AB || EH || DC, Answer the following questions given below.

a. If DE = 3cm, EA = 4cm and DG = 5cm, find GB?
b. If BH = 6cm,CB = 16cm and DB = 20cm, find DG?
c. If DA = 36cm, EA = 22cm and DG = 15cm find DB?
d. If DE = 20cm, EA = 25cm and HB = 22cm, find CH?
Answer:
a. In ΔDAB, AB || EG
∴ \(\frac{DE}{EA} = \frac{DG}{GB}; \frac{3}{4} = \frac{5}{GB} \Rightarrow GB = \frac{20}{3} = 6 \frac{2}{3}\)cm

b. In ΔBDC, DC || GH
∴ \(\frac{BH}{BC} = \frac{BG}{BD}; \frac{6}{16} = \frac{BG}{20}\);
BG = \(\frac{120}{16} = 7\frac{1}{2}\)cm
DG = BD – BG = 20 – \( 7\frac{1}{2}\) = \( 12\frac{1}{2}\)cm

c. In ΔABD, AB || EG
\(\frac{DA}{EA} = \frac{DB}{BG}; \frac{36}{22} = \frac{DB}{BG}\);
\(\frac{36}{22} \frac{DB}{DB – DG}\)
⇒ 36D(DB – DG) = DB × 22
14 DB = 540 DB = \(\frac{540}{14}\) = 38\(\frac{4}{7}\)cm

d. AB, EH and DC are 3 parallel lines.
\(\frac{DE}{EA} = \frac{CH}{HB}; \frac{20}{25} = \frac{CH}{22}\) ⇒ CH = \(\frac{88}{5}\) = 17\(\frac{3}{5}\)cm

Hss Live Guru Class 9 Maths Kerala Syllabus Question 8.
In ABC, a line parallel to BC cuts AB and AC at P and Q. Show that \(\frac{AP}{AB} = \frac{AQ}{AC}\)
Answer:

\(\frac{AB}{PB} = \frac{AQ}{QC}\) (BC || PQ)
\(\frac{AP}{PB}+1=\frac{AQ}{QC}+1; \frac{AP+PB}{PB} = \frac{AQ + QC}{QC}\)
\(\frac{AB}{PB} = \frac{AQ}{QC}\);
∴ \(\frac{AP}{AB} = \frac{AQ}{AC}\)

Question 9.
In the figure below, ABC is a right angled triangle. AB = 10cm, AC = 6cm, BC = 8cm, The midpoint of AB is M. Compute the lengths of the sides of ΔMBN
Answer:

(MN || AC)
∴ \(\frac{AM}{MB} = \frac{NC}{NB}\)
1 = \(\frac{NC}{NB}\) ∴ CN = NB = 4
∴ MB = 5, NB = 4, MN = 3

Question 10.
In the figure ABC is a right angled triangle and M is the midpoint of AB. Prove that MN = V2AC and also prove that MN = MA = MB.
Answer:

MN = \(\sqrt{x^{2} – y^{2}}\)
AC = \(\sqrt{4x^{2} – 4y^{2}} = 2\sqrt {x^{2} – y^{2}}\)
AC = 2 × MN or MN = \(\frac{1}{2}\)AC
MC = \(\sqrt{x^{2} – 4^{2} + 4^{2}}\); MC = X
∴ MC = MA = MB

You can Download New Numbers Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers

Kerala Syllabus 9th Standard Maths New Numbers Text Book Questions and Answers

Textbook Page No. 49

New Numbers Class 9 Kerala Syllabus Chapter 4 Question 1.
In the picture, the square on the hypotenuse of the top most right triangle is drawn. Calculate the area and the length of a side of the square.

Answer:
Hypotenuse of first right triangle
\(\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { {1} + {1} } = \sqrt {2} \)
Hypotenuse of second right triangle
\(\sqrt {\sqrt {{2}}^{ 2 } + 1^{2}} = \sqrt {2 + 1} = \sqrt {3} \)
Hypotenuse of third right triangle
\(\sqrt {\sqrt { { 3 } }^{ 2 } + 1^{ 2 }} = \sqrt {3 + 1} =\sqrt {4} \) = 2
Hypotenuse of fourth right triangle
\(\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { { 4 } + { 1 } } =\sqrt { 5 } \)
i. e. the length of one side of square is \(\sqrt 5 \)
Area \(\sqrt 5 \) × \(\sqrt 5 \) = 5 sq. m

Class 9 Maths Chapter 4 Kerala Syllabus Question 2.
A square is drawn on the altitude of an equilateral triangle of side 2m

(i) What is the area of the square?
(ii) What is the altitude of the triangle?
iii) What are the lengths of the other two sides of the triangle shown below?

Answer:
(i)
BD = 1 m
AD = \(\sqrt { { 2 }^{ 2 }-{ 1 }^{ 2 } } = \sqrt { { 4 }-{ 1 } } = \sqrt { 3 } \)
Area of square
= \(\sqrt 3 \) × \(\sqrt 3 \) = 3 sq. m

(ii) Height of triangle \(\sqrt 3\)cm

(iii)

Sides are 1 m and \(\sqrt 3\) m
Sides opposite to 30° angle = 1 m
Sides opposite to 60° = \(\sqrt 3\) m

Kerala Syllabus 9th Standard Maths Chapter 4 Question 3.
We have seen in Class 8 that any odd number can be written as the difference of two perfect squares. (The lesson, Identities). Using this, draw squares of areas 7 and 11 square centimetres.
Answer:
3² – 2² = 5
4² – 3² = 7
5² – 4² = 9
n² – (n – 1)² = 2n – 1
so 4² – 3² = 7
6² – 5² = 11
Draw a right angled triangle with hypotenuse 4cm and one of its side as 3 cm. It’s one side is \(\sqrt 7\) cm

Draw square BCDE with BC as the side.

Area of square BCDE = \(\sqrt 7\) × \(\sqrt 7\) = 7 sq.cm
Draw a right angled triangle with hypotenuse 6 cm and one of its side as 5 cm. It’s one side is \(\sqrt 11\) cm

Draw square BCDE with BC as the side.

Area of square QRST = \(\sqrt 11\) × \(\sqrt 11\) = 11 sq.m

Class 9 Maths Chapter 4 New Numbers Kerala Syllabus Question 4.
Explain two different methods of drawing a square of area 13 square centimetres.
Answer:
Method 1
72 – 62 = 13 Draw a right angled triangle with one side 6 cm and hypotenuse 7 cm.

Draw square QRST with QR as the side.

Method 2.

Draw a right angled triangle with 1, 3 as perpendicular sides. Hypotenuse is \(\sqrt 10\)
Draw a right angled triangle with \(\sqrt 10\), 1, as perpendicular sides. Hypotenuse is \(\sqrt 11\)
Draw a right angled triangle with \(\sqrt 11\), 1 as perpendicular sides. Hypotenuse is \(\sqrt 12\)
Draw a right angled triangle with \(\sqrt 12\), 1 as perpendicular sides. Hypotenuse is \(\sqrt 13\). If AB = \(\sqrt 13\) cm.
Draw a square ABCD with AB as side.

9th Standard Maths Chapter 4 Kerala Syllabus Question 5.
Find three fractions larger than \(\sqrt 2\) and less than \(\sqrt 3\)
Answer:
\(\sqrt 2\) = 1.41; \(\sqrt 3\) = 1.73
Numbers in between \(\sqrt 2\) and \(\sqrt 3\) are 1.5, 1.6, 1.65 So fractions are \(\frac {15}{10}\), \(\frac {16}{10}\), \(\frac {165}{100}\)

Textbook Page No. 52

Kerala Syllabus 9th Standard Maths Notes Chapter 4 Question 1.
The hypotenuse of a right triangle is 1\(\frac {1}{2}\) m and another side is \(\frac {1}{2}\) m. Calculate its perimeter correct to a centimetre.
Answer:
Square of the third side is =
\((1\frac {1}{2})^{2} – (\frac {1}{2})^{2} = (1\frac {1}{2} + \frac {1}{2}) (1\frac {1}{2} – \frac {1}{2}) \) = 2 × 1 = 2
∴ Third side is \(\sqrt 2\)
Perimeter = 1\(\frac {1}{2}\) + \(\frac {1}{2}\) + \(\sqrt 2\) = 2 + \(\sqrt 2\)
= 2 + 1.41 = 3.412 m = 341.2 cm

New Numbers Class 9 Questions And Answers Kerala Syllabus Chapter 4 Question 2.
The picture shows an equilateral triangle cut into halves by a line through a vertex.
i. What is the perimeter of a part? (See the second problem at the end of the previous section)
ii. How much less than the perimeter of the whole triangle is this?

Answer:

The sides of the new triangle is 2 cm,
1 m, \(\sqrt {2}^{2}-{1}^{2}\) m
(i) Perimeter of one of the triangle =
2 + 1 +\(\sqrt 3\) = 3 + 1.73 = 4.73 m

(ii) Perimeter of the whole triangle
= 2 + 2 + 2 = 6m
Less in the perimeter
= 6 – 4.73 = 1.27m

9th Class Maths Notes Chapter 4 Kerala Syllabus Chapter 4 Question 3.
Calculate the perimeter of the triangle shown below

Answer:
Draw BD perpendicular to AC

AB = 2;
BD= 1;
AD= \(\sqrt 3\)
BD = 1, DC = 1:
BC = \(\sqrt 2\)
AC \(\sqrt 3\) + 1; AB = 2: BC = \(\sqrt 2\)
Perimeter = \(\sqrt 3\) + 1 + 2 + \(\sqrt 2\)
= 3 + \(\sqrt 3\) + \(\sqrt 2\)
= 3 + 1.73 + 1.41 = 6.14 m

Hsslive Guru 9th Maths Kerala Syllabus Chapter 4 Question 4.
We have seen how we can draw a series of right triangles as in the picture.

(i) What are the lengths of the sides of the tenth triangle drawn like this?
(ii) How much more is the perimeter of the tenth triangle than the perimeter of the ninth triangle?
(iii) How do we write in algebra, the difference in perimeter of the n‘h triangle and that of the triangle just before it?
Answer:
i. Sides of the first triangle = 1 m, 1 m, \(\sqrt 2\) m
Sides of the second triangle = 1 m, \(\sqrt 2\) m, \(\sqrt 3\) m
Sides of the third triangle = 1 m, \(\sqrt 3\) m, \(\sqrt 4\) m
…………………………
…………………………
Sides of the tenth triangle = 1 m, \(\sqrt 10\) m, \(\sqrt 11\) m
Hypotenuse of the 10^th triangle is \(\sqrt 11\)
Perpendicular sides are \(\sqrt 10\), 1

(ii) Perimeter of tenth triangle
= 1 m + \(\sqrt 10\) m + \(\sqrt 11\) m
Perimeter of ninth triangle = 1 m + \(\sqrt 9\) m + \(\sqrt 10\) m
More in the perimeter
= (\(\sqrt 10\) + \(\sqrt 11\) + 1) – (\(\sqrt 9\) + \(\sqrt 10\) + 1)
= \(\sqrt 11\) – \(\sqrt 9\) = (\(\sqrt 11\) – 3)m

iii. Sides of the n^th triangle are
\(\sqrt n\), \(\sqrt n + 1\), 1
The sides of the (n – 1)^th triangle are
\(\sqrt n – 1\), \(\sqrt n\), 1
The difference in perimeter
= (\(\sqrt n + 1\)) – (\(\sqrt n – 1\))

Hsslive Guru Maths 9th Kerala Syllabus Chapter 4 Question 5.
What is the hypotenuse of the right triangle with perpendicular sides \(\sqrt 2\) centimetres and \(\sqrt 3\) centimetres? How much larger than the hypotenuse is the sum of the perpendicular sides?
Answer:
Hypotenuse = \(\sqrt {\sqrt 3^{2}} + {\sqrt 2^{2}}\)
= \(\sqrt 3 + 2\) = \(\sqrt 5\) cm
Sum of perpendicular sides = \(\sqrt 3\) + \(\sqrt 2\) Difference with the hypotenuse
= (\(\sqrt 3\) + \(\sqrt 2\)) – \(\sqrt 5\)
= (1.73 + 1.41) – 2.24 = 3.14 – 2.24 cm = 0.9

Textbook Page No. 57

Hsslive Guru Class 9 Maths Kerala Syllabus Chapter 4 Question 1.
Of four equal equilateral triangles, two cut vertically into halves and two whole are put together to make a rectangle.

If a side of the triangle is 1 m, what is the area and perimeter of the rectangle?
Answer:
Sides of the triangle are \(\frac {1}{2}, \frac{\sqrt 3}{2}\), 1

Perimeter = 1 + \(\frac{\sqrt 3}{2}\) + \(\frac{\sqrt 3}{2}\) + 1 + \(\frac{\sqrt 3}{2}\) + \(\frac{\sqrt 3}{2}\)
= 2 + 2\(\sqrt 3\) m
Area of the rectangle = 1 × \(\sqrt 3\) = \(\sqrt 3\) m²

Hsslive Class 9 Maths Kerala Syllabus Chapter 4 Question 2.
A square and an equilateral triangle of sides twice as long are cut and the pieces are rearranged to form a trapezium, as shown below:
If a side of the square is 2 cm, what are the perimeter and area of the trapezium?

Answer:

Perimeter = 2 + 2\(\sqrt 3\) + 2 + 2\(\sqrt 2\) + 2\(\sqrt 3\) + 2\(\sqrt 2\)
= 4 + 4\(\sqrt 3\) + 4\(\sqrt 2\) = 4(1 + \(\sqrt 3\) + \(\sqrt 2\))
Area = 2² + \(\frac {1}{2}\) × 4 × 2\(\sqrt 3\) = 4 + 4\(\sqrt 3\)
= 4(1 + \(\sqrt 3\)) = 4(1 + 1.73)
= 4 × 2.73 = 10.92 cm

9th Maths Notes Kerala Syllabus Chapter 4 Question 3.
Calculate the perimeter and area of the triangle in the picture.

Answer:
Draw a perpendicular from B to AC
AB = 4cm
AD = 2

DB = 2\(\sqrt 3\)
DB = DC = 2\(\sqrt 3\)
BC = \(\sqrt {12} + {12}\) = \(\sqrt 24\)
Perimeter =
= 4 + \(\sqrt 24\) + 2\(\sqrt 3\) + 2
= 6 + \(\sqrt 24\) + 2\(\sqrt 3\) cm
Area = \(\frac {1}{2}\) × AC × DB
= \(\frac {1}{2}\) × (2 + 2\(\sqrt 3\)) × 2\(\sqrt 3\) = (2 + 2\(\sqrt 3\)) \(\sqrt 3\)
= 2\(\sqrt 3\) + 2\(\sqrt 3\) × \(\sqrt 3\) = 2\(\sqrt 3\) + 6
= 6 + 2\(\sqrt 3\)

Kerala Syllabus 9th Standard Maths Notes Pdf Chapter 4 Question 4.
From the pairs of numbers given below, pick out those whose product is a natural number or a fraction.
i. \(\sqrt 3\), \(\sqrt 12\)
ii. \(\sqrt 3\), \(\sqrt 1.2\)
iii. \(\sqrt 5\), \(\sqrt 8\)
iv. \(\sqrt 0.5\), \(\sqrt 8\)
v. \(\sqrt {7\frac {1}{2}}\), \(\sqrt 3{\frac{1}{3}}\)
Answer:
i. \(\sqrt 3\), \(\sqrt 12\)
\(\sqrt 3\) × \(\sqrt 12\) = \(\sqrt 3 × 12\) = \(\sqrt 36\) = 6
product is natural number

ii. \(\sqrt 3\), \(\sqrt 1.2\)
\(\sqrt 3\) × \(\sqrt 1.2\) = \(\sqrt 3 × 1.2\) = \(\sqrt 3.6\)
product is neither a natural number nor a fraction

iii. \(\sqrt 5\), \(\sqrt 8\)
\(\sqrt 5\) × \(\sqrt 8\) = \(\sqrt 5 × 8\) = \(\sqrt 40\)
product is neither a natural number nor a fraction

iv. \(\sqrt 0.5\), \(\sqrt 8\)
\(\sqrt 0.5\) × \(\sqrt 8\) = \(\sqrt 0.5 × 8\) = \(\sqrt 4\) = 2
product is natural number

v. \(\sqrt {7\frac {1}{2}}\), \(\sqrt 3{\frac{1}{3}}\)
\(\sqrt {7\frac {1}{2}}\) × \(\sqrt 3{\frac{1}{3}}\) = \(\sqrt {7\frac {15}{2}}\) × \(\sqrt 3{\frac{10}{3}}\)
= \(\sqrt {7\frac {150}{6}}\) = \(\sqrt 25\) = 5
product is natural number

Textbook Page No. 60

9th Standard Maths Notes Kerala Syllabus Chapter 4 Question 1.
Calculate the length of the sides of the equilateral triangle on the right, correct to a millimetre.

Answer:

Let one side = 2x
\(\sqrt (2x)^{2} – x^{2}\) = 4
\(\sqrt 4x^{2} – x^{2}\) = 4;
\(\sqrt 3x^{2}\) = 4; \(\sqrt 3\) x = 4; x = \(\frac {4}{\sqrt 3}\) = 4;
x = \(\frac{4 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} =\frac{4}{3} \sqrt{3} =\frac{4}{3} \times 1.73 = 2.309\)cm
= 2.31 cm
Length of sides = 2 × \(\frac {4}{\sqrt 3}\) = 2 × 2.309
= 4.62 cm = 46.2 cm

Question 2.
Prove that (\((\sqrt 2 + 1)(\sqrt 2 – 1 )\)) = 1. Use this to compute \(\frac {1}{\sqrt {2} – 1}\) correct to two decimal places.
Answer:
\((\sqrt{2} + 1)(\sqrt{2} – 1) = (\sqrt{2})^{2} – 1^{2}=2 – 1 = 1\)
\((\sqrt 2 + 1)(\sqrt 2 – 1 )\) = 1
\(\frac{1}{\sqrt{2} – 1} = \frac{1}{\sqrt{2} – 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{1} = \sqrt{2} + 1\)
\(\frac{1}{\sqrt{2} – 1} = \sqrt {2} + 1 \) = 1.41 + 1 = 2.41

Question 3.
Compute \(\frac{1}{\sqrt{2} + 1} \) corrects to two decimal places.
Answer:
\(\frac{1}{\sqrt{2} + 1} = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} – 1}{\sqrt{2} – 1} = \frac{\sqrt{2} – 1}{1}\)
= \(\sqrt {2} \) – 1 = 1.41 – 1 = 0.41

Question 4.
Simplify (\(\sqrt 3 – \sqrt 2 \))(\(\sqrt 3 + \sqrt 2 \)). Use this to compute \(\frac{1}{\sqrt{3} – \sqrt {2}}\) and \(\frac{1}{\sqrt{3} + \sqrt {2}}\) correct to two decimal places.
Answer:
(a – b)(a+b) = a² – b²
\((\sqrt{3} – \sqrt{2})(\sqrt{3} + \sqrt{2}) = (\sqrt{3})^{2} – (\sqrt{2})^{2}\)
= 3 – 2 = 1
\(\frac{1}{\sqrt{3} – \sqrt{2}} = \frac{(\sqrt{3} -\sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} – \sqrt{2})} = \sqrt{3} + \sqrt{2}\)
≈ 1.732 + 1.414 = 3.146 = 3.15
\(\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} – \sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} + \sqrt{2})}=\sqrt{3} – \sqrt{2}\)
≈ 1.732 – 1.414 = 0.318 = 0.32

Question 5.
Prove that \(\sqrt{2 \frac{2}{3}} = 2 \sqrt{\frac{2}{3}}\). Can you find other numbers like this?
Answer:
\(\sqrt{2 \frac{2}{3}} = \sqrt{\frac{8}{3}} = \sqrt{4 \times \frac{2}{3}} = 2 \sqrt{\frac{2}{3}}\)
\(\sqrt{3 \frac{3}{8}} = \sqrt{\frac{27}{8}} = \sqrt{9 \times \frac{3}{8}} = 3 \sqrt{\frac{3}{8}}\)
If \(\sqrt{x \frac{x}{n}} = x \sqrt{\frac{x}{n}}\) then
\(\sqrt{x + \frac{x}{n}} = \sqrt{x^{2} \frac{x}{n}} = \sqrt{\frac{x^{3}}{n}}\)
\(x + \frac{x}{n} = \frac{x^{3}}{n}\)
nx + x = x³
n + 1 = x²
n = x² – 1
similar numbers are
\(4 \frac{4}{4^{2} – 1} = 4 \frac{4}{15}\)
\(5 \frac{5}{5^{2} – 1} = 5 \frac{5}{24}\)
\(\sqrt{4 \frac{4}{15}} = 4 \sqrt{\frac{4}{15}}\)
\(\sqrt{5 \frac{5}{15}} = 5 \sqrt{\frac{5}{15}}\) etc.

Question 6.
All red triangles in the picture are equilateral. What is the ratio of the sides of the outer and inner squares?

Answer:
Let the sides of the shaded triangles are a units. The sides of the unshaded triangles are a, a, \(4 \frac{a}{2}\sqrt 3 \) units.

Side of the outer square = \(\sqrt{3}\frac{a}{2} + {\sqrt 3 \frac{a}{2}} + a\)
\(\sqrt{3}a + a = a(\sqrt {3} + 1)\)
Side of the inner square =
\(\sqrt{3}a + a – a(a + a) = \sqrt{3}a + a – a – a\)
\(\sqrt{3}a = a (\sqrt {3} – 1)\)
Ratio of the sides = \( a(\sqrt{3} + 1):a(\sqrt{3} – 1)\)
\( \sqrt{3} + 1 : \sqrt{3} – 1\)

Kerala Syllabus 9th Standard Maths New Numbers Exam Oriented Text Book Questions and Answers

Question 1.
Find the perimeter of the given triangle.
Hypotenuse = \(\sqrt{{….}^{2} + {….}^{2}}\) = \(\sqrt{….}\)
Perimeter = 2 + 3 + ………

Answer:
Hypotenuse = \(\sqrt{13}; \sqrt{2^{2} + 5^{2}}=\sqrt{4 + 9}\)
Perimeter = 2 + 3 +\(\sqrt{13} = 5 + \sqrt{13}\)

Question 2.
Find the perimeter of the rectangle with area 10 sq. centimetres.
Answer:
Let one side of the reactable be ‘a’
Area = a²
a² =10 sq. cm a = \(\sqrt{10}\)
Perimeter = 4 × a = 4 × \(\sqrt{10}\) cm

Question 3.

Answer:
\(\sqrt{x} \times \sqrt{y} = \sqrt{x \times y} = \sqrt{x y}\)
\(\sqrt{2} \times \sqrt{7} = \sqrt{2 \times 7} = \sqrt{14}\)
\(\sqrt{8} = \sqrt{4} \times \sqrt{2} =\sqrt{4} \times \sqrt{2} = 2\sqrt{2}\)
\(\sqrt{18} = \sqrt{9} \times \sqrt{2} =\sqrt{9} \times \sqrt{2} = 3\sqrt{2}\)
\(\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}\)

Question 4.
\(\frac{\sqrt{x}}{\sqrt {y}} = \frac{\sqrt{…..}}{\sqrt {…..}}\)
Answer:
\(\frac{\sqrt{x}}{\sqrt {y}} = \frac{\sqrt{x}}{\sqrt {y}}\)

Question 5.
\(\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{…….}\)
Answer:
\(\frac{\sqrt{1}}{\sqrt{2}} = \frac{\sqrt{1}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)

Question 6.
Find the value of \(\sqrt 50\) correct to 2 decimals.
\(\sqrt{50} = \sqrt{25 \times ……} = \sqrt{25} \times \sqrt{……} = 5\sqrt{2}\)
= 5 × 1.414 = ……
Answer:
\(\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}\)
= 5 × 1.414 = 7.070 = 7.07

Question 7.
Find the perimeter of the triangle correct to two, decimal places.

Answer:
Perimeter = \(\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} = 6\sqrt{2}\)
= 6 × 1.414 = 8.484 = 8.48 cm

Question 8.
If x = \(\frac{1}{\sqrt{2}}\), find (x + \(\frac{1}{x})^{2}\)
Answer:
x = \(\frac{1}{\sqrt{2}}\)
\(\frac{1}{x} = \sqrt{2} ; \left(x + \frac{1}{x}\right)^{2} = \left(\sqrt{2} + \frac{1}{\sqrt{2}}\right)^{2}\)
2 + \(\frac{1}{2}\) + 2 = \(4\frac{1}{2}\) = 4.5

Question 9.
What is the total length of the line joining two lines of lengths \(\sqrt 2\)cm, \(\sqrt 3\). Find the length correct to 3 decimal places.
Answer:
Length of the line = \(\sqrt 2\)cm + \(\sqrt 3\)
= 1.414+ 1.732 = 3.146 cm

Question 10.
Which is greater \(\sqrt 3\) + \(\sqrt 2\), \(\sqrt 5\)cm?
Answer:
\(\sqrt 3\) + \(\sqrt 2\)
\(\sqrt 3\) + \(\sqrt 2\) = 3.146
\(\sqrt 5\) = 2.236;
\(\sqrt 3\) + \(\sqrt 2\) + \(\sqrt 3\) + \(\sqrt 2\) > \(\sqrt 5\).

Question 11.
A line of length \(\sqrt 27\)cm is cut from a line of length \(\sqrt 12\)cm. Find the length of the remaining part of the line?
Answer:
Remaining part = \(\sqrt 27\) – \(\sqrt 12\)
= \(3\sqrt 3\) – \(2\sqrt 3\) = \(\sqrt 3\) cm

Question 12.
Simplify the following.
(a) \(\sqrt 50\) × \(\sqrt 2\)
(b) \(\sqrt 27\) × \(\sqrt 3\)
(c) \(\sqrt 12\) × \(\sqrt 36\)
(d) \(\sqrt 24\) × \(\sqrt 6\)
(e) \(\sqrt 28\) × \(\sqrt 7\)
(f) \(\sqrt 32\) × \(\sqrt 2\)
(g) \(\sqrt 5\) × \(\sqrt 10\) × \(\sqrt 2\)
(h) \(\sqrt 8\) × \(\sqrt 5\) × \(\sqrt 10\)
Answer:
(a) \(\sqrt 25 × 2 \) x \(\sqrt 2\) = \(\sqrt 25\) × \(\sqrt 2\) × \(\sqrt 2\) = 5 × 2 = 10
(b) \(\sqrt 9 × 3\) × \(\sqrt 3\) = 3 × 3 = 9
(C) \(\sqrt 4 × 3\) × 6 = 2 × \(\sqrt 3\) × 6 = 12\(\sqrt 3\)
(d) \(\sqrt 6 × 4 \) × \(\sqrt 6\) = 2 × 6 = 12
(e) \(\sqrt{4 \times 7} \times \sqrt{7} = 2 \times 7 = 14\)
(f) \(\sqrt{16 \times 2} \times \sqrt{2} = 4 \times 2 = 8\)
(g) \(\sqrt{5} \times \sqrt{5} \times \sqrt{2} \times \sqrt{2} = 10\)
(h) \(\sqrt{4} \times \sqrt{2} \times \sqrt{5} \times \sqrt{2} \times \sqrt{5} = 2 \times 2 \times 5 = 20\)

Question 13.
Simplify the following.
(a) \(3\sqrt{2} + 5 \sqrt{2}\)
(b) \(7\sqrt{5} – 3 \sqrt{5}\)
(C) \(4\sqrt{7} + 5 \sqrt{7} – 3\sqrt{7}\)
(d) \(2\sqrt{3} + \sqrt{27}\)
(e) \(4\sqrt{3} – 3 \sqrt{12} + 3\sqrt{75}\)
(f) \(\sqrt{162} – \sqrt{72}\)
(g) \(4\sqrt{12 }- \sqrt{50} – 5\sqrt{48}\)
(h) \(\sqrt{8}0 + \sqrt{125}\)
(i) \(\frac{1}{\sqrt{5} – 1}\)
(j) \(\frac{3}{2 \sqrt{3} – \sqrt{2}}\)
(k) \(\frac{5}{2 \sqrt{7} – 3\sqrt{5}}\)
Answer:
(a) \(8\sqrt{2}\)

(b) \(4\sqrt{5}\)

(C) \(6\sqrt{7}\)

(d) \(\sqrt{3} + 3 \sqrt{3} = 5\sqrt{3}\)

(e) \(4\sqrt{3} – 3\sqrt{4 \times 3} + 3\sqrt{25 \times 3}\)
= 4 \(\sqrt{3}  – 6\sqrt{3} + 15 \sqrt{3}\)
= \(13\sqrt 13\)

(f) \(\sqrt{81 \times 2} – \sqrt{36 \times 2} = 9\sqrt{2} – 6\sqrt{2} = 3\sqrt{2}\)

(g) \(4\sqrt{4 \times 3} – \sqrt{25 \times 2} – 5\sqrt{12 \times 4}\)
\( = 8\sqrt{3} – 5\sqrt{2} -20 \sqrt{2}\)
\( = -12\sqrt{3} – 5\sqrt{2}\)

(h) \(\sqrt{16 \times 5}+\sqrt{25 \times 5} = 4\sqrt{5} + 5 \sqrt{5} = 9\sqrt{5}\)

Question 14.
If \(\sqrt{75} + \sqrt{363} + x\sqrt{3} = 0\). Find the value of x.
Answer:
\(x \sqrt{3} = 0 – \sqrt{75} – \sqrt{363}\)
\(= – 1 \times(\sqrt{75} + \sqrt{363})\)
\(x = \frac{-1(5 \sqrt{3} + 11 \sqrt{3})}{\sqrt{3}} = \frac{-1 \times 16 \sqrt{3}}{\sqrt{3}} = -16\)

Question 15.
Calculate \(\sqrt{3}(\sqrt{48} + \sqrt{32} – \sqrt{18})\)
Answer:

Question 16.
Write two fractions between \(\frac{1}{3}\) and \( \frac {1}{4}\)
Answer:
Fraction one = \(\frac{\frac{1}{3} + \frac{1}{4}}{2}=\frac{7}{24}\)
Another fraction =
\(\frac{\frac{1}{3} + \frac{7}{24}}{2} = \frac{8 + 7}{48} = \frac{15}{48}\)

Question 17.
In the figure given below, PQRS is a square of each side is 3 cm. Each sides of the square is divided into 3 equal parts. Joining these points to get an octagon, find its perimeter.

Answer:
ΔAPH is a right angled triangle.
AP = PH = 1 cm
∴ AH = \(\sqrt{1^{2}+1^{2}} = \sqrt{2}\)
AH = GF = ED = CB = \(\sqrt{2}\)

Perimeter of the octagon
= AB + BC + CD + DE + EF + FG + GH + HA
\(=1 + \sqrt{2} + 1 + \sqrt{2} + 1 + \sqrt{2} + 1 + \sqrt{2} = 4 + 4 \sqrt{2}\)
= 4 + 4 × 1.41 = 4 + 5.64 = 9.64cm.

Question 18.
If three points A, B,C. Such that AB = \(\sqrt{50}\)cm, BC = \(\sqrt{98}\)cm, AC = \(\sqrt{288}\)cm. Check whether the points A,B,C lie on a straight line?
Answer:
AB = \(\sqrt{50}\) = \(5\sqrt{2}\)cm
BC = \(\sqrt{98}\) = \(7\sqrt{2}\)cm
AC = \(\sqrt{288}\) = \(12\sqrt{2}\)cm
\(5\sqrt{2}\) + \(7\sqrt{2}\) = \(12\sqrt{2}\)
AB + BC = AC
The three points lie on a straight line.

Question 19.
Find the value of \(\sqrt{12} – \frac{1}{\sqrt{3}}\) correct to two decimal places.
Answer:
\(\sqrt{12} – \frac{1}{\sqrt{3}} = \frac{2 \sqrt{3}}{1} – \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} – \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}\)
= \(\frac {5}{1.73}\) = 2.89

Question 20.
Find the sum 745+7180+780
Answer:

Question 21.
A, B, C are three points such that AB = \(\sqrt {50}\) cm, BC = \(\sqrt {98}\) cm, AC =
\(\sqrt {288}\) cm. Do they lie on a straightline?
Answer:
If A, B, C are points on the same line then AB + BC = AC

Here AB + BC = AC. So the three given points are on the same straight line.

You can Download Circles Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 5 Circles

Kerala Syllabus 9th Standard Maths Circles Text Book Questions and Answers

Textbook Page No. 68

Circles Class 9 Kerala Syllabus Question 1.
Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.

Answer:

AC = AD (Radii of the same circle)
AE = AE (Common side)
BC = BD (Radii of the same circle)
ΔABC = ΔABD (Three sides are equal)
In equal triangles, angles opposite to equal sides are equal.
So, ∠CAE = ∠DAE
Consider ΔCAE and ΔEAD.
∠CAE = ∠DAE
AC = AD (Radii of the same circle)
AE = AE (Common side)
ΔAEC = ΔAED (Two sides and the angle between them)
In equal triangles, sides opposite to equal angles are equal.
So, CE = DE (∠CAE = ∠DAE)
CE = DE ………(1)
In equal triangles, angles opposite to equal sides are equal. So, ∠AEC = ∠AED
∠AEC + ∠AED = 180° (Linear pair)
∠AEC = ∠AED = 90° ………(2)
From equation (1) and (2)
The line joining the centres of the circles is the perpendicular bisector of the chord.

Kerala Syllabus 9th Standard Maths Chapter 5 Question 2.
The picture on the right shows two circles centred on the same point and a line intersecting them. Prove that the parts of the line between the circles on either side are equal.

Answer:

AD and BC are chords.
OE bisects the chords perpendicularly AD and BC
BE = CE
AE = DE
AE – BE = DE – CE
AB = CD

Circles Class 9 State Syllabus Question 3.
The figure shows two chords drawn on either sides of a diameter: What is the length of the other chord?

Answer:

PO is the perpendicular bisector of AB, OQ is the perpendicular bisector of AC.
∠OAQ = ∠OAP = 30° (Given)
∠OQA = ∠OPA = 90° (Right angles)
∴ ∠AOQ = ∠AOP (Third angle also equal)
AO = AO (Common side)
ΔOQA = ΔOPA
In equal triangles, sides opposite to equal angles are equal. So AP = AQ.
AP = ½AB (Perpendicular from the centre of a circle to a chord bisects the chord)
AQ = ½AC
Since AP = AQ
½AB = ½AC
AB = AC
So length of AC = 3 cm

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Kerala Syllabus 9th Standard Maths Notes Question 4.
A chord and the diameter through one of its ends are drawn in a circle. A chord of the same inclination is drawn on the other side of the diameter.

Prove that the chords are of the same length.
Answer:

AB is the diameter
AC and AD are the chords
Given that ∠OAC = ∠OAD
In triangle OAC,
∠OAC = ∠OCA
In triangle OAD,
∠OAD = ∠ODA
Consider the ΔOAC and ΔOAD
∠OAC = ∠OAD; ∠OCA = ∠ODA
∠AOC = ∠AOD; AO = AO
Triangles are equal.
Sides opposite to equal angles are also equal.
∴ AC = AD

9th Class Maths Notes Kerala Syllabus Question 5.
The figure shows two chords drawn on either sides of a diameter. How much is the angle the other chord makes with the diameter?

Answer:

AD is the diameter and O is the centre of the circle.
∠OAB = 40°
Consider ΔOAB and ΔOAC
AB = AC = 3cm
OC = OB (Radius of the circle)
OA = OA (Common side)
Three sides ΔOAB of are equal to three sides of ΔOAC
In equal triangles, angle opposite to equal sides are equal.
∴ ∠OAB = ∠OAC
∴ ∠OAC = 40°

Hss Live Guru 9th Maths Kerala Syllabus Question 6.
Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.
Answer:
AB, AC are the chords of same length AD is the diameter of the circle.

When we consider ΔAOB, ΔAOC
AB = AC (Given)
OB = OC (Radius)
OA = OA (Common side)
Three sides ΔOAB of are equal to three sides of ΔOAC.
In equal triangles, angle opposite to equal sides are equal.
∠BAO = ∠CAO
∴ the diameter AD bisects ∠A.

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 7.
Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.

Prove that this polygon is a regular octagon.
Answer:
The diameters are parallel to the sides

∠ADC = ∠BDC = 90°
Consider ΔADC & ΔBDC
AD = BD (Perpendicular from the centre of a circle to a chord bisects the chord)
DC = DC (Common side)
∠ADC = ∠BDC (90° each)
Two sides and the angle between them of ΔADC are equal to two sides and the angle between them of ΔBDC.
So ΔADC & ΔBDC are equal.
∴ AC = BC
In the same way we can see that other sides of the octagon are also equal.
So it is a regular octagon.

Textbook Page No. 72

Scert Class 9 Maths Solutions Kerala Syllabus Question 1.
Prove that chords of the same length in a circle are at the same distance from the centre.
Answer:

AB, CD are the chords of same length.
AB = CD
AP = ½AB (Perpendicular from the centre of a circle to a chord bisects the chord)
Similarly CQ = ½CD
AP = CQ [Since AB = CD]
Consider the right angled triangle ΔAOP and ΔCOQ
OP² = OA² – AP²
OP² = OB² – CQ² [Since OA = OB, AP = CQ]
OP² = OQ²
∴ OP = OQ
So, the chords of the same length in a circle are at the same distance from the centre.

Kerala Syllabus 9th Standard Maths Guide Question 2.
Two chords intersect at a point on a circle and the diameter through this point bisects the angle between the chords. Prove that the chords have the same length.
Answer:
OA = OC (radius of the same circle)
OB = OB (common side)

∠OBA = ∠OBC (given)
∠BAO = ∠BCO [Base angle of isosceless triangle ΔOCB & ΔOCA]
∠AOB = ∠BOC;
∴ ΔAOB = ΔBOC
So the sides AB and BC opposite to equal angles are also equal.

Kerala Syllabus 9th Standard Notes Maths Question 3.
In the picture on the right, the angles between the radii and the chords are equal. Prove that the chords are of the same length.

Answer:
Perpendiculars are drawn from the centre of the circle to the chords.

Consider ΔAOM, ΔCON;
OM = ON
OA = OC (radii)
∠AMO = ∠CNO = 90°
∠A = ∠C (given)
ΔAOM ≅ ΔCON (A.A.S)
In equal triangles, angle opposite to equal sides are equal.
AM = CN
½AB = ½CD
∴ AB = CD

Textbook Page No. 73

Kerala Syllabus 9th Standard Maths Guide In Malayalam Question 1.
In a circle, a chord I cm away from the centre is 6 cm long. What is the length of a chord 2 cm away from the centre?
Answer:

Radius of the circle =
\(\sqrt{3^{2} + 1^{2}} = \sqrt{9 + 1} = \sqrt{10}\)
AB = \(\sqrt{\sqrt{10}^{2} – 2^{2}} = \sqrt{10 – 4} = \sqrt{6}\)
Length of the chord \(\sqrt 6 + \sqrt 6 = 2\sqrt 6\)

Kerala Syllabus 9th Std Maths Notes Question 2.
In a circle of radius 5cm, two parallel chords of lengths 6cm and 8cm are drawn on either side of a diameter. What is the distance between them? If parallel chords of these lengths are drawn on the same side of a diameter, what would be the distance between them?
Answer:

OM = \(\sqrt{5^{2} – 3^{2}}\)
= \(\sqrt{25 – 9}\)
=\(\sqrt {16}\) = 4 cm
ON = \(\sqrt{5^{2} – 4^{2}}\)
= \(\sqrt{25 – 16}\)
=\(\sqrt 9\) = 3 cm
The distance between the chords = 4 + 3 = 7cm
If the chords are on same side = 4 – 3 = 1cm

Kerala Syllabus Maths Class 9 Question 3.
The bottom side of the quadrilateral in the picture is a diameter of the circle and the top side is a chord parallel to it. Calculate the area of the quadrilateral.

Answer:

AB = \(\sqrt{2.5^{2} – 1.5^{2}}\)
= \(\sqrt{6.25 – 2.25}\)
=\(\sqrt 4\) = 2 cm
The quadrilateral is a trapezium.
The distance between the parallel sides = 2 cm
Area = \(\frac{1}{2}\) × 2 × (5 + 3) = 8cm²

Circles Questions And Answers Kerala Syllabus 9th Question 4.
In a circle, two parallel chords of lengths 4 and 6 centimetres are 5 centimetres apart. What is the radius of the circle?
Answer:

MN = 5
ON = x
OM = 5 – x
x² + 3² = (5 – x)² + 2²
x² + 9 = 25 – 10x + x² + 4
9 = 25 – 10x + 4
10x = 25 + 4 – 9
10x = 20
x = 20/10 = 2
Radius = \(\sqrt{2^{2} + 3^{2}}\) = \(\sqrt {4 + 9}\) = \(\sqrt {13}\)cm

Textbook Page No. 78

9th Standard Maths Notes Kerala Syllabus Question 1.
Draw three triangles with lengths of two sides 4 cm and 5 cm and angle between them 60°, 90° and 120°. Draw the circumcircle of each . (Note how the position of the circumcentre changes).
Answer:

In this triangle all the angles are less than 90°. The circum centre ‘O’ is inside the triangle.

In the triangle with one angle is 90°. The circum centre is the midpoint of the hypotenuse.

In this triangle with an angle greater than 90°. The circumcentre ‘O’ is outside the triangle.

Question 2.
The equal sides of an isosceles triangle are 8 cm long and the radius of its circumcircle is 5 cm. Calculate the length of its third side.
Answer:
ΔABC is an isosceles triangle The bisector of ∠A bisects BC
OM = x; BM = \(\sqrt{5^{2} – x^{2}}\)
When we consider ΔAMB,
AB² = AM² + BM²

82 = (5 + r)² + \((\sqrt {5^{2} – x^{2}})^{2}\)
64 = 25 + 10x + x² + 25 – x²
64 = 10x + 50
14 = 10x
x = 14/10
= 1.4
BM = \((\sqrt {5^{2} – 1.4^{2}}\)
BC = \(2(\sqrt {5^{2} – 1.4^{2}} = 2\sqrt {23.04} = \sqrt {92.16}\)
= 9.6 cm

Question 3.
Find the relation between the length of a side and the circumradius of an equilateral triangle.
Answer:

In ΔABC,
AB = BC + AC (sides of an equilateral triangle)
∠DAO = 30°, ∠ADO = 90°
∴ ∠AOD = 60°
By using the properties of angles 30°, 60°, 90°.
If OA = r
OD = \(\frac{r}{2}\)
AD = \(\frac{\sqrt 3}{2}\)r
AB = 2 × \(\frac{\sqrt 3}{2}\)r = \(\sqrt 3\)r
One side of an equilateral triangle is \(\sqrt 3\) times its circumradius.

Kerala Syllabus 9th Standard Maths Circles Exam Oriented Text Book Questions and Answers

Question 1.
Draw a circle which passes through the points A, B and radius 5 cm.
Answer:

The 5 cm line drawn from A meet the perpendicular bisector of AB at point O. The a circle drawn with O as centre and OA as radius will pass through A and B.

Question 2.
The distance between the points A and B is 3 cm. Find out the radius of the smallest circle which passes through these points? What is AB about this circle?
Answer:
1.5 cm. Diameter.

Question 3.
Draw circles which passes through the points A and B and radius 3 cm, 4 cm and 5 cm.
Answer:
Centres of the circle lie on the same straight line.

Question 4.
Two circles in the diagram have same radius. Prove that AC = BD.

Answer:
OP is drawn perpendicular to the chord. OP bisect CD and AB.

Question 5.
A and B are the centres of two circles in the diagram. Circles meet at points O and P. MN || AB. Then prove that MN = 2 × AB.

Answer:

Draw perpendicular lines AX, BY
∴ XM = XO similarly YN = YO
MN = MX + XY + YN
= OX + XY + OY = XY + XY
= 2XY = 2AB

Question 6.
AB is he diameter of the circle with centre C. PQ || AB, AB = 50cm, PQ = 14cm. Find BQ.

Answer:

Draw CM perpendicular PQ.
PQ = 14cm,
∴ MQ = 7 cm, CN = 7;
CB = 25 cm
CQ = 25 cm
NB = 25 – 7 = 18
CM = \(\sqrt{25^{2} – 7^{2}} = \sqrt{625 – 49}\)
=\(\sqrt{576}\) = 24
∴ NQ = 24
BQ = \(\sqrt{NQ^{2} – NB^{2}} = \sqrt{24^{2} + 18^{2}}\)
= \(\sqrt {576 + 524} = \sqrt {900}\) = 30 cm

Question 7.
AB, AC are two chords of a circle and the bisector of ∠BAC is a diameter of the circle. Prove that AB = AC.
Answer:
OA = OA (common side)
∠OPA = ∠OQA = 90°
(OP ⊥ AB, OQ ⊥ AC) ∠PAO
= ∠QAO (AE bisector)
∴ ΔOAP = ΔOAQ
OP = OQ therefore AB = AC (Equal chords of a circle are equidistant from the centre).

Question 8.
In the question above instead of assuming ∠OAB = ∠OCD assuming that AB = CD and then prove that ∠OAB = ∠OCD.

Answer:
∠OAB = ∠OCD (Given)
OA = OC (radius).
∠P = ∠Q = 90°
∴ (OP ⊥ AB, OQ ⊥ CD ) ∴ ΔOAP ≅ ΔOCQ;
∴ OP = OQ, Therefore AB = CD
[Equal chords of a circle are equidistant from die centre]

Question 9.
What is the distance from the centre of a circle of a circle of radius 5 cm to a chord of length 8 cm.
Answer:

Distance from the centre = cp = \(\sqrt{5^{2} – 4^{2}}\)
= \(\sqrt{25 – 16} = \sqrt{9}\) = 3cm

You can Download Area Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 1 Area

Kerala Syllabus 9th Standard Maths Area Text Book Questions and Answers

Textbook Page No. 11

9th Standard Maths Notes Kerala Syllabus Question 1.
Draw a triangle of sides 3,4 and 6 centimeters. Draw three different right triangles of the same area.
Answer:
Draw ∆ ABC having sides 3 cm, 4 cm, and 6 cm.

Draw a line parallel to AB through C and a line from B perpendicular to the parallel line, both lines intersect at point E. Now ∆ABE is a right angled triangle. Also draw a line perpendicular to the parallel line through A, which makes a point Don parallel line, join BD. ∆ ABD is a right angled triangle.

Similarly make right-angled triangle by drawing line parallel to AC through B, and a line from A perpendicular to the parallel line, both lines intersect at point F.

The right traingle ∆ DAB, ∆ ABE, ∆ CAF have the same area as that of ∆ ABC.

Kerala Syllabus 9th Standard Maths Solution Chapter 1 Area In English Question 2.
Draw the triangle shown below in your notebook. Draw triangles ABP, BCQ and CAR of the same area with measurements given below.

1. ∠BAP = 90°
2. ∠BCQ = 60°
3. ∠ACR = 30°

Answer:
1.

Draw a line parallel to AB through C. Draw a perpendicular from A and the point of intersection of perpen-dicular and parallel line is P. Then draw ∆ABP then ∠BAP = 90°. Areas of ∆ABC and ∆ABP are equal.

2.

Draw a line parallel to BC through A from C make angle 60° both lines intersect at point Q. Then draw ∆ BCQ. Areas of ∆ ABC and ∆ BCQ are equal.

3.

Draw a line parallel to AC through B from C make angle 30° both lines intersect at point R. Then draw ∆ACR. Areas of ∆ABC and ∆ ACR are equal.

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Maths 9th Class Chapter 1 Area English Medium Kerala Syllabus Question 3.
Draw a circle and a triangle with one vertex at the centre of the circle and the other two on the circle. Draw another triangle of the same area with all three vertices on the circle.
Answer:
Draw a line with centre C and AB is the chord of the circle. Draw ∆ ABC then draw a line parallel to AB through C this line touches the circumference at a point D then complete ∆ ABD. Area of ∆ ABC = Area of ∆ ABD

9th Standard Maths Notes Malayalam Medium Kerala Syllabus Question 4.
How many different (non-congruent) triangles can you draw with two sides 8 and 6 centimetres and area 12 square centimetres? What if the area is to be 24 square centimetres?
Answer:
For the triangle of area of 12 sq.cm., the height from the 8 cm side should be 3cm.
Then the area = \(\frac{1}{2}\) x 8 x 3 =12 sq.cm
Draw a line 8 cm long. Draw another line parallel to it at a distance of 3 cm. Draw a circle of radius 6 cm with one end of the first line as the center. The points where this circle cuts the second line are the third vertex of the triangle.

We can draw two triangles ∆ OBC and ∆ OCA having area 12 sq.cm.
For the triangle of area 24 sq cm, the height from the 8 cm side should be 6 cm.

We can draw only one triangle of this type.

Draw A Square Of Area 10 Square Centimeters Malayalam Question 5.
In the picture below, the lines parallel to each side of the blue triangle through the opposite vertex are drawn to make the big triangle.

How many triangles in the picture have the same area as that of the blue triangle? How many of them have the same measures of the blue triangle?
Answer:

AB is parallel to PQ. So ∆ PAQ and ∆ QPB have the same area as that of ∆ PQR.
BC is parallel to RP, So ∆ CPB and ∆ CRB have the same area as that of ∆ PQR.
AC is parallel to RQ, So A CRA and ∆ CAQ have the same area as that of ∆ PQR.
Also ∆ PCQ, ∆ PAR and ∆ BRQ have the same area as that of ∆ PQR.
Total = 9 triangles
Also ∆ PCQ, ∆ PAR and ∆ BRQ have the same measurement as that of ∆ PQR.
Total = 3 triangles.

Kerala Syllabus 9th Standard Notes Maths Question 6.
Prove that the two triangles shown below have the same area:

Answer:

∠ A = 50°
∠B = 130°
∠A + ∠B = 50 + 130 = 180°
These angles are supplementary. So lines AD and BC are parallel. Their lengths are equal.
Since these triangles have the same base and their third vertices on a line parallel to the base, their areas are equal.

Textbook Page No. 15

Maths Notes Class 9 Kerala Syllabus Question 1.
Draw the two quadrilaterals shown below, in your note book. Draw tri-angles of the same area and calcu-late the areas (The lengths needed may be measured).

Answer:
1.

In the given figure, a line parallel to BD through C and AB produced intersect at E. Now area of ∆AED is equal to the area of quadrilateral. Measure the required lengths and then find the area.
AE = 10.8 cm
Height from D to AE = 2.9 cm
Area of ∆ AED = \(\frac{1}{2}\) x 10.6 x 2.9
= 15.66 sq.cm

2.

In the given figure, a line parallel to BD through C and AB produced intersect at E. Now area of ∆ AED is equal to the area of quadrilateral. Measure the required lengths and then find the area. AE = 12.6 cm DP = 5.2 cm Area of ∆ AED = \(\frac{1}{2}\) x AE X PD
= \(\frac{1}{2}\) x 12.6 x 5.2 = 32.76 sq cm

Kerala Syllabus Class 9 Maths Solutions Question 2.
Draw a rhombus of sides 6 centimetres and one angle 60°; then draw a right triangle of the same area.
Answer:

Draw rhombus ABCD.
∠A = ∠C = 60° and ∠B = ∠D = 120°
Draw diagonal BD. Draw line parallel to BD through C and AB produced intersect at point E. Now area of rhombus is same as the area of ∆ ADE and ∆ ACE. Lines AC bisect ∠A and ∠C.
∴ Angles of ∆ADE is equal to 30°, 60°, 90°
∴ The triangles drawn are right-angled triangles.

Kerala Syllabus 9th Standard Maths Guide Pdf Question 3.
Draw a regular pentagon and then a triangle of the same area. Calculate the area.
Answer:

Draw a regular pentagon ABCDE. Draw a line parallel to BD through C, line parallel to AD through E, AB produced to both sides intersects points F and G. Now area of AFDG is same as area of pentagon. Measure the required lengths and then find the area.

Kerala Syllabus 9th Standard Maths Notes Question 4.
The picture shows a rectangle divided into two parts. Instead of the broken line separating these parts, draw a straight line to divide the rectangle into two other parts of the same area. Calculate the areas of these parts.

Answer:

Join PQ, draw a line parallel to PQ through O intersecting points are M and N line PN or MQ can make 2 parts with same area.

Proof:
Area of rectangle ABCD = 5 x 3 = 15 sq.cm
Area of trapezium JOPD = \(\frac{1}{2}\) x 1 x (1+3) = 2 sq cm
Area of trapezium AQOJ = \(\frac{1}{2}\) x 2 x (2 + 3) =5 sqcm
Area of part AQOPD = Area of trapezium JOPD + Area of trapezium AQOJ
= 2 + 5 = 7 sqcm
Area of the remaining part = 15 – 7 = 8 sq cm
Now area of trapezium ANPD is equal to the area of pentagon AQOPD.

Textbook Page No. 19

9th Class Maths Notes Malayalam Medium Kerala Syllabus Question 1.
In the picture below, two lines are drawn from the top vertex of a tri-angle to the bottom side:

Prove that the ratio in which these lines divide the length of the bot¬tom side is equal to the ratio of the area of the three smaller triangles in the picture.
Answer:

If AB : BC : CD
x : y : z and height = h then area of triangles are \(\frac{1}{2}\) x h, \(\frac{1}{2}\) yh, \(\frac{1}{2}\)zh
then ratio of areas are \(\frac{1}{2}\) xh : \(\frac{1}{2}\) yh \(\frac{1}{2}\) zh = x : y : z

Maths Class 9 Kerala Syllabus Question 2.
In the picture below, the top vertex of a triangle is joined to the mid point of the bottom side of the tri-angle and then the mid point of this line is joined to the other two vertices.

Prove that the areas of all four tri-angles obtained thus are equal to a fourth of the area of the whole triangle.
Answer:

Mid point of BC is D.
Line AD divides ∆ ABC into 2 triangles with equal areas.
E is the midpoint of AD.
Line BE divide ∆ ADB into 2 triangles with equal areas. Similarly CE divides ∆ ADC.
4 triangles of equal area.

Class 9 Maths Chapter 1 Kerala Syllabus Question 3.
In the picture below, the top vertex of a triangle is joined to the mid point of the opposite side and then the point dividing this line in the ratio 2 :1 is joined to the other two vertices:

Prove that the areas of all three triangles in the picture on the right are equal to a third of the area of the whole triangle.
Answer:

Mid point of AB is Q
∴ ∆ AQC and ∆ BQC have equal area’s P divides AQ in the ratio 2 : 1.
Area of ∆ APC is twice the area of ∆ AQP.
∴ Sum of the areas of ∆ AQP and ∆ BQP is equal to the sum of the area’s of ∆ APC and ∆ BPC.
The area of the APB, PCB and CPA are same. The area of three triangles is equal to the third of area of the whole triangle.

Question 4.
Prove that the lengths of the per-pendiculars from any point on the bisector of an angle to the sides are equal.
Answer:

OC is the bisector of ∠AOB
C ia a point on this bisector.
CA is perpendicular to OA.
CB is perpendicular to OB.
We have to prove that CA = CB Consider the two triangle ∆ OAC and ∆ OBC.
OC = OC (Common side)
∠OAC = ∠OBC = 90°
∠AOC = ∠BOC (OC is the bisector of AOB)
One side and two angles on it of OAC are equal o one side and two angles on it of OBC.
So these triangles are equal.
Then sides opposite to equal angles are equal.
CA = CB
∴ The lengths of the perpendiculars from any point on the bisector of an angle to the sides are equal.

Question 5.
In the picture below, the side AC of the triangle ABC is extended to D, by adding the length of the side CB. Then the line through C parallel to DB is drawn to meet AB at E.

1. Prove that CE bisects ∠C.
2. Describe how this can be used to divide an 8 centimeters long line in the ratio 4 : 5.
3. Can we use it to divide an 8 centimeters long line in the ratio 3 : 4? How?

Answer:
1. ∆ BCD is an equilateral triangle If∠CBD = x then ∠CDB = x
∠BCD = 180 – 2x
∠BCE = x
(line BC passing through parallel lines EC and BD makes equal angle)
∴ ∠ACE = x
∴ line CE bisects ∠C.

2. If AB = 8, AC = 4, and CB = 5
then the line CE divides AB in the ratio 4 : 5.

3. No.
We can’t draw a triangle with sides 8cm, 3cm and 4cm. Since 3 : 4 = 6 : 8, take 6cm and 8cm as two sides then find third side.

Question 6.
In the figure, the diagonals of a quadrilateral split it into four tri-angles. The areas of three of them are shown in the picture:

Calculate the area of the whole quadrilateral.
Answer:
A line from the vertex of a triangle divides the length of the opposite side and the area of triangle in the same ratio.

In ∆ ACD, a line DE is drawn from D to AC.
∴ AE : EC = Area ∆ AED : Area of ∆ CED
AE : EC = 60 : 30 = 2 : 1
In ∆ ACB, a line BE is drawn from B to AC.
∴ AE : EC = Area of ∆ AEB : Area of ∆ CEB
2 : 1= Area of ∆ AEB : 40 = 2 : 1
∴ Area of ∆ AEB = \(\frac{40×20}{1}\) =80 sq.cm
∴ Total area of the quadrilateral = 60 + 30 + 80 + 40 = 210 sq.cm

Question 7.
In this picture the hori∠ontal lines at the top and bottom are parallel.

Prove that the yellow and red tri-angles are of the same area.
Answer:

AB ∥ DC
∴ Area of ∆ ABC = Area of ∆ ABD (Same base and the third vertex on a line parallel to the base)
Area of ∆ ABC – Area of ∆ ABE = Area of ∆ ABD – Area of ∆ ABE
∴ Area of ABEC = Area of ∆ AED ∆ AED ↔ ∆ BEC

Question 8.
In the figure, the diagonals of a trapezium split it into four triangles.

The area of the yellow triangle is 10 square centimetres and the area of the green triangle is 20 square centi – metres. What is the area of the whole trapezium?
Answer:
Since quadrilateral ABCD is a trapezium, lines AB and CD are parallel.

So area of ∆ ABD = 20 + 10 = 30 sq.cm
Area of ∆ ABC = 30 sq.cm
(Same base and the third vertex on a line parallel to the base)
Area of ∆ ABC = 30 sq.cm
Area of ∆ ABE = 20 sq.cm
Area of ∆ BEC = 30 – 20 = 10 sq.cm
From ∆ ABC,
AE : EC = area of ∆ ABE : Area of ∆ BEC
AE : EC = 20 : 10 = 2 : 1
From ∆ ACD,
AE : EC = 10 : area of ∆ DEC 2 : 1 = 10: area of ∆ DEC
Area of ∆ DEC = \(\frac{1×10}{2}\) sq.cm
Total area of the trapezium
= 20 + 10 + 10 + 5
= 45 sq.cm

Question 9.
The picture below shows a trapezium divided into four parts by the diagonals.

The area of the blue triangle is 4 square centimetres and the area of the green triangle is 9 square centimetres. What is the total area of the trapezium?
Answer:

Area of ∆ AMD and ABMC are equal.
If area = A sq. cm then
9 : A = BM : MD
A : 4 = BM : MD;
9 : A = A : 4
A² = 36
A = 6 sq. cm
∴ Area of trapezium = 6 + 4 + 6 + 9 = 25 sq. cm

Kerala Syllabus 9th Standard Maths Area Exam Oriented Text Book Questions and Answers

Question 1.
a. Where should be the position P in CD to set the isosceles triangle with half the area of the rectangle.

b. Where should be P in CD to get ∠ABP a right angle.
c. If ∠APB is a right angle how can you find the position P?
d. Can such right angled triangle be drawn in any of the rectangle?
Answer:
a. P is the centre of CD
b. It should be in C.
c. The distance from centre of AB to P should be half the length of AB.
d. No.

Question 2.
The lines in the diagram are parallel. Construct ∆ ABC with ∠C right angled

Answer:

AB is the diameter.

Question 3.
Construct ∆ ABC with AB = 8cm, BC = 6cm, AC = 5cm. Draw an isosceles triangle with same area and one side as AB itself.
Answer:

In figure AB ∥ PQ. D is the center point of AB.
∴ AD = BD.
Hence ∆ ABD is an isosceles triangle.

Question 4.
Find the area of the triangles with the given sides.
(a) 21 cm, 17 cm, 10 cm
(b) 12 cm, 16 cm, 20 cm
(c) 34 cm, 30 cm, 16 cm
(d) 20 cm, 21 cm, 29 cm
Answer:

Question 5.
ABCD is a rectangle in the diagram. P is a point on the side CD. Show that the area of ∆ AQB is equal to the sum of the areas of ∆ PQD and ∆ PCB

Answer:
⇒ area of ∆ AQB = area of ∆ APB – area of ∆ BQP
⇒ area of ∆ ADB = area of ∆ DBC area of (∆ ADQ + ∆ AQB) = area of (∆ DQP + ∆ QBP + ∆ BPC ) …….(1)
(∆ DAP = ∆ DBP) ⇒ (∆ ADQ = ∆ QBP) ⇒ area of (∆ ADQ + ∆ AQB)
= area of (∆ DQP + ∆ ADQ + ∆ BPC)
⇒ area of ∆ AQB = area of ∆ DQP + area of ∆ BPC.

Question 6.
In the figure XY, RS are parallel. The area of ∆ ABD is 20 cm²
a) Find the area of the parallelogram ABCD.
b) Find two triangles in the diagram with same area

Answer:
a.
20 + 20 = 40 sq. cm
b.
i. ABD, ABC
ii. DCB, DCA

Question 7.
Find the area of triangle in the dia-gram.

Answer:
\(\frac{1}{2}\) x 6 x 4 = 12 cm

Question 8.
In the figure BP : PQ : QC = 1 : 2 : 1. Area of triangle APQ is 8 sq.cm.
a. Find the area of triangle ABP.
b. Find the area of triangle ABC.

Answer:
a. BP : PQ = 1:2
Area of ∆ ABP is half the area of ∆ APQ.
Area of ∆ ABP = \(\frac{1}{2}\) x 8 = 4 sq.cm

b. Area of ∆ ABC = 4 + 8 + 4 = 16 sq.cm

Question 9.
In the figure line AB is parallel to CD. If AB = 5cm, AC = 4cm and ∠CAB = 90°.
a. Calculate the area of triangle ABC.
b. How much is the area of triangle ∆ ABD? Write reason

Answer:
a. Area of ∆ ABC = \(\frac{1}{2}\) x 5 x 4 = 10 sq.cm
b. Area of ∆ ABD = Area of ∆ ABC
Triangles with the same base and the third vertex on a line parallel to the base, have the same area.
Area of ∆ ABD = 10 sq.cm

Question 10.
Draw triangle ABC with AB = 8cm, BC = 6cm and B = 30°. Draw a right angled triangle of the same area. Measure its perpendicular sides. Calculate the area of the triangle.
Answer:
Draw ∆ ABC with the given measures.

Draw a line through C, parallel to AB.
Mark D on this parallel line such that ∠ BAD = 90°.
∆ ADB is the required triangle.
Measure the required lengths and then find the area.
AB = 8 cm
AD = 2.9 cm
Area of ∆ ADB = \(\frac{1}{2}\) x 8 x 2.9 = 11.6 sq.cm

Question 11.
Draw a quadrilateral with given measures. Draw a triangle of the same area as that of the quadrilateral.

Answer:
‘Draw a line through C parallel to DB. Which cuts AB at E.

∆ AED has the same area as that of quadrilateral ABCD.

Question 12.
If AB = 8cm, AC = 6cm, BC = 9cm,
AD is the bisector of A , then
a. BD : DC = ………….
b. What is the ratio of areas of triangle ABD and triangle ACD?
c. Draw a line of length 9 cm and divide it in the ratio 3 : 4.

Answer:
a. BD : DC = 8 : 6 = 4 : 3
b. Ratio of the areas of ∆ ABD and ∆ ACD = 8 : 6 = 4 : 3
c. AB = 9 cm
AC = 3 cm
CD = 4 cm
Draw CE through C and parallel to DB.

AC : CD = 3 : 4
AE : EB = 3 : 4

You can Download Pairs of Equations Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 3 Pairs of Equations

Kerala Syllabus 9th Standard Maths Pairs of Equations Text Book Questions and Answers

Textbook Page No. 36

Do each problem below either in your head, or using an equation with one letter, or two equations with two letters:

Pairs Of Equations Class 9 Questions And Answers Kerala Syllabus Question 1.
In a rectangle of perimeter one metre, one side is five centimetres longer than the other. What are the lengths of the sides?
Answer:
Shortest side = x
Longest side = x + 5
Perimeter = 1 m = 100 cm
2(x + x + 5) = 100
2x + 5 = 50; 2x = 45;
x = 22.5
∴ Shortest side = 22.5
Longest side = 22.5 + 5 = 27.5

Pairs Of Equations Questions And Answers Kerala Syllabus Question 2.
A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?
Answer:
Number of boys = x
Number of girls = x + 4
2(x – 8) = x + 4
2x – 16 = x + 4
2x – x = 4 +16; x = 20
∴ Number of boys = 20
Number of girls = 24

Pairs Of Equations Problems Kerala Syllabus Question 3.
A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?
Answer:
If one part is x then
the remaining part is 10000 – x
\(x\times \frac { 8 }{ 100 } +\left( 10000-x \right) \times \frac { 9 }{ 100 } =100\)
8x + 90000 – 9x = 87500
90000 – 87500 = x
2500 = x
one part = 2500 and
remaining part = 7500

Kerala Syllabus 9th Standard Maths Chapter 3  Question 4.
A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?
Answer:
Total length = 3½ m
Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.
∴ Length of one side
\( 3\frac { 1 }{ 2 } \div 7=\frac { 7 }{ 2 } \div 7=\frac { 1 }{ 2 } \)m
Length of the rod for the square
\(= 4\times \frac { 1 }{ 2 } \) = 2m
Length of the rod for the equilateral triangle = \(3\times \frac { 1 }{ 2 } \) = \(1\frac { 1 }{ 2 } \)m

Class 9 Maths Chapter 3 Kerala Syllabus Question 5.
The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + \(\frac { 1 }{ 2 } \) at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?
Answer:
If t = 2
ut + \(\frac { 1 }{ 2 } \)at²= 10
2u + 2a= 10
u + a = 5 — (1)
If t = 4
4u + 8a = 28
u + 2a = 7 — (2)
from (1) and (2)
a = 2
∴ u = 3

Textbook Page No. 40

Pairs Of Equations Class 9 Questions And Answers Pdf  Question 1.
Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?
Answer:
Cost of 200 page note book = x
Cost of 100 page note book = y
7x + 5y= 107 …………(1)
5x + 7y = 97 …………(2)
(1) × 5 \(\Rightarrow \) 35x + 25y = 535 …………(3)
(2) × 7 \(\Rightarrow \) 35x + 49y = 679 …………(4)
(4) – (3) \(\Rightarrow \) 24y = 144
y = \(\frac{144}{24} \) = 6
Substitute y = 6 in equation (1)
7x + 30 = 107; 7x = 77
x = \(\frac{77}{7} \) = 11
Price of the 200 pages notebook = Rs. 11
Price of the 100 pages notebook = Rs. 6

Pairs Of Equations Class 9 Extra Questions And Answers Question 2.
Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?
Answer:
Let the first number = x and
the second number = y
4x + 3y = 43 …………(1)
3x – 2y = 11 …………(2)
(1) × 3 \(\Rightarrow \) 12x + 9y= 129 …………(3)
(2) × 4 \(\Rightarrow \) 12x – 8y = 44 …………(4)
(3) -(4) \(\Rightarrow \) 17y = 85; y = \(\frac{85}{17} \) = 5
Substitute y = 5 in equation (1)
4x + 3y = 43
4x + 15 = 43
4x = 43 – 15 = 28
∴ x = \(\frac{28}{4} \) = 7, y = 5
First number = 7
Second number = 5

9th Standard Maths Chapter 3 Kerala Syllabus Question 3.
The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
Answer:
If the numbers are x and y
x + y = 11 …………(1)
10x + y + 27 = 10y + x
9x – 9y = -27
X – y = -3 …………(2)
(1) + (2) 2x = 8; x = 4
x + y = 11
4 + y = 11
y = 7
∴ Required number is 47.

Kerala Syllabus 9th Standard Maths Notes Question 4.
Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?
Answer:
Ramu’s present age = x
Rahim’s present age = y
4 years back,
Ramu’s age = x – 4
Rahim’s age = y – 4
3(x – 4) = y – 4
3x – 12 = y – 4
3x – y = 8 ……….(1)
After 2 years,
Ramu’s age = x + 2
Rahim’s age = y + 2
2(x + 2) = y + 2
2x + 4 = y + 2
2x – y = -2 ……….(2)
(1) – (2) \(\Rightarrow \) x = 10
3x – y = 8; 30 – y = 8; y = 22
x = 10, y = 22
Ramu’s present age = 10
Rahim’s present age = 22

Pair Of Equations Class 9 Kerala Syllabus Question 5.
If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?
Answer:
length = x; breadth = y
(x + 5)(y – 3) = xy – 5
xy – 3x + 5y – 15 = xy – 5
– 3x + 5y = + 10
3x – 5y = -10 ………..(1)
(x + 3)(y + 2) = xy + 50
xy + 2x + 3y + 6 = xy + 50
2x + 3y = 44 ………..(2)
(2) × 1 \(\Rightarrow \) 6x-10y = -20 ……….(3)
(3) × 2 \(\Rightarrow \) 6x + 9y = 132 …………(4)
(3)- (4) \(\Rightarrow \) -19y = -152
y = \(\frac{-152}{-19} \) = 8, 2x + 3y = 44
2x + 24 = 44; 2x = 20; x = 10
∴ x = 10, y = 8
Length of the rectangle = 10 m
Breadth of the rectangle = 8m

Textbook Page No. 42

Hsslive Guru Maths 9th Kerala Syllabus Question 1.
A 10 metre long rope is to be cut into two pieces and a square is to be made using each. The difference in the areas enclosed must be 1\(\frac{1}{4} \) square metres. How should it be cut?
Answer:
Length of one piece = x m
Length of other piece = (10 – x) m

∴ Rope is divided into 6 m and 4 m.

9th Maths Notes Kerala Syllabus Question 2.
The length of a rectangle is 1 metre more than its breadth. Its area is 3\(\frac{3}{4} \) square metres. What are its length and breadth?
Answer:
Length = x
Breadth = y
x = y + 1; x – y = 1
\(x y = 3 \frac{3}{4} =\frac{15}{4}\)
(x + y)² = (x – y)² + 4xy
1² + 4 × \(\frac{15}{4}\) = 1 + 15 = 16
x – y = 1; x + y = 4
2x = 5; x = 5/2 = 2.5
y=1.5
∴ Length = 2.5 m
Breadth = 1.5 m

Hsslive Maths Class 9 Kerala Syllabus Question 3.
The hypotenuse of a right triangle is 6\(\frac{1}{2} \) centimetres and its area is 7\(\frac{1}{2} \) square centimetres. Calculate the lengths of its perpendicular sides.
Answer:
The perpendicular sides are x and y. Given that,

From (3) and (4)
2x = 24/2 = 12
∴ x = 6
6 – y = 7/2
∴ y = 5/2 = 2.5
∴ Perpendicular sides = 6 and 2.5

Kerala Syllabus 9th Standard Maths Pairs of Equations Exam Oriented Text Book Questions and Answers

Kerala Syllabus 9th Standard Notes Maths Question 1.
There are some oranges in a bag. When 10 oranges more added in the bag; the numbers become 3 times of the oranges initially had. Then how many oranges were there in the bag initially.
Let the number of oranges initially taken = x
X + …….. = 3X
3X – X = ……..
2X = ……..
X = ……… /2 = ……..
Answer:
x + 10 = 3x; 3x – x = 10 ; 2x = 10;
x = \(\frac{10}{2} \) = 5

Kerala Syllabus 9th Standard Maths Notes Pdf Question 2.
A box contains some white balls and some black balls. The number of black balls is 8 more than the number of white balls. The total number of balls is 4 times the number of white balls. Find out the number of white balls and the number of black balls.
Number of white balls = x
Number of black balls = ……… + 8
Total number of balls = ……. ‘ x;
i. e. (x) + (x + 8) = ………. x;
2x + 8 = ……. x;
8 = …… x – 2x
= …… x; x = 8/ …….
white balls = ……….
black balls = ……… + 8 = ……..
Answer:
Number of white balls = x
Number of black balls = x + 8
Total number of balls = 4 ‘ x;
(x) + (x + 8) = 4x ; 2x + 8 = 4x;
8 = 4x – 2x = 2x ; x = \(\frac{8}{2} \) = 4
white balls = 4
black balls = 4 + 8 = 12

Pairs Of Equations Questions Kerala Syllabus Question 3.
The sum of two numbers is 36 and the difference is 8. Find the numbers.
Let x, y be the numbers
x + y = 36
x – y = 8
(x + y) + (x – y) = ……. + ……
2x = …..
x = …… /2 = ……..
x – y = 8
……. – y = 8
…….. – 8 = y
Answer:
x + y = 36; x – y = 8
(x + y) + (x – y) = 36 + 8
2x = 44; x = \(\frac{44}{2} \) = 22
x – y = 8; 22 – y = 8;
22 – 8 = y; y = 14
numbers 22, 14

Kerala Syllabus 9th Std Maths Notes  Question 4.
The cost of 2 pencils and 5 pens is Rs 17, two pencils and 3 pens of the same rate is Rs 11. Find out the prices of a pencil and a pen.
Let the price of pencil = x;
Price of pen = y;
∴ 2x + 5y = …….. 2x +….. y = …..
2x + …… y = 11
(2x + 5y) (-2x + ….. y) = -11
…… y = …….
y = \(\frac{…..}{……} \)
2x + 5x ….. = 17;
2x = 17 – ……..; x = ……. /2; = ……..
Answer:
2x + 5y = 17; 2x + 3y = 11;
(2x + 5y) – (2x + 3y) = 17 – 11; 2y = 6
y = \(\frac{6}{2} \) = 3; 2x + 5 × 3 = 17;
2x = 17 – 15;
x = \(\frac{2}{2} \) = 1
Price of pencil = Rs 1
Price of pen = Rs 3

Question 5.
Twice of a number added with thrice of another number gives 23. Four times the first number and 5 times the second number when added gives 41. Find out the numbers.
First number = x
Second number = y
∴ 2x + 3y = ……
4x + 5y = ……
2(2x + 3y) = 2 …….
4x + 6y = …….
(4x + 6y) – (4x + 5y) = ( …… ) – ( ….. )
y = …….; 2x + 3x ……. = …….
2x = ( …… ) – ( ……. ); x = ………./2
Answer:
2x + 3 y = 23
4x + 5y = 41
2(2x + 3y) = 2 × 23; 4x + 6y = 46
(4x + 6y) – (4x + 5y) = 46 – 41;
y = 5
2x + 3 × 5 = 23
2x = 23 – 15;
x = \(\frac{8}{2} \) = 4
Price of pencil = Rs 4
Price of pen = Rs 5

Question 6.
Rama spends Rs 97 to buy 4 two hundred page note books and 5 hundred page note books. Geetha spends Rs 101 to buy 5 two hundred page note book and 4 one hundred page note books. What is the prices of two types of note books?
Let the cost of two hundred page note books = x
The cost of one hundred page note books = y
(1) 4x + 5y = 97
(1) × 5 \(\Rightarrow \) 20x + ……..y = ……..
(2) 5x + 4y = 101
(2) × 4 \(\Rightarrow \) 20x + …….y = …….
……y = ( ….. ) – ( ……. );
y = …….. /……..
4x + 5x( ……. ) = 97
4x = 97 – ( ……. )
x = ……../4
Answer:
(1) 4x + 5y = 97
(2) 5x + 4y = 101
(1) × 5 → 20x + 25y = 485
(2) × 4 → 20x + 16y = 404
9y = 485 – 404
y = \(\frac{81}{9} \) = 9
4x + 5 × 9 = 97
4x = 97 – 45 = 52
Cost of two hundred page note book = Rs 13
Cost of one hundred page note book = Rs 9

Question 7.
6 years back the age of Muneer was 3 times the age of Mujeeb. After 4 years the age of Muneer becomes twice the age of Mujeeb. Find the age of two of them now.
Answer:
Age of Mujeeb 6 years back = x
Age of Muneer 6 years back = 3x
After 4 years
3x + 4 = 2(x + 4)
3x + 4 = 2x + 8
3x – 2x = 8 – 4;
x = 4
Age of Mujeeb 6 years back = 4 + 6 = 10
Age of Muneer 6 years back = 3(4 + 6) = 18 years.

Question 8.
The cost of 4 chairs and 5 tables is Rs 6600 and the cost of 5 chairs and 3 tables is Rs 5000 at the same prices. What are the prices of a table and a chair?
Answer:
Cost of a chair = Rs a
Cost of a table = Rs b
4a + 5b = 6600 ¾ …………(1)
5a + 3b = 5000 ¾ …………(2)
(1) × 5 → 20a + 25b = 33000
(2) × 4 → 20a + 12b = 20000
(20a + 25b) – (20a + 12b) = 33000 – 20000
(1) \(\Rightarrow \) 13b = 13000
b = \(\frac{13000}{13} \) = Rs 1000
4a + 5b = 6600;
4a + 5 × 1000 = 6600
4a = 6600 – 5000 = 1600
a = \(\frac{1600}{4} \) = Rs 400
Cost of a table = Rs 1000
Cost of a chair = Rs 400