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Shinjini

· started a discussion

· 1 Months ago

No shortcut for this ??? (Question: 119796.0 )

This seems quite tricky.

Question:

Three circles of radius √2 + 1, √2 + 1 and 1 unit, touch each other externally, then find the perimeter of the surrounded part by three circles.

Options:
A) \(\cfrac{π}{2}\) (2√2 +2)       
B) \(\cfrac{π}{2}\) (√2 +2)          
C)  π (√2 +2) 
D) \(\cfrac{π}{3}\) (2√2 +2)       
E) None of these 
Solution:

AB = 2(√2 + 1) and AC = BC

= 2 + √2

∴ AB2 = 4(√2 + 1)2

And AC2 = BC2 = (2 + √2)2

= 2 (√2 + 1)2 

So, it is clear

AB2 = AC2 + BC2

i.e. ABC is an isosceles right angled triangle.

∴ ∠ACB = 90° and ∠CAB

= ∠ABC = 45°

∴ required perimeter

= 2 π (√2+ 1) \(\cfrac{45°}{360° }\) + 2 π (√2+ 1) \(\cfrac{45°}{360° }\) +2 π × 1 × \(\cfrac{90°}{360° }\)

= 2 × 2 π (√2+ 1) \(\cfrac{1}{8}\) + 2 π(1/8)

= \(\cfrac{2π}{4}\) (√2 + 1 + 1)

= \(\cfrac{π}{2}\) (√2 + 2)

Knowledge Expert

· commented

· 1 Months ago

it looks like tricky.....but its not.......detailed solution has been given......please see by clicking report....
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