SOLUTION: One leg of a right triangle is 7in longer than the smaller leg, and the hypotenuse is 8in longer than the smaller leg. Find the lengths of the sides of the triangle.

Algebra ->  Pythagorean-theorem -> SOLUTION: One leg of a right triangle is 7in longer than the smaller leg, and the hypotenuse is 8in longer than the smaller leg. Find the lengths of the sides of the triangle.      Log On


   



Question 400983: One leg of a right triangle is 7in longer than the smaller leg, and the hypotenuse is 8in longer than the smaller leg. Find the lengths of the sides of the triangle.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = the smaller leg
Let b = longer leg
Let c = hypotenuse
given:
b+=+a+%2B+7 in
c+=+a+%2B+8 in
It's a right tringle, so
a%5E2+%2B+b%5E2++=+c%5E2
By substitution:
a%5E2+%2B+%28a+%2B+7%29%5E2++=+%28a+%2B+8%29%5E2
a%5E2+%2B+a%5E2+%2B+14a+%2B+49+=+a%5E2+%2B+16a+%2B+64
a%5E2+-+2a++=+15
I can complete the square to find a
a%5E2+-+2a+%2B+%28-2%2F2%29%5E2+=+15+%2B+%28-2%2F2%29%5E2
a%5E2+-+2a+%2B+1+=+15+%2B+1
Both sides are now a perfect square
%28a+-+1%29%5E2+=+4%5E2
Take the square root of both sides
a+-+1+=+4
a+=+5
It is also true that
a+-+1+=+-4, but I can't use a negative result
Since
b+=+a+%2B+7
b+=+12
and, since
c+=+a+%2B+8
c+=+13
The sides are 5, 12, and 13
check answer:
5%5E2+%2B+12%5E2+=+13%5E2
25+%2B+144+=+169
169+=+169