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In an equilateral triangle of side . 24 ...

In an equilateral triangle of side . 24 cm., a circle is inscribed touching its sides. The area of the remaining portion of the triangle is `(sqrt(3) = 1.732)`

A

98.55 sq cm

B

100 sq cm

C

101 sq cm

D

95 sq cm

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The correct Answer is:
To solve the problem step by step, we will calculate the area of the equilateral triangle, the area of the inscribed circle, and then find the area of the remaining portion of the triangle. ### Step 1: Calculate the area of the equilateral triangle The formula for the area \( A \) of an equilateral triangle with side length \( a \) is given by: \[ A = \frac{\sqrt{3}}{4} a^2 \] For our triangle, the side length \( a = 24 \) cm. Plugging in the value: \[ A = \frac{\sqrt{3}}{4} \times (24)^2 \] \[ A = \frac{\sqrt{3}}{4} \times 576 \] \[ A = 144\sqrt{3} \text{ cm}^2 \] ### Step 2: Calculate the radius of the inscribed circle The radius \( r \) of the inscribed circle in an equilateral triangle can be calculated using the formula: \[ r = \frac{a \sqrt{3}}{6} \] Substituting \( a = 24 \): \[ r = \frac{24 \sqrt{3}}{6} \] \[ r = 4\sqrt{3} \text{ cm} \] ### Step 3: Calculate the area of the inscribed circle The area \( A_c \) of the circle is given by: \[ A_c = \pi r^2 \] Substituting \( r = 4\sqrt{3} \): \[ A_c = \pi (4\sqrt{3})^2 \] \[ A_c = \pi \times 16 \times 3 \] \[ A_c = 48\pi \text{ cm}^2 \] ### Step 4: Calculate the area of the remaining portion of the triangle The area of the remaining portion of the triangle is given by: \[ \text{Remaining Area} = \text{Area of Triangle} - \text{Area of Circle} \] Substituting the areas we calculated: \[ \text{Remaining Area} = 144\sqrt{3} - 48\pi \] Using the approximate value \( \sqrt{3} \approx 1.732 \) and \( \pi \approx 3.14 \): \[ \text{Remaining Area} \approx 144 \times 1.732 - 48 \times 3.14 \] \[ \text{Remaining Area} \approx 249.408 - 150.72 \] \[ \text{Remaining Area} \approx 98.688 \text{ cm}^2 \] ### Final Answer The area of the remaining portion of the triangle is approximately \( 98.688 \text{ cm}^2 \). ---
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