Class 9 Maths MCQ – Circles – Perpendicular from the Centre to a Chord

This set of Class 9 Maths Chapter 10 Multiple Choice Questions & Answers (MCQs) focuses on “Circles – 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

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

Answer: d
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, OQ² = OB² + BQ²
⇒ 10² = 6² + OB²
⇒ OB² = 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, OS² = OA² + AS²
⇒ 10² = 8² + OA²
⇒ OA² = 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

Answer: a
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, OP² = OB² + BP²
⇒ r² = x² + (4)²
⇒ r² = x² + 16  ——————- (i)
Similarly, In ΔAOR, ∠OAR = 90°
By Pythagoras theorem, OR² = OA² + AR²
⇒ r² = (7 – x)² + (3)²
⇒ r² = (7 – x)² + 9  ———————— (ii)
From equation i and ii, x² + 16 = (7 – x)² + 9
⇒ x² + 16 = 7² – 2 x 7 x x + x² + 9
⇒ 2 x 7 x x = 49 – 16 + 9
⇒ x = 3
Substituting value of x in equation i, r² = 3² + 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
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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
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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

Answer: d
Explanation:

Since line joining the centres of the circles bisects the common chord, AC = BC = 12cm.
Also, In ΔAOC, ∠OCA = 90°
By Pythagoras theorem, OA² = OC² + AC²
⇒ 15² = OC² + 12²
⇒ OC² = 225 – 144
⇒ OC = 9cm
Similarly, In ΔAO’C, ∠O’CA = 90°
By Pythagoras theorem, O’A² = O’C² + AC²
⇒ 13² = O’C² + 12²
⇒ OC² = 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

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

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