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jsackmann
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An isosceles right triangle has sides of x, x, and x(root2). So, the perimeter is 2x + x(root2).
To solve for x:
2x + x(root2) = 16 + 16(root2)
x(2+root2) = 16+16(root2)
x = (16+16(root2)) / (2+root2)
(yes, I know the math gets hairy here)
Multiply top and bottom by (2-root2):
= (32 + 32(root2) - 16(root2) - 32) / 4 - 2
= 16(root2) / 2
= 8(root2)
That's x, and the hypotenuse is x(root2)
= 8(root2)(root2) = 8(2) = 16
That's the textbook approach. It's probably faster to try the answer choices once you realize that 2x + x(root2) needs to equal 16 + 16(root2).
To solve for x:
2x + x(root2) = 16 + 16(root2)
x(2+root2) = 16+16(root2)
x = (16+16(root2)) / (2+root2)
(yes, I know the math gets hairy here)
Multiply top and bottom by (2-root2):
= (32 + 32(root2) - 16(root2) - 32) / 4 - 2
= 16(root2) / 2
= 8(root2)
That's x, and the hypotenuse is x(root2)
= 8(root2)(root2) = 8(2) = 16
That's the textbook approach. It's probably faster to try the answer choices once you realize that 2x + x(root2) needs to equal 16 + 16(root2).
