Geometry Problem: Find the area of $ABCD$ square

Question

My Problem is:

The $E-$ point on the $AC$ diagonal is marked on the $ABCD$ square.If $BE= 13$, $CE=17$ find the square area.

My way:

$$a=\frac{k+17}{\sqrt2}$$

$$13^2=k^2+\frac {(k+17)^2}{2}-\frac {2k(k+17)}{\sqrt2}×\frac{\sqrt2}{2} \Rightarrow k^2+289=338 \Rightarrow k^2=49 \Rightarrow k=7$$

$$S_{ABCD}=\frac {(k+17)^2}{2}=\frac {(7+17)^2}{2}=288 $$

It looks ugly solution. Is this solution correct?

Answer 1

Let $F$ be the projection of $E$ on $BC$.

$CE=17$ implies $EF=FC=\frac{17}{\sqrt{2}}$. $BE=13$ and the Pythagorean theorem imply $$ BF = \sqrt{13^2-\frac{17^2}{2}}=\frac{7}{\sqrt{2}} $$ hence $$ BC=BF+FC = \frac{17+7}{\sqrt{2}} = 12\sqrt{2} $$ and the area of $ABCD$ is clearly $288$. You do not really need the cosine theorem, just the trick behind one of its proofs.

Answer 2

Let $ABE_1E$ be a parallelogram.

Thus, $BECE_1$ is isosceles trapezoid ($BE=E_1C=13$) and since $EE_1\perp BC$,

we obtain: $$BE_1^2+EC^2=BE^2+E_1C^2$$ or $$BE_1^2+17^2=13^2+13^2,$$ which gives $$BE_1=7.$$ Id est, $$AC=AE+EC=7+17=24$$ and $$S_{ABCD}=\frac{1}{2}\cdot24^2=288.$$