Question:

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is \(12\sqrt{2}\) cm, then area of the equilateral triangle is :

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1. {Square's side:} Diagonal \(d = s\sqrt{2}\). Given \(d = 12\sqrt{2}\), so side \(s=12\) cm. 2. {Square's perimeter:} \(P_s = 4s = 4 \times 12 = 48\) cm. 3. {Triangle's side:} Perimeters are equal, so \(P_t = 48\) cm. For equilateral triangle, \(P_t = 3a\). So \(3a = 48 \implies a = 16\) cm. 4. {Triangle's area:} Area \(A_t = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(16)^2 = \frac{\sqrt{3}}{4}(256) = 64\sqrt{3} \text{ cm}^2\).
  • \(24\sqrt{2} \text{ cm}^2\)
  • \(24\sqrt{3} \text{ cm}^2\)
  • \(48\sqrt{3} \text{ cm}^2\)
  • \(64\sqrt{3} \text{ cm}^2\)
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The Correct Option is D

Solution and Explanation

Concept: This problem involves relating the dimensions and perimeters of a square and an equilateral triangle, and then calculating the area of the equilateral triangle. Formulas needed:
For a square with side \(s\): Diagonal \(d = s\sqrt{2}\), Perimeter \(P_s = 4s\).
For an equilateral triangle with side \(a\): Perimeter \(P_t = 3a\), Area \(A_t = \frac{\sqrt{3}}{4}a^2\). Step 1: Find the side of the square Let the side of the square be \(s\). The diagonal of the square is given as \(d = 12\sqrt{2}\) cm. We know \(d = s\sqrt{2}\). So, \(s\sqrt{2} = 12\sqrt{2}\). Dividing by \(\sqrt{2}\), we get \(s = 12\) cm. Step 2: Calculate the perimeter of the square Perimeter of the square \(P_s = 4s = 4 \times 12 = 48\) cm. Step 3: Find the side of the equilateral triangle Let the side of the equilateral triangle be \(a\). The problem states that the square and the equilateral triangle have equal perimeters. So, Perimeter of equilateral triangle \(P_t = P_s = 48\) cm. We also know \(P_t = 3a\). Therefore, \(3a = 48\). \(a = \frac{48}{3} = 16\) cm. Step 4: Calculate the area of the equilateral triangle The area of an equilateral triangle with side \(a\) is \(A_t = \frac{\sqrt{3}}{4}a^2\). Substitute \(a = 16\) cm: \[ A_t = \frac{\sqrt{3}}{4}(16)^2 \] \[ A_t = \frac{\sqrt{3}}{4}(256) \] \[ A_t = \sqrt{3} \times \frac{256}{4} \] \[ A_t = \sqrt{3} \times 64 \] \[ A_t = 64\sqrt{3} \text{ cm}^2 \] The area of the equilateral triangle is \(64\sqrt{3} \text{ cm}^2\).
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