Question

a chord pq of a circle with centre o and radius 15 cm is bisected at m by a diameter ab if om = 9 cm dind the lenght of pq ap bp​

Answer 1

Answer:

AB bisects chord PQ at point C.

Therefore,

PR=QR=8cm

RB=4cm

Let, radius OA=OB=r

OB=OR+RB

=>r=OR+4

=>OR=r−4 (i)

Now,

OP

2

=OR

2

+PR

2

=>r

2

=(r−4)

2

+8

2

(using (i)

=>r

2

=r

2

−8r+16+64

=>8r=80

=>r=10cm

The radius of the circle is 10cm

Answer 2

Given, radius = 15 cm

OA = OB = OP = OQ = 15 cm

Also, OM = 9 cm

MB = OB – OM = 15 – 9 = 6 cm

AM = OA + OM =15 + 9 cm = 24 cm

In ∆OMP, By using Pythagoras Theorem,

OP2 = OM 2 + PM2

152 = 92 + PM2

PM2 = 255 – 81

PM = √144 = 12 cm

Also, In ∆OMQ

By using Pythagoras Theorem

OQ2 = OM2 + QM2

152 = OM2 + QM2

152 = 92 + QM2 (QM2 = 225 – 81)

QM = √144 = 12 cm

PQ = PM + QM

(As radius is bisected at M)

PQ = 12 + 12 cm = 24 cm

(ii) Now in ∆APM

AP2 = AM2 + OM2

AP2 =242 + 122

AP2 = 576 + 144

AP = √720 = 12 √5 cm

(iii) Now in ∆BMP

BP2 = BM2 + PM2

BP2 = 62 + 122

BP2 = 36 + 144

BP = √180 = 6 √5 cm

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