The hypotenuse of a right triangle is 52 in. One leg of the triangle is 8 in. more than twice the length of the other. What is the perimeter of the triangle?

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The hypotenuse of a right triangle is 52 in. One leg of the triangle is 8 in. more than twice the length of the other. What is the perimeter of the triangle?

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Answer 1 · 2016-12-01T11:06:50

The perimeter of the triangle is $120$ $120$ cm.

Explanation:

Let hypotenuse of the right triangle be $h = 52$ $h=52$ in.
One leg of the right triangle be $l_{1}$ $l_1$ in.
Other leg of the right triangle be $l_{2} = 2 \cdot l_{1} + 8$ $l_2=2*l_1+8$ in.
We know $h^{2} = l_{1}^{2} + l_{2}^{2} or 52^{2} = l_{1}^{2} + {(2 l_{1} + 8)}_{1}^{2} or 2704 = l_{1}^{2} + 4 l_{1}^{2} + 32 l_{1} + 64 or 5 l_{1}^{2} + 32 l_{1} - 2640 = 0$ $h^2 = l_1^2+l_2^2 or 52^2=l_1^2+(2l_1+8)^2 or 2704 = l_1^2+4l_1^2+32l_1+64 or 5l_1^2+32l_1-2640=0$ . This is a quadratic eqqution of which $a = 5; b = 32; c = - 2640; l_{1} = \frac{- 32 \pm \sqrt{32^{2} - 4 \cdot 5 \cdot (- 2640)}}{2 \cdot 5} or l_{1} = 20.0 or l_{1} = - 26.4$ $a=5;b=32;c=-2640 ; l_1= (-32+-sqrt(32^2-4*5*(-2640)))/(2*5) or l_1=20.0 or l_1= -26.4$ Length can not be negative number. So $l_{1} = 20; l_{2} = 2 \cdot 20 + 8 = 48; h = 52$ $l_1=20 ; l_2=2*20+8=48 ;h =52$
The perimeter of the triangle is $P = l_{1} + l_{2} + h = 20 + 48 + 52 = 120$ $P= l_1+l_2+h=20+48+52=120$ cm [Ans]

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The hypotenuse of a right triangle is 52 in. One leg of the triangle is 8 in. more than twice the length of the other. What is the perimeter of the triangle?

1 Answer

Explanation:

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