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?

[tex]52^2=x^2+(2x+8)^2\\\\\Rightarrow\ 2704=x^2+4x^2+32x+64\\\\\Rightarrow\ 2704=5x^2+32x+64\\\\\Rightarrow\5x^2+32x-2640=0\\\\\Rightarrow\ (x-20)(5x+132)\\\\\Rightarrow\ x=20\ or\ x=\dfrac{-132}{5}\ \ \ \text{[side cannot be negative]}\\\\\Rightarrow\ x=20[/tex]

Perimeter of triangle will be:

[tex]52+x+(2x+4)\\\\=52+20+(2(20)+8)\\\\=120\text{ in.}[/tex]

Pythagoras theorem of right triangle says that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

keshavgandhi04 keshavgandhi04 · Answer 2 · 2022-02-07T17:10:46+01:00

The hypotenuse of a right triangle is 52inches One leg of the triangle is 8inches the perimeter of the triangle will be 120 in.

Given:

The hypotenuse of a right triangle is 52 in.

One leg of the triangle is 8 in.

According to the question,

Assume the length of one leg be x, then the length of other leg will be 2x+8

What is pythagoras theorem ?

Pythagoras theorem of right triangle derieves that the square of the hypotenuse will be equal to sum of the squares of the other two sides i.e.perpendicular & base

We will be using pythagoras theorem of right triangle,

[tex]\rm h^2=p^2+b^2\\h=hypotenuse\\p=perpendicular\\b=base[/tex]

[tex]\rm 52^2=x^2=(2x+8)^2\\\\2704=x^2+4x^2+32x+64\\\\x^2=32x-2640=0\\\\(x-20)(5x+132)\\\\x=20 \;or\; x=\dfrac{-132}{5}\\\\We\;will\;take\;only\;positive.\\\\Therefore\;\;x=20[/tex]

Now perimeter of the triangle will be:

[tex]\rm 52+x+(2x+4)\\\\=52+20+(2\times 20+8)\\\\=120 inches[/tex]

Therefore the perimeter of the triangle 120 inches.

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