Use algebra to calculate the perimeter of a right angle triangle where the hypotenuse is $(x+4)$, the base is $(x+3)$ and the remaining side is $(x-4)$.
Thanks to the earlier comments I know that I need to use Pythagoras.
So far I know that the Theorem says that $a^2 + b^2 = c^2$
I can write my problem as $(x-4)^2 + (x+3)^2 = (x+4)^2$
What should I do next?
$\endgroup$ 81 Answer
$\begingroup$You're almost done. $$\begin{align*}(x - 4)^2 + (x + 3)^2 &= (x + 4)^2 \\ \implies x^2 - 8x + 16 + x^2 + 6x + 9 &= x^2 + 8x + 16 \\ \implies x^2 - 10x + 9&= 0 \\ \implies (x - 9)(x - 1) &= 0 \end{align*}$$ So either $x = 9$ or $x = 1$. We know $x \neq 1$ because that would make a side of the triangle $x - 4 = -3$ which is negative. Hence $x = 9$.
Now the perimeter is $(x + 4) + (x + 3) + (x - 4) = 13 + 12 + 5 = 30$.
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