CAT 2024 Slot 3 · Quant · Question 21

A regular octagon has all eight vertices on one circle, so let that circumradius be R.

Each side subtends \frac{{360}^{∘}}{8}={45}^{∘} at the centre.

The vertices A, C, E and G skip one vertex each time, so each of AC, CE, EG and GA subtends 2\times{45}^{∘}={90}^{∘} at the centre.

Four equal chords at {90}^{∘} apart form a square, confirming that ACEG is indeed a square.

A chord subtending {90}^{∘} has length R\sqrt{2}, so the square's side is R\sqrt{2} and its area is 2{R}^{2}.

Now find R from the side of the octagon: a side subtends {45}^{∘}, so 6=2R\sin{22.5}^{∘}.

Rather than evaluate \sin{22.5}^{∘}, use the chord-length form directly: {6}^{2}=2{R}^{2}\left( 1-\cos{45}^{∘}\right).

Since \cos{45}^{∘}=\frac{1}{\sqrt{2}}, this gives 36=2{R}^{2}\left( 1-\frac{1}{\sqrt{2}}\right).

So 2{R}^{2}=\frac{36}{1-\frac{1}{\sqrt{2}}}=\frac{36\sqrt{2}}{\sqrt{2}-1}.

Rationalising by multiplying top and bottom by \sqrt{2}+1, 2{R}^{2}=36\sqrt{2}\left( \sqrt{2}+1 \right)=36\left( 2+\sqrt{2}\right).

Since the square's area is exactly 2{R}^{2}, it is 36\left( 2+\sqrt{2}\right) — the third option.

Shortcut. Build the octagon as a square with its four corners cut off. If the cut is a on each leg, the hypotenuse a\sqrt{2} equals the side 6, so a=3\sqrt{2}. Then AC is the hypotenuse of a right triangle with legs a and a+6, giving A{C}^{2}=18+{\left( 3\sqrt{2}+6 \right)}^{2}=18+18+36\sqrt{2}+36=72+36\sqrt{2}=36\left( 2+\sqrt{2}\right), which is the area at once.