Tadpoles
The diagram shows a shaded shape bounded by circular arcs. What is the difference in area betweeen this and the equilateral triangle shown?
Age
14 to 16
Challenge level
Problem
The diagram shows a shaded shape bounded by circular arcs with the same radius. The centres of three arcs are the vertices of an equilateral triangle; the other three centres are the midpoints of the sides of the triangle. The sides of the triangle have length 2.
What is the difference between the area of the shaded shape and the area of the triangle?
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If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.
Student Solutions
Before you read through the full solution, have a look at the following diagram.
Can you see how one might use it in the solution?
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The length of the side of the triangle is equal to four times the radius of the arcs. So the arcs have radius $2\div 4 = \frac{1}{2}.$ In the diagram above, three semicircles have been shaded dark grey. The second diagram shows how these semicircles may be placed inside the triangle so that the whole triangle is shaded.
Therefore, the difference between the area of the shaded shape and the area of the triangle is the sum of the areas of three sectors of a circle. The interior angle of an equilateral triangle is $60^\circ$, so the angle at the centre of each sector is $180^\circ-60^\circ=120^\circ.$
Therefore, each sector is equal in area to one-third of the area of a circle. Their combined area is equal to the area of a circle of radius $\frac{1}{2}.$ So the required area is $\pi\times\left(\frac{1}{2}\right)^2 = \frac{\pi}{4}.$
Therefore, the difference between the area of the shaded shape and the area of the triangle is the sum of the areas of three sectors of a circle. The interior angle of an equilateral triangle is $60^\circ$, so the angle at the centre of each sector is $180^\circ-60^\circ=120^\circ.$
Therefore, each sector is equal in area to one-third of the area of a circle. Their combined area is equal to the area of a circle of radius $\frac{1}{2}.$ So the required area is $\pi\times\left(\frac{1}{2}\right)^2 = \frac{\pi}{4}.$
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