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
Answers
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:
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: