(adsbygoogle = window.adsbygoogle || []).push({}); "Solving Equations by Factoring" - Find the length of the hypotenuse

1. The problem statement, all variables and given/known data

One leg of a right triangle is 2 feet more than twice the other leg. The hypotenuse is 1 foot more than the longer leg. Find the length of the hypotenuse of the right triangle.

* I am not sure how to calculate the square of the hypotenuse [tex](2n+2)+1 [/tex]

Whether it's supposed to be [tex]4n^2+4+1 = 4n^2+5[/tex]

or [tex]4n^2+9[/tex]

Maybe this is why I don't know how to get the right answer?

Given formulas:

Hypotenuse denoted by c = [tex](2n+2)+1[/tex]

Leg 1 denoted by a = [tex]2n+2[/tex]

Leg 2 denoted by n = n

The book's answer: The length of the hypotenuse is 13 feet.

3. The attempt at a solution

Attempt 1:

[tex]c^2-a^2=n^2[/tex]

[tex](4n^2+9)-(4n^2+4)=n^2[/tex]

[tex]5=n^2[/tex]

[tex]\sqrt{5}=n[/tex]

[tex]n=2.2[/tex] I don't think they want decimals so I tried another way. Also, if I were to use [text] c=4n^2+5[/text] I would get n=1 again.

Attempt 2:

[tex]c^2=a^2+b^2[/tex]

[tex](4n^2+4)+1=(4n^2+4)+(n^2)[/tex]

[tex](4n^2+4)+1=5n^2+4[/tex]

[tex](-n^2)+1=0[/tex]

[tex]-(n^2+1)=0[/tex]

[tex]n^2-1=0[/tex]

[tex]n^2=1[/tex]

[tex]n=1[/tex]

Are there any hints that would help me a lot? Thanks folks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Solving Equations by Factoring - Find the length of the hypotenuse

**Physics Forums | Science Articles, Homework Help, Discussion**