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Question
Two circles touch each other externally. The sum of their areas is 490 π cm2. Their centres are separated by 28 cm. The difference of their perimeters is ______.
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Solution
Two circles touch each other externally. The sum of their areas is 490 π cm2. Their centres are separated by 28 cm. The difference of their perimeters is 28π cm.
Explanation:
Let the radii be r1, r2.
Since the circles touch externally, r1 + r2 = 28.
Given π(r12 + r22) = 490π, so r12 + r22 = 490.
Then (r1 + r2)2 = r12 + r22 + 2r1r2
⇒ 282 = 490 + 2r1r2
⇒ r1r2 = 147
Now (r1 – r2)2 = r12 + r22 – 2r1r2
= 490 – 294
= 196
⇒ r1 – r2 = 14
Difference of perimeters = 2π(r1 – r2)
= 2π·14
= 28π cm
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