One leg of a right triangle is is 4 inches longer than the other. The hypotenuse of the triangle is 8 inches longer than the shorter leg. What are the lengths of the three sides of the triangle?
2006-07-26 16:15:18 · 5 answers · asked by Anonymous in Education & Reference ➔ Homework Help
Short leg: x
Long leg: x+4
Hypot: x+8
Pythag. Thm:
x^2 + (x+4)^2 = (x+8)^2
Multiply out (FOIL)
x^2 + x^2 + 8x + 16 = x^2 + 16x + 64
2x^2 + 8x + 16 = x^2 + 16x + 64
x^2 - 8x - 48 = 0
(x-12) (x + 4 ) = 0
x = 12 or x = -4 (impossible)
Short leg: 12 in
Long leg: 16 in
Hypot: 20 in
2006-07-26 16:21:58 · answer #1 · answered by jenh42002 7 · 0⤊ 0⤋
The legs are x and x+4 and the hypotenuse is x+8. You use the Pythagorean Theorem and solve x^2 +(x+4)^2 = (x+8)^2. After you foil you get x^2 + x^2 + 8x + 16 = x^2 +16x+64 which simplifies to x^2 - 8x - 48=0. That factors into (x-12)(x+4)=0 which means x=12, making the legs 12 and 16 and the hypotenuse 20.
2006-07-26 23:26:56 · answer #2 · answered by MollyMAM 6 · 0⤊ 0⤋
short=x hyp=x=8 other=x=4 (8+x)^2=(4+x)^2+x^2
x^2-8x+48=0 x=12 short=12 hype=20 other=16
2006-07-26 23:40:34 · answer #3 · answered by ? 1 · 0⤊ 0⤋
[x+8]^2 = [x+4]^2 + x^2
x^2 + 16x + 64 = x^2 + 8x + 16 + x^2
x^2 + 16x + 64 = 2x^2 + 8x + 16
2x^2 + 8x + 16 - x^2 - 16x - 64 = 0
x^2 - 8x - 48 = 0
[x +8] [ x-12] = 0
x+8 = 0; or x-12 = 0;
x = - 8 ; x = 12;
the value of x is 12
2006-07-27 01:17:28 · answer #4 · answered by fzaa3's lover 4 · 0⤊ 0⤋
The shortest side is 12, the other side is 16, and the hypotenuse is 20.
2006-07-26 23:24:11 · answer #5 · answered by heatherou02 1 · 0⤊ 0⤋