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:  If we can move all the terms to one side and factor, we can use the zero product principle to solve.  Since this is the only method we know – let’s hope it works!

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 -6 and 2:

[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 -2, 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

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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 6, 8, and 10