# Math

The shorter leg of a right triangle is 14 feet shorter than the longer leg. The hypotenuse is 26 feet. How long is each leg?

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1. (X-14)^2+X^2 = 26^2
solve for x and x-14

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posted by hy
2. a=shorter leg
b=longer leg
c=hypotenuse

a=b-14

c^2=a^2+b^2
26^2=(b-14)^2+b^2
676=b^2-2*b*14+14^2+b^2
676=2b^2-28b+196
676-196=2(b^2-14b)
480=2(b^2-14b) Divide with 2
240=b^2-14b
b^2-14b=240
b^2-14b+7^2=240+7^2

Becouse: (b-7)^2=b^2-14b+7^2

(b-7)^2=240+49
(b-7)^2=289
b-7=sqroot(289)

You have two solutions:
b-7=17 and
b-7=-17

Becouse sqroot(289)=+/- 17

First solution: b-7=17
b=17+7=24

Second solution:
b-7=-17
b=17+7=-10

leg of triangle can not be negative number so:
b=24
a=b-14=24-14=10

a=10 b=24

c=sqroot(a^2+b^2)
c=sqroot(10^2+24^2)
c=sqroot(100+576)
c=sqroot(676)

c=26

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posted by Bosnian

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