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A semi-circular shaped window has diameter of 63 cm

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by sachin_yadav » Fri Apr 26, 2013 9:33 pm
Need help on this geometry concept question. At first it seems to be very easy, but i am missing something. Don't know what it is.

A semi-circular shaped window has diameter of 63 cm. Its perimeter equals:

(A) 126
(B) 162
(C) 198
(D) 251
(E) 221

Answer is B

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by Brent@GMATPrepNow » Fri Apr 26, 2013 9:42 pm
sachin_yadav wrote:Need help on this geometry concept question. At first it seems to be very easy, but i am missing something. Don't know what it is.

A semi-circular shaped window has diameter of 63 cm. Its perimeter equals:

(A) 126
(B) 162
(C) 198
(D) 251
(E) 221

Answer is B

Thanks
Sachin
The perimeter of a semi-circular shaped window involves adding the diameter of the semi-circle AND the circumference of the semi circle.

So, perimeter = diameter + (1/2)(pi)(diameter) [since it's half a circle]
= 63 + (1/2)(pi)(63)

IMPORTANT: At this point, before we perform tedious calculations, we should check the answer choices. Since they are nicely spread apart, we can estimate aggressively to get:

perimeter = 63 + about 100
= 163 (approx)
= B

Cheers,
Brent
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by Anju@Gurome » Fri Apr 26, 2013 9:47 pm
sachin_yadav wrote:A semi-circular shaped window has diameter of 63 cm. Its perimeter equals:
Image

Perimeter of the window = length of diameter + length of the circular arc = length of diameter + half of the perimeter of the circle = d + (Ï€d)/2, where d is the diameter of the circle

SO, perimeter = d + (Ï€d)/2
= 63 + π*63/2
= 63 + (22/7)*63/2 [As, π ≈ 22/7]
= 63 + 11*9
= 63 + 99
= 162

The correct answer is B.
Anju Agarwal
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Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by sachin_yadav » Fri Apr 26, 2013 9:57 pm
Brant and Anju, thank you for your replies.

Now i understand where i was making mistake. I wasn't adding the diameter with (1/2)(pi)(diameter)

Thank you once again :D

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