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

P= 15.544 ft

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

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=(-5 \pm√(25-4*1*-12) )/(2*1) \\=(-5 \pm√(25+48) )/(2) \\=(-5 \pm√(73) )/(2) \\taking ~positive~sign~only,as ~x~is~positive.\\x \approx(-5 \pm 8.54)/(2) \\or~x \approx (3.54)/(2)\\or~x \approx 1.77\\x+5=6.77\\perimeter=7+1.77+6.77=15.54 ~ft`