Two equilateral triangles of side 12 cm, sharing the same centre with one…
2026
Two equilateral triangles of side 12 cm, sharing the same centre with one triangle inverted relative to the other, are placed one on top of the other so that a 6-pointed star (hexagram) is formed. The star's six outer vertices lie on a circle. What is the area of the circle not enclosed by the star? (Give the answer to the nearest sq cm.)
- A.
68
- B.
83
- C.
57
- D.
61
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Show answer & explanation
Correct answer: A
Concept: For an equilateral triangle of side a, the circumradius — the radius of the circle passing through all three vertices — is R = a/√3, from the general circumradius relation R = a/(2 sin A) with A = 60°. When two congruent equilateral triangles are overlaid (rotated 60° about a common centre, so one points up and the other down) to form a six-pointed star, the star's area equals twice one triangle's area minus the area of the regular hexagon where the two triangles overlap, and that hexagon's side is exactly one-third of the triangles' side.
Applying it here:
Circle area — the side is a = 12 cm, so the circumradius is R = 12/√3 = 4√3 cm. Then R2 = 48, so the circle's area is πR2 = 48π.
Each triangle's area — for side a = 12, the area is (√3/4)a2 = (√3/4) × 122 = 36√3.
Overlap hexagon — its side is one-third of 12, i.e. 4, so its area is (3√3/2) × 42 = 24√3.
Star's area — the union of the two triangles is 2 × 36√3 − 24√3 = 48√3.
Circle area not enclosed by the star = 48π − 48√3 = 48(π − √3) ≈ 48 × 1.4095 ≈ 67.66, which rounds to 68 sq cm.
Cross-check: The star's area, 48√3 ≈ 83.14 sq cm, is comfortably less than the circle's area, 48π ≈ 150.80 sq cm; their difference, ≈ 67.66 sq cm, matches the algebraic result 48(π − √3), confirming the rounded answer of 68 sq cm.