AB and CD are two parallel chords of a circle such that AB = 10 cm, CD= 4√30 cm. If the distance between AB and CD is 5 cm. Find the diameter of the circle.
Answers
Answer:
r
2
=x
2
+(5)
2
also, r
2
=(17−x)
2
+(12)
2
So, x
2
+(5)
2
=(17−x)
2
+(12)
2
x
2
+25=(17−x)
2
+144
⇒x=12
∴r
2
=144+25=169
⇒r=13cm
Answer:
The diameter is 26cm.
Step-by-step explanation:
Given:
AB || CD
AB = 10cm
CD = 4√30cm
Distance between AB and CD = 5cm
To find the diameter
Construction:
Draw OX perpendicular to AB and CD
Join the radii OA and OC
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
CM = 1/2 CD = 2√30cm
AN = 1/2 AB = 5cm
let OM = x
MN = 5cm (given)
OC = OA = r (construction)
ON = OM + MN
ON = (x + 5) cm
Using the Pythagoras Theorem:
OC^2 = OM^2 + CM^2
= x^2 + (2√30)^2
r^2 = x^2 + 120
OA^2 = ON^2 + AN^2
= (x + 5)^2 + 5^2
r^2 = x^2 + 10x + 50
x^2 + 120 = x^2 + 10x + 50
120 = 10x + 50
10x = 70
x = 7
r^2 = x^2 + 120 = 7^2 + 120 = 169
r = 13cm
The diameter is 26cm.
