Mathematics Questions and Answers – Perpendicular from the Centre to a Chord

«
»

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

Answer: b
Explanation: Since radius of circle = 5cm, OA = OC = 5cm
In ΔBOC, ∠OBC = 90° and OB = 3cm
By Pythagoras theorem, OC2 = OB2 + BC2
⇒ 52 = 32 + BC2
⇒ BC2 = 52 – 32
⇒ BC2 = 25 – 9 = 16
⇒ BC = 4cm
As perpendicular from centre to a chord bisects the chord, AB = BC
⇒ AC = 2BC = 8cm.
advertisement

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.
advertisement
advertisement

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

Answer: c
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

Answer: d
Explanation: Since O1B and O2B are perpendicular from centre to the chord, O1O2CB is a rectangle.
⇒ O1O2 = 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)
advertisement

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

Answer: b
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°.
advertisement

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

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.