Can someone walk me through this problem. thanks.
OA = B
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- MAAJ
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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
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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- smackmartine
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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
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
- MAAJ
- Master | Next Rank: 500 Posts
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- Joined: Sun Jul 12, 2009 7:12 am
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My bad, there's a typo in my post. 2 - 1/2 = 3/2! I corrected my previous post.
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."