A right triangle has one leg that is 8 inches shorter than the other leg, the hypotenuse is 30 inches long, how do you find the length of each leg?

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Answer 1 · 2017-06-03T17:03:43

The legs are $24.83 and 16.83$ inches long.

The lengths of the two short sides (the legs of the right-angle) are:

$x and (x-8)$ inches

The length of the hypotenuse is $30$ inches

Write an equation using Pythagoras' Theorem:

$x^2 + (x-8)^2 = 30^2$

$x^2 +x^2 -16x +64 = 900" "larr$ make =0

$2x^2 -16x -836 =0" "larr div 2$

$x^2 -8x-418 =0" "larr$ does not factorise

Solve for $x$ by completing the square method:

$x^2 -8x +16 = 418+16$

$(x-4)^2 = 434$

$x -4 =+-sqrt434" "larr$ ignore the negative root.

$x = sqrt434+4$

$x = 24.83$

The legs are $24.83 and 16.83$ inches long.

Check:

$sqrt(24.83^2 +16.83^2) = 30$