The square on the diagonal is double the square

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Let $ABCD$ be a square of side $\ell$ and diagonal $AC$.

• (a) Prove that the square built on the diagonal $AC$ is equivalent to twice the square $ABCD$.
• (b) Taking $\ell=7$, compute the length of the diagonal $\overline{AC}$.

Solution

(a) Proof. The diagonal $AC$ splits the square into the right triangle $ABC$, right-angled at $B$, whose legs $AB$ and $BC$ both measure $\ell$. By the Pythagorean theorem: $\overline{AC}^{\,2}=\overline{AB}^{\,2}+\overline{BC}^{\,2}=\ell^2+\ell^2=2\ell^2.$ The square built on the diagonal has area $\overline{AC}^{\,2}=2\ell^2$, while the square $ABCD$ has area $\ell^2$. Hence the former is double the latter. $\;\blacksquare$

(b) Numerical computation. With $\ell=7$: $\overline{AC}=\sqrt{2\ell^2}=\ell\sqrt2=7\sqrt2\approx 9{,}899.$ $\boxed{\overline{AC}=7\sqrt2\approx 9{,}899}$