Normal tendency to assume that the side of the isosceles traingle = a, thus hypotenuse = a√2airan wrote:The perimeter of a certain isosceles right triangle is 16 +16 sqrt(2). What is the length of the hypotenuse of the triangle
1. 8
2. 16
3. 4 sqrt(2)
4. 8 sqrt(2)
5. 16 sqrt(2)
=> Perimeter = a+a+a√2 = 2a + a√2 = a(2 + √2)
=> a(2 + √2) = 16 + 16√2
=> a(2 + √2) = 16(1 + √2)
Most have stuck at this stage, and few committed mistakes.
The challenge with a(2 + √2) = 16(1 + √2) is that (2 + √2) and (1 + √2) are not the same and thus, cannot be cancelled.
However, we can make them equal.
We are at: a(2 + √2) = 16(1 + √2)
=> a[(√2)^2 + √2] = 16(1 + √2); we can write √2 = (√2)^2
=> a√2(√2 + 1) = 16(1 + √2); taking √2 common
=> a√2 = 16 = hypotenuse
OA: B
Hope this helps!
-Jay
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