Example

A right triangle has one leg with length x, another whose length is greater by two,  and the length of the hypotenuse is greater by four.  Find the lengths of the sides of the triangle. Use the image below.

Answer: Read and understand: We know the lengths of all the sides of a triangle in terms of one side. We also know that the Pythagorean theorem will give us a relationship between the side lengths of a right triangle. Translate: 

[latex]\begin{array}{l}a^2+b^2=c^2\\x^2+\left(x+2\right)^2=\left(x+4\right)^2\end{array}[/latex]

Solve:  To solve this equation, we need to start by simplifying the equation and moving all the terms to one side.  If we can factor it, then we can use the zero product principle to solve. First, multiply the binomials and simplify so we can see what we are working with.

[latex]\begin{array}{l}x^2+\left(x+2\right)^2=\left(x+4\right)^2\\x^2+x^2+4x+4=x^2+8x+16\\2x^2+4x+4=x^2+8x+16\end{array}[/latex]

Now move all the terms to one side and see if we can factor.

[latex]\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\,\,\,\,\,\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\\x^2-4x-12=0\end{array}[/latex]

This went from a messy looking problem to something promising. We can factor using the shortcut:

[latex]-6\cdot{2}=-12,\text{ and }-6+2=-4[/latex]

So we can build our binomial factors with [latex]-6[/latex] and [latex]2[/latex]:

[latex]\left(x-6\right)\left(x+2\right)=0[/latex]

Set each factor equal to zero:

[latex]x-6=0, x=6[/latex]

[latex]x+2=0, x=-2[/latex]

Interpret: Ok, it doesn't make sense to have a length equal to [latex]-2[/latex], so we can safely throw that solution out.  The lengths of the sides are as follows:

[latex]x=6[/latex]

[latex]x+2=6+2=8[/latex]

[latex]x+4=6+4=10[/latex]

Check: Since we know the relationship between the sides of a right triangle we can check that we are correct. Sometimes it helps to draw a picture

.

We know that [latex]a^2+b^2=c^2[/latex], so we can substitute the values we found:

[latex]\begin{array}{l}6^2+8^2=10^2\\36+64=100\\100=100\end{array}[/latex]

Our solution checks out.

Answer

The lengths of the sides of the right triangle are [latex]6, 8[/latex], and [latex]10[/latex]