Geometry Class 9th Mathematics Term 3 Tamilnadu Board Solution

Class 9^th Mathematics Term 3 Tamilnadu Board Solution

Exercise 3.1

1. The radius of a circle is 15 cm and the length of one of its chord is 18 cm.…
2. The radius of a circles 17 cm and the length of one of its chord is 16 cm. Find…
3. A chord of length 20 cm is drawn at a distance of 24 cm from the centre of a…
4. A chord is 8 cm away from the centre of a circle of radius 17 cm. Find the…
5. Find the length of a chord which is at a distance of 15 cm from the centre of a…
6. In the figure at right, AB and CD are two parallel chords of a circle with…
7. AB and CD are two parallel chords of a circle which are on either sides of the…
8. In the figure at right, AB and CD are two parallel chords of a circle with…
9. Find the value of x in the following figures.
10. Find the value of x in the following figures.
11. Find the value of x in the following figures.
12. Find the value of x in the following figures.
13. Find the value of x in the following figures.
14. Find the value of x in the following figures.
15. In the figure at right, AB and CD are straight lines through the centre O of a…
16. In the figure at left, PQ is a diameter of a circle with centre O. If ∠PQR =…
17. In the figure at right, ABCD is a cyclic quadrilateral whose diagonals…
18. In the figure at left ,ABCD is a cyclic quadrilateral in which AB || DC. If…
19. In the figure at right, ABCD is a cyclic quadrilateral in which ∠BCD = 100°and…
20. In the figure at left, O is the centre of the circle, ∠AOC = 100° and…

Exercise 3.2

1. O is the centre of the circle. AB is the chord and D is mid-point of AB. If the…
2. ABCD is a cyclic quadrilateral. Given that ∠ADB + ∠DAB = 120°and ∠ABC + ∠BDA =…
3. In the given figure, AB is one of the diameters of the circle and OC is…
4. In the given figure, AB is a diameter of the circle and points C and D are on…
5. Angle in a semi circle isA. obtuse angle B. right angle C. an acute angle D.…
6. Angle in a minor segment isA. an acute angle B. an obtuse angle C. a right angle…
7. In a cyclic quadrilateral ABCD, ∠A = 5x, ∠C = 4x the value of x isA. 12° B. 20°…
8. Angle in a major segment isA. an acute angle B. an obtuse angle C. a right…
9. If one angle of a cyclic quadrilateral is 70°, then the angle opposite to it…

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Exercise 3.1

Question 1.

The radius of a circle is 15 cm and the length of one of its chord is 18 cm. Find the distance of the chord from the centre.

Answer:

The figure is attached above

A is the centre of the circle.

AB is the radius = 15 cm

CD is the chord = 18 cm.

We need to find the distance of the chord from the centre i.e. AE

In this circle, we draw the perpendicular

We know that perpendicular drawn from the centre to the chord, will bisect the chord, such that CE = ED =  = 9 cm

Now,

In ΔAEC,

Applying Pythagoras theorem,

AC² = AE² + EC²

⇒AE² = AC² – EC²

⇒ AE² = (15cm)²–(9cm)²

⇒ AE² = 225 – 81 = 144

⇒ AE = √144

⇒ AE = 12 cm

∴The chord is 12cm away from the center of the circle.

Question 2.

The radius of a circles 17 cm and the length of one of its chord is 16 cm. Find the distance of the chord from the centre.

Answer:

In the given figure:

Radius AB = 17cm

Chord CD = 17cm

Now we draw a perpendicular bisector of CD from A which is AB

Hence CE = DE = 8cm

We have to find the distance of chord from centre i.e. AE

In the triangle ACE

AC² = AE² + EC²

Therefore AE² = 17²– 8²

AE² = 225

AE = 15cm

Hence the distance of chord from the centre is 15cm

Question 3.

A chord of length 20 cm is drawn at a distance of 24 cm from the centre of a circle. Find the radius of the circle.

Answer:

The figure is given below:

CD = 20cm = length of chord

AE = 24cm = distance of chord from centre

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

From here we get,

r² = 100 + 576 = 676

r = 26cm

Hence radius is 26cm.

Question 4.

A chord is 8 cm away from the centre of a circle of radius 17 cm. Find the length of the chord.

Answer:

The figure is given below:

AE = 8cm

AB = AC = 17cm

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

d = 8cm

r = 17cm

Hence the length of the chord is 30cm.

Question 5.

Find the length of a chord which is at a distance of 15 cm from the centre of a circle of radius 25 cm.

Answer:

The figure is given below:

AE = 15cm

AB = AC = 25cm

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

Here d = 15cm and r = 25cm

l = 40cm

Hence the length of the chord is 40cm.

Question 6.

In the figure at right, AB and CD are two parallel chords of a circle with centre O and radius 5 cm such that AB = 6 cm and CD = 8 cm. If OP ⊥ AB and CD ⊥ OQ determine the length of PQ.

Answer:

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

For Triangle COQ

OC = 5cm ,

CQ = half of chord length = 4cm

From Pythagoras theorem

OQ = 3cm

For triangle POA

OA = 5cm, AP = half of chord length = 3cm

From Pythagoras theorem

OP = 4cm

From figure PQ = OP – OQ

Therefore PQ = 1cm

Question 7.

AB and CD are two parallel chords of a circle which are on either sides of the centre. Such that AB = 10 cm and CD = 24 cm. Find the radius if the distance between AB and CD is 17 cm.

Answer:

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

In the figure, AB = 10cm = l

 .....(1)

And CD = 24cm

 .....(2)

As Distance between the two chords = PQ = 17cm

We have OP + OQ = 17....(3)

Equating equations 1 and 2 we get,

From identity we get

(OP + OQ)(OP–OQ) = 119

Using equation 3 we get

OP – OQ = 7 ....(4)

Solving equation 3 and 4 we get

OP = 12cm

OQ = 5cm

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

Using this in Triangle APO we get

Hence the radius of the triangle is 14cm

Question 8.

In the figure at right, AB and CD are two parallel chords of a circle with centre O and radius 5 cm. Such that AB = 8 cm and CD = 6 cm. If OP ⊥ AB and OQ ⊥ CD determine the length PQ.

Answer:

From Theorem: A circle with radius r and a chord of length l which is at distance d from the centre, follows the equation,

In triangle APO

OP = 3cm.

In triangle COQ

OQ = 4cm

From figure PQ = OP + OQ = 7cm.

Hence PQ = 7cm.

Question 9.

Find the value of x in the following figures.

Answer:

Triangle AOC is a right angled triangle at O and sides AO = OC.

Hence ∠OAC = ∠OCA = y

As the sum of all angles = 180°

90 + 2y = 180

y = 45°

∠OAC = 45°

Similarly in triangle AOB

Sides AO = OB and hence

∠OBA = ∠OAB = a

As sum of all angles of a triangle = 180°

120 + 2a = 180

a = 30°

In the figure,

∠BAC = ∠CAO + ∠BAO = 45 + 30 = 75°

Question 10.

Find the value of x in the following figures.

Answer:

According to theorem, angle subtended by the diameter on the circle is 90°.

∠ACB = 90° ; ∠ABC = 35°

As the sum of all angles = 180°

∠CAB = 180– 35 = 145°

Question 11.

Find the value of x in the following figures.

Answer:

According to theorem that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference.

∠AOB = 2 × 25 = 50°

∠AOC = 2 × 30 = 60°

At centre O

∠AOB + ∠AOC + x° = 360°

x = 360–60–50 = 250°

Question 12.

Find the value of x in the following figures.

Answer:

According to theorem that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference.

Hence Angle by chord AC on the circumference = (angle subtended at centre)

∠ABC =  = 65°

Question 13.

Find the value of x in the following figures.

Answer:

As the sum of all angles of a triangle is 180°

Angles of triangle DBC are 90°, 50° and x°

Question 14.

Find the value of x in the following figures.

Answer:

According to theorem that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference.

Here angle subtended by a chord at circumference = 48°

Hence angle subtended at centre = 2x 48°

Therefore x = 96°

Question 15.

In the figure at right, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 98° and ∠CDE = 35°

find (i) ∠DCE (ii) ∠ABC

Answer:

According to theorem the angle subtended by a diameter on the circumference of a circle is 90°

Hence ∠CED = 90°

As the sum of all angles of a triangle CDE = 180°

∠DCE = 180 – 90 – 35 = 55°

As the sum of all angles of triangle OCB = 180°

As AB is a straight line ∠COB = 180–98 = 82°

Hence ∠ABC = 180 – 82 – 55 = 43°

Question 16.

In the figure at left, PQ is a diameter of a circle with centre O. If ∠PQR = 55°, ∠SPR = 25°and ∠PQM = 50°.

Find (i) ∠QPR, (ii) ∠QPM and (iii) ∠PRS.

Answer:

According to theorem the angle subtended by a diameter on the circumference of a circle is 90°

∠PRQ = 90°

As the sum of all angles of a triangle is 180°

In triangle PRQ,

∠QPR = 180 – 55 – 90 = 35°

∠PMQ = 90°

As sum of all angles of triangle = 180°

In triangle PQM,

∠QPM = 180– 90– 50 = 40°

Triangle formed by POS is an isosceles triangle as two of sides are radius OP and OS.

Hence ∠OPS = ∠OSP = 35 + 25 = 60°

As sum of all angles of a triangle is 180°.

∠POS = 180 –60–60 = 60°

According to theorem that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference.

Angle subtended by chord SP on centre = 60°

Hence angle subtended by chord SP on circumference is 

Hence ∠PRS = 30°

Question 17.

In the figure at right, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠DBC = 30°and ∠BAC = 50°.

Find (i) ∠BCD (ii) ∠CAD

Answer:

As ABCD is a cyclic quadrilaterals, sum of all angles = 360°

As the angle subtended by chord BC on circumference = 50°

Therefore BDC = 50°

Sum of all angles of triangle BCD = 180°

Hence BCD = 180– 30–50 = 100°

As the angle subtended by chord DC on circumference = 30°

Therefore CAD = 50°

Question 18.

In the figure at left ,ABCD is a cyclic quadrilateral in which AB || DC. If ∠BAD = 100°

find (i) ∠BCD (ii) ∠ADC (iii) ∠ABC.

Answer:

As ABCD is a cyclic quadrilaterals, sum of all angles = 360°

As AB is parallel to CD

∠ADC = 180 – ∠DAB = 80° as they are interior angles on the same side.

As angle subtended by chord BD on circumference = 100°

Therefore ∠BCD = 100°

As ∠BCD and ∠ABC are interior angles

∠ABC = 180 – ∠BCD = 80°

Question 19.

In the figure at right, ABCD is a cyclic quadrilateral in which ∠BCD = 100°and ∠ABD = 50°. Find ∠ADB

Answer:

As ABCD is a cyclic quadrilaterals, sum of all angles = 360°

As angle subtended by chord BD on circumference = 100°

Therefore ∠DAB = 100°

As sum of all angles of triangle ADB is 180°

Hence ∠ADB = 180–50–100 = 30°

Question 20.

In the figure at left, O is the centre of the circle, ∠AOC = 100° and side AB is produced to D.

Find (i) ∠CBD (ii) ∠ABC

Answer:

As the angle subtended by chord AC on center O is 100°.

Angle subtended by chord AC on major segment = 50°

As the sum of angle subtended by chord AC on major and minor segments is 180°

Hence ∠ABC = 180–50 = 130°

As line AD is straight

Hence ∠CBD = 180 – 130 = 50°

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Exercise 3.2

Question 1.

O is the centre of the circle. AB is the chord and D is mid-point of AB. If the length of CD is 2cm and the length of chord is12 cm, what is the radius of the circle


A. 10cm

B. 12cm

C. 15cm

D. 18cm

Answer:

In triangle AOD

AD = 6cm

AO = r

OD = OC–CD = r–2

Applying Pythagoras theorem we get

r = 10cm

Question 2.

ABCD is a cyclic quadrilateral. Given that ∠ADB + ∠DAB = 120°and ∠ABC + ∠BDA = 145°. Find the value of ∠CDB


A. 75°

B. 115°

C. 35°

D. 45°

Answer:

As ADB + DAB = 120°

In triangle ABD as sum of all angles = 180°

ABD = 180–120 = 60°

As ABCD is cyclic, opposite angles have sum of 180°

Let BAD = x°

Hence BCD = 180–x

Let BDC = a°

Let DBC = z°; ABC = 60 + z

Let ADB = y°

As sum of angles of triangle BDC = 180°

180–x + a + z = 180

a = x–z ...(1)

According to question,

x + y = 120 ..(2)

and

60 + z + y = 145...(3)

Subtracting (3) from (2)

We get

x + y–60–z–y = –25

x–z = 35

Equating from (1) we get

a = 35°

hence CDB = 35°

Question 3.

In the given figure, AB is one of the diameters of the circle and OC is perpendicular to it through the center O. If AC is 7√2 cm, then what is the area of the circle in cm²


A. 24.5

B. 49

C. 98

D. 154

Answer:

In triangle AOC AO = OC = radius of circle = r

Using Pythagoras theorem,

As area of a circle is given by  = 

Question 4.

In the given figure, AB is a diameter of the circle and points C and D are on the circumference such that ∠CAD = 30°and ∠CBA = 70° what is the measure of ACD?


A. 40°

B. 50°

C. 30°

D. 90°

Answer:

According to theorem the angle subtended by a diameter on the circumference of a circle is 90°

ACB = 90°

Angle subtended by chord AC on major segment = ABC = 70°

Angle subtended on minor segment = 180–70 = 110° = ADC

As sum of angles of triangle ADC = 180°

ACD = 180–30–110 = 40°

Question 5.

Angle in a semi circle is
A. obtuse angle

B. right angle

C. an acute angle

D. supplementary

Answer:

As a semicircle is formed by a diameter and the angle subtended on the circumference by the diameter is 90°.

Question 6.

Angle in a minor segment is
A. an acute angle

B. an obtuse angle

C. a right angle

D. a reflexive angle

Answer:

The angle formed by a chord in the major segment is acute whereas in the minor segment is obtuse and the sum of both angles is 180°.

Question 7.

In a cyclic quadrilateral ABCD, ∠A = 5x, ∠C = 4x the value of x is
A. 12°

B. 20°

C. 48°

D. 36°

Answer:

In a cyclic quadrilateral sum of opposite angles is 180°

Question 8.

Angle in a major segment is
A. an acute angle

B. an obtuse angle

C. a right triangle

D. a reflexive angle

Answer:

The angle formed by a chord in the major segment is acute whereas in the minor segment is obtuse and the sum of both angles is 180°.

Question 9.

If one angle of a cyclic quadrilateral is 70°, then the angle opposite to it is
A. 20°

B. 110°

C. 140°

D. 160°

Answer:

As the sum of opposite angles in a cyclic quadrilateral is 180°

Let opposite angle be x°