Posted by Megan on Saturday, December 4, 2010 at 12:20am.
(X-14)^2+X^2 = 26^2
solve for x and x-14
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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