Total area of the two crescents?
A hard-level "Area" problem.Found by Hippocrates 2,200 years ago. It is one of the rare curved shapes whose area comes out exactly equal to a straight-sided figure.
GivenRight triangle with legs 6 cm and 8 cm, hypotenuse 10 cm / π = 3.14
Hints
- Look first at the semicircle on the 10 cm hypotenuse. Notice that its arc passes exactly through the right-angle vertex.
- So that semicircle splits into the triangle in the middle plus the two lens-shaped pieces left on either side (arc against side).
- Now look at the two small semicircles. Each one is "one lens piece + one crescent" — and those lens pieces are exactly the ones we just saw.
- Work out the semicircles. The radius is half the diameter. The 6 cm one is 3×3×3.14÷2 = 14.13 cm², the 8 cm one is 25.12 cm². Together 39.25 cm².
- Do the same for the 10 cm semicircle: 5×5×3.14÷2 = 39.25 cm² — exactly the same as the two small ones together. That is no coincidence.
- The two small semicircles are two lens pieces + two crescents; the hypotenuse semicircle is two lens pieces + the triangle. Since the totals are equal, remove the same two lens pieces from both and what is left must be equal too.
- In other words the two crescents equal the right triangle. π vanishes completely — that is the beauty of this problem. 6 × 8 ÷ 2
Answer 24cm²
More "Area" problems
- Area of the inner triangle?
- Area of triangle DOA?
- Area of quadrilateral ABCD?
- Area of the overlap?
- Area swept by the square?
- What area of the table is covered by paper?
Browse by idea
- Add, then subtract — an overlap is a counting problem
- Circles and sectors — multiply by π once, at the very end