CAT 2021

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Mensuration — Area of Regular Hexagon and Equilateral Triangle
• Formula: $$\text{Area of Regular Hexagon} = \frac{3\sqrt{3}}{2}x^2 \quad \text{and} \quad \text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4}a^2$$

2. Calculation

Let the side of the regular hexagon be $x$ cm. The side of the equilateral triangle is given as $a = 12$ cm.

A regular hexagon can be divided into 6 equilateral triangles, each with side equal to $x$. So its area is $6 \times \frac{\sqrt{3}}{4}x^2 = \frac{3\sqrt{3}}{2}x^2$.

The area of the equilateral triangle with side 12 cm is $\frac{\sqrt{3}}{4}(12)^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}$.

Setting the two areas equal: $$\frac{3\sqrt{3}}{2}x^2 = 36\sqrt{3}$$

Divide both sides by $\sqrt{3}$: $$\frac{3}{2}x^2 = 36$$

Multiply both sides by $\frac{2}{3}$: $$x^2 = 36 \times \frac{2}{3} = 24$$

Taking the square root: $$x = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$$

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3. Solution

Answer = Option D ✅

The final calculated value is $2\sqrt{6}$ cm.