This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Perpendicular from the Centre to a Chord”.

1. What is the value of AC if radius of circle is 5cm and OB = 3cm?

a) 5cm

b) 8cm

c) 4cm

d) 2cm

View Answer

Explanation: Since radius of circle = 5cm, OA = OC = 5cm

In ΔBOC, ∠OBC = 90° and OB = 3cm

By Pythagoras theorem, OC

^{2}= OB

^{2}+ BC

^{2}

⇒ 5

^{2}= 3

^{2}+ BC

^{2}

⇒ BC

^{2}= 5

^{2}– 3

^{2}

⇒ BC

^{2}= 25 – 9 = 16

⇒ BC = 4cm

As perpendicular from centre to a chord bisects the chord, AB = BC

⇒ AC = 2BC = 8cm.

2. Find the value of AB if radius of circle is 10cm and PQ = 12cm and RS = 16cm?

a) 14cm

b) 16cm

c) 4cm

d) 2cm

View Answer

Explanation:

Since radius of circle = 10cm, OS = OQ = 10cm

In ΔBOQ, ∠OBQ = 90°

and BQ = PQ/2 = 6cm (perpendicular from centre to a chord bisects the chord)

By Pythagoras theorem, OQ

^{2}= OB

^{2}+ BQ

^{2}

⇒ 10

^{2}= 6

^{2}+ OB

^{2}

⇒ OB

^{2}= 100 – 36 = 64

⇒ OB = 8cm

Similarly, In ΔAOS, ∠OAS = 90°

and AS = RS/2 = 8cm (perpendicular from centre to a chord bisects the chord)

By Pythagoras theorem, OS

^{2}= OA

^{2}+ AS

^{2}

⇒ 10

^{2}= 8

^{2}+ OA

^{2}

⇒ OA

^{2}= 100 – 64 = 36

⇒ OA = 6cm

Now, AB = OB – OA

⇒ AB = 8 – 6 = 2cm.

3. Find the radius of circle if AB = 7cm, RS = 6cm and PQ = 8cm.

a) 5cm

b) 8cm

c) 4cm

d) 20cm

View Answer

Explanation:

Let r be the radius of circle.

Since perpendicular from centre to a chord bisects the chord,

AR = RS/2 = 3cm and BP = PQ/2 = 4cm

In ΔBOP, ∠OBP = 90°

By Pythagoras theorem, OP

^{2}= OB

^{2}+ BP

^{2}

⇒ r

^{2}= x

^{2}+ (4)

^{2}

⇒ r

^{2}= x

^{2}+ 16 ——————- (i)

Similarly, In ΔAOR, ∠OAR = 90°

By Pythagoras theorem, OR

^{2}= OA

^{2}+ AR

^{2}

⇒ r

^{2}= (7 – x)

^{2}+ (3)

^{2}

⇒ r

^{2}= (7 – x)

^{2}+ 9 ———————— (ii)

From equation i and ii, x

^{2}+ 16 = (7 – x)

^{2}+ 9

⇒ x

^{2}+ 16 = 7

^{2}– 2 x 7 x x + x

^{2}+ 9

⇒ 2 x 7 x x = 49 – 16 + 9

⇒ x = 3

Substituting value of x in equation i, r

^{2}= 3

^{2}+ 16

⇒ r = 5cm.

4. What is the ratio of PQ/RS if OT is perpendicular to line l?

a) 0.5

b) 2

c) 1

d) 1.5

View Answer

Explanation: Since perpendicular from centre to a chord bisects the chord,

QT = TR —————–(i)

and PT = TS —————– (ii)

Subtracting equation i from ii,

PT – QT = TS – TR

⇒ PQ = RS.

5. What is the length of AD if distance between centres of the two circles is 9cm?

a) 9cm

b) 20cm

c) 15cm

d) 18cm

View Answer

Explanation: Since O

_{1}B and O

_{2}B are perpendicular from centre to the chord, O

_{1}O

_{2}CB is a rectangle.

⇒ O

_{1}O

_{2}= BC = 9cm ———————– (i)

Also, perpendicular from centre to a chord bisects the chord

AB = BE —————–(ii)

and CE = CD —————– (iii)

Now, AD = AB + BE + CE + CD

⇒ AD = 2BE + 2CE (from equation ii and iii)

⇒ AD = 2(BE + CE)

⇒ AD = 2(BC)

⇒ AD = 2 x 9 = 18cm. (from equation i)

6. Two circles of radius 13cm and 17cm intersect at A and B. What is the distance between the centres of the circles if AB = 24cm?

a) 15cm

b) 25cm

c) 22cm

d) 14cm

View Answer

Explanation:

Since line joining the centres of the circles bisects the common chord, AC = BC = 12cm.

Also, In ΔAOC, ∠OCA = 90°

By Pythagoras theorem, OA

^{2}= OC

^{2}+ AC

^{2}

⇒ 15

^{2}= OC

^{2}+ 12

^{2}

⇒ OC

^{2}= 225 – 144

⇒ OC = 9cm

Similarly, In ΔAO’C, ∠O’CA = 90°

By Pythagoras theorem, O’A

^{2}= O’C

^{2}+ AC

^{2}

⇒ 13

^{2}= O’C

^{2}+ 12

^{2}

⇒ OC

^{2}= 169 – 144

⇒ OC = 5cm

Now, OO’ = OC + O’C

⇒ OO’ = 9 + 5 = 14cm.

7. What is the value of ∠PSQ in the given figure?

a) 90°

b) 60°

c) 30°

d) 45°

View Answer

Explanation: In ΔPRQ and ΔPQS,

PR = PS (Radii of same circle)

QR = QS (Radii of same circle)

PQ = PQ (Common side)

⇒ ΔPRQ ≅ ΔPSQ (by SSS congruence criterion)

⇒ ∠PRQ = ∠PSQ (Corresponding parts of congruent triangles)

⇒ ∠PSQ = 60°.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 9**.

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