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Two circles touch each other externally. The sum of their areas is 490 π cm^2. Their centres are separated by 28 cm. The difference of their perimeters is ______.

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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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Chapter 13: Areas Related to Circles - FILL IN THE BLANK TYPE QUESTIONS (FBQs) [Page 13.47]

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R.D. Sharma Mathematics [English] Class 10
Chapter 13 Areas Related to Circles
FILL IN THE BLANK TYPE QUESTIONS (FBQs) | Q 10. | Page 13.47
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