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

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!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter