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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

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.

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

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

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

Which among Mex, RCH2X, R2CHX , and R3CX is most reactive towards sn2 reaction.

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°

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.

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

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.

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.

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

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 \(\sqrt 6 + \sqrt 6 = 2\sqrt 6\)

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

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

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

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