SOLUTION: The hypotenuse of a right triangle is 2 units longer than the longer leg which is 14 units longer than the shorter leg. Determine the lengths of the three sides. How do i begin s

Algebra ->  Triangles -> SOLUTION: The hypotenuse of a right triangle is 2 units longer than the longer leg which is 14 units longer than the shorter leg. Determine the lengths of the three sides. How do i begin s      Log On


   



Question 246102: The hypotenuse of a right triangle is 2 units longer than the longer leg which is 14 units longer than the shorter leg. Determine the lengths of the three sides. How do i begin solving this question?
Found 2 solutions by richwmiller, oberobic:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
The hypotenuse of a right triangle is 2 units longer than the longer leg which is 14 units longer than the shorter leg. Determine the lengths of the three sides. How do I begin solving this question?
What is the formula for the hypotenuse?
a^2+b^2=c^2 where c is the hypotenuse and a and b are the legs

Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
The Pythagorean Theorem is fundamental to the solution: c%5E2+=+a%5E2+%2B+b%5E2
, where
c = hypotenuse
a = longer leg
b = other leg
.
The problem setup gives you 3 important relationships.
c+=+a+%2B+2
and
a+=+b+%2B+14
.
So,
c+=+b+%2B+14+%2B+2+=+b+%2B+16
.
Of course,
b+=+b
.
Going back to the Pythagorean Theorem:
c%5E2+=+a%5E2+%2B+b%5E2
.
By substitution, we can cast the entire problem in terms of b, the shorter leg.
%28b%2B16%29%5E2+=+%28b%2B14%29%5E2+%2B+b%5E2
.
b%5E2+%2B+32b++%2B+256+=+b%5E2+%2B+28b+%2B+196+%2B+b%5E2+=+2b%5E2+%2B+28b+%2B+196
.
Subtract b^2 from both sides
32b+%2B+256+=+b%5E2+%2B+28b+%2B+196
.
Subtract 32b from both sides
256+=+b%5E2+-+4b+%2B+196
.
Subtract 256 from both sides
0+=+b%5E2+-+4b+-+60
so,
b%5E2+-+4b+-+60+=+0
.
Can we factor 60 such that the factors are 4 apart? 2*30? no 3*20? no. 4*15? no. 5*12? no. 6*10? yes!
%28b+-+10%29%28b+%2B+6%29+=+0
.
Thus we have two candidate values for b: 10 and -6. Since the leg of a triangle cannot be negative, the proposed solution is b = 10.
.
Substituting back into what we were told in the problem setup:
a+=+b+%2B+14+=+24
c+=+a+%2B+2+=+26
.
We always check our proposed solution. Here we just use the Pythagorean Theorem again.
a%5E2+=+24%2A24+=+576
b%5E+2+=+10%2A10+=+100
a%5E2+%2B+b%5E2+=+676
c%5E2+=+26%2A26+=+676
That checks, so we have the correct answer.
Done.