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## 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

Which among MeX, RCH2X , R2CHX and R3CX is most reactive towards SN2 reaction Haloalkanes and Haloarenes.

**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