IPMAT Indore 2025 · Quantitative Ability · Question 31

Step 1 — slice the octagon from the centre. Joining the centre to all 8 vertices cuts the octagon into 8 identical triangles. Each has two sides equal to the radius, 1 and 1.

Step 2 — find the apex angle. The 8 angles at the centre fill a full turn:

\frac{{360}^{∘}}{8}={45}^{∘}

Step 3 — area of one triangle using 1/2ab\sin C:

\frac{1}{2}\times 1\times 1\times \sin{45}^{∘}=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}

Step 4 — multiply by 8.

8\times \frac{\sqrt{2}}{4}=2\sqrt{2}

General formula worth remembering: a regular n-gon in a circle of radius R has area \frac{n}{2}{R}^{2}\sin \frac{{360}^{∘}}{n}.

Answer: Option 1 — 2\sqrt{2}