Question

The hypotenuse of a right triangle is 7 feet long. One leg of the triangle is 5 feet longer than the other leg. Find the perimeter of the triangle.

Answer 1

Answer:

P= 15.544 ft

Step-by-step explanation:

l$a^{2} +b^{2} =c^{2} \\(5+ Leg)^{2} +Leg^{2} =7^{2} \\(5+L)(5+L)+L^{2} =49\\25+5L+5L+L^{2} +L^{2} =49\\2L^{2}+10L+25=49 \\2L^{2}+10L-24=0$

Solve this using quadratic formula (if you don't know how to solve using quadratic formula then just ask me)

You get 1.772 (length of one leg)

(1.772+5) + 1.772+ 7= 15.544

I hope this helps!

Answer 2

Answer:

Step-by-step explanation:

let the legs of triangle be x  and x+5

then x²+(x+5)²=7²

x²+x²+10x+25=49

2x²+10x-24=0

x²+5x-12=0

$x=\frac{-5 \pm\sqrt{25-4*1*-12} }{2*1} \\=\frac{-5 \pm\sqrt{25+48} }{2} \\=\frac{-5 \pm\sqrt{73} }{2} \\taking ~positive~sign~only,as ~x~is~positive.\\x \approx\frac{-5 \pm 8.54}{2} \\or~x \approx \frac{3.54}{2}\\or~x \approx 1.77\\x+5=6.77\\perimeter=7+1.77+6.77=15.54 ~ft$