Practice Set 3.2 Circle Class 10th Mathematics Part 2 MHB Solution

Practice Set 3.2

1. Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the…
2. Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance…
3. If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each…
4. In fig 3.27, the circles with centres P and Q touch each other at R. A line passing…
5. In fig 3.28 the circles with centres A and B touch each other at E. Line is a common…

Practice Set 3.2

Question 1.

Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.

Answer:

Given: Two circles are touching each other internally.

∵The distance between the centres of the circles touching internally is equal to the difference of their radii.

⇒distance between their centres = 4.8 cm – 3.5 cm = 1.3 cm

Question 2.

Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.

Answer:

Given: Two circles are touching each other externally

We know that if the circles touch each other externally, distance between their centres is equal to the sum of their radii.

⇒distance between their centres = 5.5 cm + 4.2 cm = 9.7 cm

Question 3.

If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other –

(i) externally

(ii) internally.

Answer:

(i) 

Steps of construction:

1. Draw a circle with radius 4cm and centre A.

2. Draw another circle with radius 2.8 cm and centre B such that they touch each other externally.

(ii)

Steps of construction:

1. Draw a circle with radius 4cm and centre A.

2. Draw another circle with radius 2.8 cm and centre B such that they touch each other internally.

Question 4.

In fig 3.27, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively. Prove that -

(1) seg AP || seg BQ,

(2) ∆ APR ~ ∆ RQB, and

(3) Find ∠ RQB if ∠ PAR = 35°

Answer:

(1)In ΔAPR,

AP = RP {Radius of the circle with centre P}

∠PAR = ∠PRA … (1)

In ΔRQB,

RQ = QB {Radius of the circle with centre Q}

∠QRB = ∠QBR …. (2)

⇒ ∠PRA = ∠QRB {Vertically Opposite Angle} ….(3)

⇒ ∠PAR = ∠QBR {From (1), (2) and (3)}

⇒ Alternate interior angles are equal.

⇒ AP || BQ

Hence, proved.

(2) In ∆ APR and ∆ RQB,

∠PAR = ∠QBR and ∠PRA = ∠QRB {From (1) and (2)}

⇒ ∆ APR ~ ∆ RQB {AA}

Hence, proved.

(4) Given: ∠ PAR = 35°

⇒ ∠QBR = 35° = ∠QRB {Proved previously}

In ∆ RQB,

⇒ ∠ RQB + ∠ QRB + ∠QBR = 180° {Angle sum property of the triangle}

⇒∠ RQB + 35° + 35° = 180°

⇒∠ RQB = 180°- 70° = 110°

Question 5.

In fig 3.28 the circles with centres A and B touch each other at E. Line is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii Fig. 3.28 of the circles are 4 cm, 6 cm.

Answer:

Given that two circles with centre A and B touch each other externally. We know that if the circles touch each other externally, distance between their centres is equal to the sum of their radii.

⇒AB = (4 + 6) cm = 10 cm

In ∆ABC right-angles at A,

BC2 = CA2 + AB2 {Using Pythagoras theorem}

⇒BC2 = 42 + 102

⇒BC2 = 16 + 100

⇒ BC = √116 cm

In ∆DBC,

∠ BDC = 90° because D is the point of contact of tangent CD to circle centred B

BC2 = CD2 + DB2 {Using Pythagoras theorem}

⇒CD2 = 116 - 62

⇒CD2 = 116 - 36

⇒ CD = √80 cm = 4√5