Geometry problem has me stumped.

[rofl]

A piece of wire 16 cm long is bent into a right triangle. If one leg of the triangle is 5 cm long, findthe lengths of the other leg and the hypotenuse of the right triangle.

In the answer it says the leg is 48/11 and the hypotenuse is 73/11 I have no idea how they came to these answers. I tried making two functions b+c=11 and 25+bsquared=csquared and then using substitution but it doesnt work. Any help would be appreciated!!!

 

[soroban]

Hello, rofl!

Did you make a sketch?

A piece of wire 16 cm long is bent into a right triangle.
If one leg of the triangle is 5 cm long, find the lengths of the other leg
and the hypotenuse of the right triangle.

      *
      |  *
      |     *
    5 |        * 11-x
      |           *
      |              *
      * - - - - - - - - *
               x

Let \(\displaystyle x\) = length of the other leg.
Then the hypotenuse is: \(\displaystyle \,16\,-\,5\,-\,x\:=\:11\,-\,x\) cm.

From Pythagorus, we have: \(\displaystyle x^2\,+\,5^2\:=\11\,-\,x)^2\)

Now solve for \(\displaystyle x.\)

 

[tkhunny]

Too bad you didn't show us what didn't work. Mathematics ALWAYS works. Normally, it would be User Error that is the problem.

You have:

5 + b + c = 16

5^2 + b^2 = c^2

This leads to:

c + b = 11

c^2 - b^2 = 25

One could notice that:

c^2 - b^2 = (c+b)(c-b)= 25

Substituting:

(c+b)(c-b)= 25 ==> (11)(c-b)= 25 ==> c - b = 25/11

Summarize:

c + b = 11
c - b = 25/11

Adding:

2c = 146/11

Solving:

c = 73/11

Conclusion: Yeah!! Mathematics still works!

 

[rofl]

thanks you guys cleared it all up