Question

In the given figure, O is the centre of a circle of radius r сm, OP and OQ are perpendiculars to AB

and CD respectively and PQ = 1cm. If AB || CD, AB = 6cm and CD = 8cm, determine r.
Plz answer it fast​

Answer 1

Step-by-step explanation:

in ΔAOP, by Pythagoras theorem

r

2

=3

2

+(1+x)

2

....(1)

In ΔCOQ,r

2

=4

2

+x

2

...(2)

eq

n

(1)−eq

n

(2)O=9−16+(1+x)

2

−x

2

⇒2x=6→x=3

By, equation (1) r

2

=9+16⇒r

2

=25⇒r=5cm

Answer 2

Answer:

Since the perpendicular drawn from the centre of the circle to a chord bisects the chord. Therefore, P and Q are mid-points of AB and CD respectively.

Consequently, AP = BP = 12AB = 3 cm

and CQ = QD = 12CD = 4 cm

In right-angled AQAP, we have

OA2 = OP2 + AP2

r2 = OP2 + 32

r2 = OP2 + 9

In right-angled ∆OCQ, we have

OC2 = OQ2 + CQ2

r2 = OQ2 + 42

p2 = OQ2 + 16 … (ii)

From (i) and (ii), we have

OP2 + 9 = OQ2 + 16

OP2 – OQ2 = 16 – 9

x2 – (x – 1)2 = 16 -9 [where OP = x and PQ = 1 cm given]

x2 – y2 – 1 + 2x = 7

2x = 7 + 1

x = 4

⇒ OP = 4 cm

From (i), we have

r2 = (4)2 + 9

r2 = 16 + 9 = 25

r = 5 cm

Step-by-step explanation: