A circular disc of radius 7 cm is inscribed inside an equilateral triangle. What is the approximate area of the remaining portion of the triangle?

Correct — A. Remaining area = triangle − disc, and the disc's radius is the triangle's inradius.

Inradius of an equilateral triangle: r = a⁄(2√3)
Solve for the side: a = 2√3 × 7 = 14√3 cm
Triangle area: (√3⁄4) × (14√3)² = (√3⁄4) × 588 = 147√3 ≈ 254.61 cm²

Disc area: π × 7² = 49π ≈ 153.94 cm²
Subtract: 254.61 − 153.94 ≈ 100.67 cm²
Match: 100.66 cm² differs only in the rounded last digit → option (a)

Why the others are wrong

• (b)
  148.2 cm² — 148.2 cm² would leave the disc only 254.61 − 148.2 ≈ 106.4 cm². A disc of radius 7 cm covers 49π ≈ 153.94 cm², so the leftover is about 100.67 cm².
• (c)
  200.3 cm² — 200.3 cm² would leave the disc only about 54.3 cm² of the 254.61 cm² triangle, far less than the 153.94 cm² a radius-7 disc covers.
• (d)
  155.6 cm² — 155.6 cm² would leave the disc about 99.0 cm² (254.61 − 155.6). That swaps the two parts: the disc is the larger piece at 153.94 cm², and the leftover is about 100.67 cm².

An equilateral triangle's incircle touches all three sides, and its centre is also the centroid, so the inradius is one-third of the height.

Height h = (√3⁄2)a, so r = h⁄3 = a⁄(2√3), which gives a = 2√3r.

Substituting into (√3⁄4)a² gives a triangle area of 3√3 r². The leftover is then r²(3√3 − π) = 49 × 2.0546 ≈ 100.67 for r = 7, with no need to find the side.

Key facts

• Inradius of an equilateral triangle: r = a⁄(2√3), one-third of the height (√3⁄2)a.
• Area of an equilateral triangle in terms of its inradius: 3√3 r².
• Area left outside the incircle: r²(3√3 − π) ≈ 2.05 r².

Common traps

• Taking 7 cm as the circumradius. An inscribed disc's radius is the inradius, and the circumradius of this triangle is 14 cm.
• Expecting an exact match with π = 22⁄7 (100.61 cm²). The options are rounded, so take the nearest value.

No directly related past PYQ was found.