In the figure, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).
Step-by-step solution
Idea: These are the ‘lunes of Hippocrates’. The semicircle on the hypotenuse passes through the right-angle corner, so it covers the triangle plus two segments, and those same segments sit inside the small semicircles.
- Let the legs be a, b and the hypotenuse c; a2 + b2 = c2. The semicircle on a side of length d has area 12π(d2)2 = πd28. So πa28 + πb28 = πc28.1 mark
- The midpoint of the hypotenuse is the same distance from all three corners, so the semicircle on the hypotenuse passes through the right-angle corner. It is made up of triangle C and the two segments Sa, Sb cut off by the legs: πc28 = C + Sa + Sb.1 mark
- The semicircle on leg a is the crescent A plus the segment Sa: πa28 = A + Sa. Similarly πb28 = B + Sb.1 mark
- Add: A + B + Sa + Sb = πc28 = C + Sa + Sb. Cancel the segments: A + B = C.1 mark
Check: Legs 6, 8, hypotenuse 10: semicircles 4.5π + 8π = 12.5π = π × 1008 ✓, and C = 24, so A + B = 24 although A and B are curved ✓.
Answer to write in the exam
Semicircle on side d = πd28; a2 + b2 = c2 ⇒ πa28 + πb28 = πc28
πc28 = C + Sa + Sb (semicircle on hypotenuse passes through right-angle vertex)
πa28 = A + Sa, πb28 = B + Sb
A + B + Sa + Sb = C + Sa + Sb
∴ A + B = C
Common mistakes that cost marks
- Using πd2/2 for a semicircle on a side d; the radius is d/2, so the area is πd2/8.
- Forgetting that the big semicircle covers the triangle (it passes through the right-angle corner).
- Mixing up which region is C (the triangle) and which are crescents.
How this can come in the exam
A right triangle has legs 20 cm and 21 cm. Semicircles are drawn as in this question. Area(A) + Area(B) equals
- 210 cm2
- 420 cm2
- 841π8 cm2
- 290 cm2
Show answer
(A) 210 cm2
A + B = area of the triangle = 12 × 20 × 21 = 210 cm2.
Try one yourself
Semicircles are drawn on the sides 9, 12 and 15 of a right triangle as in this question. Find Area(A) + Area(B).
Show answer
Equal to the triangle: 12 × 9 × 12 = 54 sq. units.
More questions like this
- The figure shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius r.
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- In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.
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