Greetings!
To find the length of any side of a right triangle, you can use the Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
[tex]a^2+b^2=c^2[/tex]
Input the information from the diagram into the formula:
[tex](x)^2+(x+7)^2=(13)^2[/tex]
Expand each term:
[tex](x)^2+(x+7)^2=(13)^2[/tex]
[tex]x^2+((x+7)(x+7))=169[/tex]
[tex]x^2+(x(x+7)+7(x+7))=169[/tex]
[tex]x^2+x^2+7x+7x+49=169[/tex]
Combine like terms:
[tex]2x^2+14x+49=169[/tex]
Add -169 to both sides:
[tex](2x^2+14x+49)+(-169)=(169)+(-169)[/tex]
[tex]2x^2+14x-120=0[/tex]
Factor out the Common Term (2):
[tex]2(x^2+7x-60)=0[/tex]
Factor the Complex Trinomial:
[tex]2(x^2-5x+12x-60)=0[/tex]
[tex]2(x(x-5)+12(x-5))=0[/tex]
[tex]2(x-5)(x+12)=0[/tex]
Set Factors to equal 0:
[tex]x-5=0[/tex]
[tex]x=5[/tex]
or
[tex]x+12=0[/tex]
[tex]x=-12[/tex]
However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)
The Solution Is:
[tex]\boxed{x=5} [/tex]
I hope this helped!
-Benjamin
