A friend sent me this.
ABCD is a rectangle with sides 8cm and 4cm. Taking each of the 8cm sides as diameter, two semicircles are drawn as shown in picture above. Find the area of the common shaded portion. I have modified the text in the picture above slightly to make the problem clearer.
I marked points E, F, G, H on the common area's perimeter and point O which is at the center of the area.
DFC is a semicircle with CD as the diameter and AHB is another semicircle with AB as its diameter.
Solution:
AD = 4cm
DC = 8cm
FO = 1/2 of FH (because of symmetry) = 2cm
The shaded area EOH (with EH being a part of the circumference) is 1/4 of the total common or shaded area.
FE and FH are both equal to radius of circle = FA = 4cm
In right angled triangle FEO, FE = 4cm, FO = 2cm, EO can be found out.
Area of right angled triangle EOF can be determined. If the shaded area to the north west of the chord EH can be determined then the shaded area EOH can be determined which is the area of the right angled triangle EOH and the shaded area to the NW of EH. And the total shaded area = 4 * shaded area EOH.
So all that remains is to find out the area to the NW of EH.
In triangle FEH, FE and FH are both 4 cm (both being radius) and hence it's an isosceles triangle. EH is also a radius and hence equals 4cm. EFH becomes an equilateral triangle of side 4cm. And hence its area can be determined. Area of EFH which is a part of the circle = 60/360 * area of circle. 60 being each of the angles in the triangle = 1/6 area of circle whose radius =4cm
Now subtract the area of the equilateral triangle from 1/6 part of circle and the reminder gives the area to the NW of EH. Now add the area of right angled triangle EOH. The sum gives 1/4 of the common area. Multiply by 4 to get the total shaded area.