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Triangle Practices

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Triangle Practices

by tonebeeze » Mon May 16, 2011 4:10 pm
Can someone walk me through this problem. thanks.

OA = B
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by MAAJ » Mon May 16, 2011 6:24 pm
IMO answer is [spoiler](B)[/spoiler]

y² = 2 -> y = sqrt(2)

a² + b² = c²
(sqrt(2)/2)² + b² = sqrt(2)²
(2/4) + b² = 2
(1/2) + b² = 2
b² = 2 - (1/2)
b² = 3/2
b = sqrt(3)/sqrt(2)

Area = a * b * 1/2
(sqrt(2)/2) * sqrt(3)/sqrt(2) * 1/2
sqrt(3)/4
Last edited by MAAJ on Tue May 17, 2011 5:04 am, edited 2 times in total.
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by smackmartine » Mon May 16, 2011 6:41 pm
IMO B

Given y² = 2

The base would be sqrt [y^2 - (y/2)^2] = sqrt [y^2 - (y^2/4)]= sqrt [2 - (2/4)] = sqrt [2 - (1/2)] = sqrt [3/2]

Area = (1/2)*Base * Height
= (1/2)* [sqrt(3)/sqrt(2)]* y/2
= (1/2)* [sqrt(3)/sqrt(2)]* sqrt(2)/2

Cancelling sqrt(2) from both numerator and denominator

= sqrt(3)/4
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by MAAJ » Tue May 17, 2011 5:02 am
My bad, there's a typo in my post. 2 - 1/2 = 3/2! I corrected my previous post.
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