In an equilateral triangle of side 24 centimetres, a circle is inscribed so that it touches all three sides. Find the area, in square centimetres, of the remaining portion of the triangle that lies outside the circle but inside the triangle.

Introduction / Context:
This question combines properties of an equilateral triangle with those of its incircle. You are asked to find the area of the region inside the triangle but outside the circle, which is a common type of shaded region problem. It tests your ability to compute triangle area, find the inradius, and compute circle area, then subtract.

Given Data / Assumptions:

• The triangle is equilateral.
• Side length a = 24 centimetres.
• A circle is inscribed, touching all three sides (incircle).
• We must find area of triangle minus area of circle.
• Use the standard formulas for equilateral triangle area and inradius.
• Use pi = 22/7 for circle area approximation.

Concept / Approach:
For an equilateral triangle of side a, the area Atriangle is (sqrt(3) / 4) * a^2. The inradius r of an equilateral triangle is given by r = a * sqrt(3) / 6. The area of the incircle is Acircle = pi * r^2. The required shaded area is Atriangle - Acircle. We compute both areas using a = 24, then subtract, and finally match the numerical result to the closest option.

Step-by-Step Solution:
Step 1: Compute area of the equilateral triangle. Atriangle = (sqrt(3) / 4) * a^2. Here a = 24, so a^2 = 576. Atriangle = (sqrt(3) / 4) * 576 = 144 * sqrt(3) square centimetres. Step 2: Find the inradius r = a * sqrt(3) / 6. Substitute a = 24: r = 24 * sqrt(3) / 6 = 4 * sqrt(3) centimetres. Step 3: Compute area of incircle: Acircle = pi * r^2. r^2 = (4 * sqrt(3))^2 = 16 * 3 = 48. Using pi = 22/7, Acircle = (22/7) * 48 = 1056 / 7 square centimetres. 1056 / 7 is approximately 150.857 square centimetres. Step 4: Approximate the triangle area numerically. Take sqrt(3) ≈ 1.732. Atriangle ≈ 144 * 1.732 ≈ 249.408 square centimetres. Step 5: Shaded area = Atriangle - Acircle ≈ 249.408 - 150.857 ≈ 98.551 square centimetres. Rounded to one decimal place, the shaded area is about 98.5 sq cm.

Verification / Alternative check:
An approximate check is to see that the circle should occupy a bit more than half of the triangle's area. Our triangle area is about 249.4 sq cm, and the circle area is about 150.9 sq cm. The ratio 150.9 / 249.4 is a bit over 0.6, which seems reasonable, since the incircle is fairly large inside an equilateral triangle. The remaining area of about 98.5 sq cm is therefore realistic and consistent with these approximate calculations.

Why Other Options Are Wrong:
Option A (36.6 sq cm): This is far too small and would imply the circle covers almost the entire triangle, which is not possible. Option B (54.2 sq cm): This still underestimates the shaded region and does not match the triangle and circle areas. Option C (72.8 sq cm): This is closer than the smaller options but still significantly below the calculated value. Option E (120 sq cm): This is larger than the computed shaded area and would require the circle to be much smaller relative to the triangle than it actually is.

Common Pitfalls:
Students may confuse the formulas for inradius and circumradius or misremember the area formula for an equilateral triangle. Another common issue is to use radius a/2 instead of a * sqrt(3)/6 for the incircle. Rounding errors in sqrt(3) and pi can also affect the final result. Writing each formula clearly and doing the algebra symbolically before substituting approximate values helps minimise mistakes.

Final Answer:
The area of the remaining portion of the triangle outside the circle is approximately 98.5 sq cm.