SOLUTION: One leg of a right triangle has length 7. The lengths of the other two sides are whole numbers. The length of the other leg is ? and the length of the hypotenuse is ?

Algebra ->  Triangles -> SOLUTION: One leg of a right triangle has length 7. The lengths of the other two sides are whole numbers. The length of the other leg is ? and the length of the hypotenuse is ?      Log On


   



Question 934943: One leg of a right triangle has length 7. The lengths of the other two sides are whole numbers. The length of the other leg is ? and the length of the hypotenuse is ?
Found 2 solutions by KMST, MathLover1:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The length of the other leg is highlight%2824%29 and the length of the hypotenuse is highlight%2825%29 .
Sets of 3 integers that could be lengths of the sides of a right triangle are called Pythagorean triples.
The only one I remember is {3, 4, 5} , but I looked up a list online and {7, 24, 25} was the only set with a 7 in it.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
let sides be a, b and c
If it is a right angle triangle, then the relationship between the sides must follow the Pythagorean theorem:
a%5E2+%2B+b%5E2+=+c%5E2
Obviously, the side called c must be the longest one (because squares can never be negative).
Let's get rid of the "degenerate" solution of
a+=+7
b+=+0
c+=+7
(this is called a "degenerate" triangle because one side measures 0)
(mathematically, it could still be a triangle, but in real life, it is simply a line segment)
You are looking for a solution where one side is 7 and the other two sides are whole numbers.
We know that there are no solutions if you use c+=+7

Next, you can check values where a+=+7
(you don't need to check b=7 because it will be the same exercise)
7%5E2+%2B+b%5E2+=+c%5E2
49+=+c%5E2+-+b%5E2
The "formula" to find the difference between two squared integers goes like this
c%5E2+-+b%5E2+=+c+%2B+b+%2B+%28twice_+each_+number_+between_+b_+and_+c%29
for example
17%5E2+-+16%5E2+=+17+%2B+16+=+33
let's check
289+-+256+=+33
20%5E2+-+17%5E2+=+20+%2B+17+%2B+18%2B18+%2B+19%2B19+=+111
400+-+289+=+111
(this is not a coincidence, there is a true reason why this "trick" works).
So we need two numbers such that the difference of their squares is 49
c%5E2+-+b%5E2+=+49
because we need the two end numbers, the smallest difference would be 24 and 25
because 24+%2B+25+=+49
25%5E2+-+24%5E2+=+625+-+576+=+49

let's try
7%5E2+%2B+24%5E2+=+25%5E2
49%2B+576+=+625
625+=+625
There is a triangle 7, 24, 25 would be a right-angle triangle with three sides being whole+numbers and one side being 7.