Points in a square, surface

[ghostfirefox]

Let E, F be such points inside the ABCD square that | ∢AEF | = | ∢EFC | = 90∘ and | AE | = 2, | CF | = 6 and | EF | = 6. Calculate the surface of ABCD.

[shyamjayakannan]

Screenshot 2026-03-12 230907.png (77.67 KiB) Viewed 88 times

Construct GH and IJ as shown in the image. [tex]\angle DAE=\angle BCF[/tex] because they are angles between pairs of parallel lines (AD, BC) and (AE, CF). Let this angle be [tex]\theta[/tex]. Now, [tex]\angle FEK={180}^\circ-{90}^\circ-\angle AEG={90}^\circ-({90}^\circ-\theta)=\theta[/tex].

Now, [tex]AB=GE+EK+KH=2\sin\theta+6\cos\theta+6\sin\theta=6\cos\theta+8\sin\theta[/tex] and [tex]BC=AG+FJ-FK=2\cos\theta+6\cos\theta-6\sin\theta=8\cos\theta-6\sin\theta[/tex].

Since AB=BC, [tex]6\cos\theta+8\sin\theta=8\cos\theta-6\sin\theta\Rightarrow14\sin\theta=2\cos\theta\Rightarrow\tan\theta=\frac{1}{7}[/tex].
From this, [tex]\displaystyle\sin\theta=\frac{1}{\displaystyle\sqrt{1+7^2}}=\frac{1}{\sqrt{50}}[/tex] and [tex]\displaystyle\cos\theta=\frac{7}{\displaystyle\sqrt{1+7^2}}=\frac{7}{\sqrt{50}}[/tex].

So, area=[tex]{AB}^2=(6\cos\theta+8\sin\theta)^2=\left(\frac{6\times7}{\sqrt{50}}+\frac{8}{\sqrt{50}}\right)^2=\boxed{50}[/tex]