Problem of the Week
Problem E and Solution
A Shady Space
A
Problem
4ABC is an equilateral triangle with sides of length 2 cm.
ALB, BM C, and CKA are arcs of circles having centres
C, A, and B, respectively. Determine the total area of the
shaded regions in the diagram.
L
B
K
D
M
C
Solution
First, determine the area of equilateral 4ABC. Construct altitude AD, shown
above. Since the triangle is equilateral, each of the angles are
60◦ . Therefore,
√
4ABD is a 30◦ - 60◦ - 90◦ triangle with sides in the ratio 1 : 3 : 2.
√
Since AB = 2 cm and BD = 1 cm, it follows that AD = 3 cm, and
area 4ABC = (BC)(AD) ÷ 2
√
= (2)( 3) ÷ 2
√
=
3 cm2
Each sector ABM C, BCKA and CALB has the same radius, 2 cm, and a 60◦
central angle. Therefore each sector has the same area, 60 ÷ 360 or one-sixth
the area of a circle of radius 2 cm.
1
2
1
area ABM C = πr2 = π(2)2 = π cm2
6
6
3
The shaded part of each sector is equal to the area of the sector minus the area
of the equilateral 4ABC.
∴ Total Shaded Area = 3 (area of any whole sector - area of the equilateral triangle)
= 3(area of sector ABM C − area 4ABC)
√
2
= 3
π− 3
3
√
= (2π − 3 3) cm2