Question 1040514
The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimension of the triangle, correct to one decimal place.
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Area = b*h/2 = 24
Using 12 for the base, h = 4.
The altitude from the right angle = 4.
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Label the 2 parts of the base c & d, and the 2 sides e & f
c + d = 12
e^2 - c^2 = 16
f^2 - d^2 = 16
e^2 + f^2 = 12^2
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Sub for c: c = 12 - d
e^2 - d^2 + 24d - 144 = 16
e^2 = d^2 - 24d + 160
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Sub for f: f^2 = d^2 + 16
e^2 + d^2 + 16 = 144
--> e^2 = 128 - d^2
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128 - d^2 = d^2 - 24d + 160
2d^2 - 24d + 32 = 0
d^2 - 12d + 16 = 0
*[invoke solve_quadratic_equation 1,-12,16]
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The 2 solutions are c & d.  Their sum is 12, the hypotenuse.
6 + sqrt(20) and 6 - sqrt(20)
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e^2 = c^2 + 16
= 36 + 20 + 12sqrt(20)
{{{e = sqrt(56 + 12sqrt(20))}}}
{{{f = sqrt(56 - 12sqrt(20))}}}