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.
Answer: Let OA = OB = R. Triangle AOB = R22. AB = R√2, so the semicircle on AB has area πR24. Segment AFB = quarter disc − triangle = πR24 − R22. Crescent AEBF = semicircle − segment = R22 = area of △AOB.
Step-by-step solution
Idea: This is Hippocrates’ lune: the semicircle on AB has the same area as the quarter circle AOB, so taking away their shared segment leaves equal pieces.
- Let OA = OB = R (radii) with ∠AOB = 90°. Area(△AOB) = 12R × R = R22. By Baudhāyana–Pythagoras, AB = √(R2 + R2) = R√2.1 mark
- Semicircle AEB (diameter AB, radius R√22): area = 12π × 2R24 = πR24, the same as the quarter circle OAFB.1 mark
- Segment AFB (between chord AB and arc AFB) = quarter circle − △AOB = πR24 − R22.1 mark
- Crescent AEBF = semicircle AEB − segment AFB = πR24 − (πR24 − R22) = R22 = area(△AOB). The two shaded regions are equal.1 mark
Both shaded regions have area R²/2, where R = OA: the crescent equals the semicircle on AB (πR²/4) minus the segment AFB (πR²/4 − R²/2).
Check: R = 14 cm: △AOB = 98 cm2; semicircle on AB = 12 × 227 × 98 = 154; segment = 154 − 98 = 56; crescent = 154 − 56 = 98 ✓.
Answer to write in the exam
OA = OB = R, ∠AOB = 90° ⇒ ar(△AOB) = R22, AB = R√2
Semicircle on AB = 12π(R√22)2 = πR24
Segment AFB = πR24 − R22
Crescent = πR24 − (πR24 − R22) = R22
∴ ar(crescent) = ar(△AOB)
Common mistakes that cost marks
- Taking the radius of the small semicircle as R/2; it is half of AB = R√22.
- Using the whole semicircle on AC instead of the quarter circle OAB.
- Subtracting the triangle from the small semicircle instead of the segment.
How this can come in the exam
MCQ (1 mark)
In a figure like this one, OA = 10 cm. The area of the crescent is
- 25 cm2
- 78.5 cm2
- 100 cm2
- 50 cm2
Show answer
(D) 50 cm2
Crescent = area of △AOB = 12 × 10 × 10 = 50 cm2.
Try one yourself
In a figure like this one, AB = 8 cm. Find the area of the crescent.
Show answer
R√2 = 8 ⇒ R2 = 32; crescent = R22 = 16 cm2.
More questions like this
- In the figure, you see athletes assembled at the start of a 4 × 100 m relay race. The tracks are laid out, and the athletes are all set to go racing down the tracks. Do you notice that the athletes are not at the same starting line? Those in the outer lanes seem to be starting ahead of those in the inner lanes while the finish line is the same for all of them. What could be the reason for this? The distance between the starting points of adjacent lanes is called the ‘stagger’. Notice that the stagger continues all the way to the outermost lane. Do you think the stagger gives anyone (those in the outer lanes or in the inner lanes) an unfair advantage? Why or why not? On what basis can the organisers work out the length of the stagger between lanes?
- In my school, the playground is too small to have a 400 m track, so the school constructed a 200 m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (i.e., smaller than the stagger used in the Olympics), for the same 4 × 100 m relay race?
- Here we see a circle with radius r units. What is its perimeter? How do we find out?
- What is the connection between this question and the one about the 400 m athletics track?
- What happens to the perimeter of a square if we double its side?