Library / /460 of 482

, 2026

Geometry Difficulty 3.8 AMC 10/12 Find the answer Canada

On each side of the square shown, a semi-circle is drawn inside
the square. The side length of the square is $10$ and is equal to the diameter of each semi-circle. The four semi-circles overlap to form the shaded four-petal flower.If $n$ is the closest integer to the area of the shaded flower, what is the value of $n$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Label the centre of the square $A$ , the midpoint of the base $B$ , and the bottom-right corner of the square $C$ . Then $\triangle ABC$ has a right angle at $B$ and $AB=CB$ , as shown. [[IMAGE0]] The flower is made up of four "petals", each of which is made up of two copies of the region formed by taking a sector of the circle of angle $90\degree$ and removing the right-isosceles triangle formed by the two radii. The radius of each circle is $\dfrac{10}{2}=5$ , so the area of each sector is $\dfrac{\pi(5)^2}{4}=\dfrac{25\pi}{4}$ . The area of the triangle being removed is $\dfrac{1}{2}(5)^2=\dfrac{25}{2}$ . Therefore, the area of each petal is $\$2\times\left(\dfrac{25\pi}{4}-\dfrac{25}{2}\right) = \dfrac{25\pi-50}{2}$ . The area of the flower is

Figure for this problem 4 times the area of each petal, which is

Figure for this problem $\text{4 ( 25 -50 2 ) = 2 (25 - 50) = 50 -100 57.08}$ The integer closest to the area of the flower is

Figure for this problem 57$.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.