The legs of a right triangle have lengths of x + 4 and x + 7. The hypotenuse length is 3x. How do you find the perimeter of the triangle?

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Answer 1 · 2016-01-03T19:50:57

$36$

The perimeter is equal to the sum of the sides, so the perimeter is:

$(x+4)+(x+7)+3x=5x+11$

However, we can use the Pythagorean theorem to determine the value of $x$ since this is a right triangle.

$a^2+b^2+c^2$

where $a,b$ are legs and $c$ is the hypotenuse.

Plug in the known side values.

$(x+4)^2+(x+7)^2=(3x)^2$

Distribute and solve.

$x^2+8x+16+x^2+14x+49=9x^2$

$2x^2+22x+65=9x^2$

$0=7x^2-22x-65$

Factor the quadratic (or use the quadratic formula).

$0=7x^2-35x+13x-65$

$0=7x(x-5)+13(x-5)$

$0=(7x+13)(x-5)$

$x=-13/7,5$

Only $x=5$ is valid here, since the hypotenuse's length would be negative if $x=-13/7$ .

Since $x=5$ , and the perimeter is $5x+11$ , the perimeter is:

$5(5)+11=36$