The side of an equilateral triangle is 28 cm. Taking each vertex as the centre, a circle is described with a radius equal to half the length of the side of the triangle. Find the area of that part of the triangle which is not included in the circles (use = and ).

Correct — B. Take the whole triangle, then subtract the three corner sectors.

Triangle area = √3⁄4 × 28² = √3⁄4 × 784 = 196√3
= 196 × 1.73 = 339.08

Radius = half the side = 28 ÷ 2 = 14 cm.
Each vertex angle of an equilateral triangle is 60°, so the three sectors span 60° + 60° + 60° = 180° — exactly half a circle of radius 14.

Sector total = ½ × 22⁄7 × 14² = ½ × 22⁄7 × 196 = 308

Uncovered area = 339.08 − 308 = 31.08 → option (b).

Why the others are wrong

• (a)
  30.89 cm — 30.89 leaves 308.19 for the sectors. They come to exactly 308 — half a circle of radius 14 with π = 22⁄7 — so the gap is an arithmetic slip, not a rounding of √3.
• (c)
  39.08 cm — 39.08 uses 300 for the three sectors. Their combined angle is 180°, and half a circle of radius 14 is 308, not 300. That missing 8 is the entire distance between this option and the key.
• (d)
  38.08 cm — 38.08 leaves 301 for the sectors, seven short of 308. Compute the three corners as one half-circle in a single step rather than as three separate sixths.

The idea worth carrying away is that the three angles of any triangle add to 180°.

So three sectors drawn with the same radius at the three vertices always merge into exactly half a circle, whatever the triangle's shape. You never need each sector on its own.

What the equilateral condition adds is the area formula √3⁄4 × side², and a radius of half the side that keeps the circles from overlapping — each one reaches precisely to the midpoints of its two sides, so the three touch and never cross.

Key facts

• Area of an equilateral triangle = √3⁄4 × side².
• Equal-radius sectors at the three vertices of any triangle total 180°, so they form half a circle.
• Half a circle of radius 14 cm with π = 22⁄7 has area 308 cm².
• A radius of half the side makes the three circles touch at the side midpoints without overlapping.

Common traps

• Working three 60° sectors separately and slipping on the sixths instead of taking half a circle at once.
• Reading the radius as 28 cm, the full side, rather than 14 cm.
• Substituting √3 = 1.732 where the paper prints 1.73, which lands on 31.47 and no option.

No directly related past PYQ was found.