Answer:
[tex]x=10[/tex]
Step-by-step explanation:
We can use the Pythagorean Theorem:
[tex]a^2+b^2=c^2[/tex]
The two legs are x and (x + 14) and the hypotenuse is (x + 16). Therefore:
[tex](x)^2+(x+14)^2=(x+16)^2[/tex]
Expand:
[tex]\displaystyle x^2+(x^2+28x+196)=x^2+32x+256[/tex]
Combine like terms:
[tex]2x^2+28x+196=x^2+32x+256[/tex]
Subtract the right from the left:
[tex](2x^2+28x+196)-(x^2+32x+256)=0[/tex]
Subtract:
[tex]x^2-4x-60=0[/tex]
Factor:
[tex](x-10)(x+4)=0[/tex]
Zero Product Property:
[tex]x-10=0\text{ or } x+4=0[/tex]
Solve:
[tex]\displaystyle x = 10 \text{ or } x = -4[/tex]
The x cannot be negative since a side cannot measure -4. So, our only answer is:
[tex]x=10[/tex]
So, the legs are 10 and 24 with the hypotenuse being 26.
