The figure is made up of a right-angled triangle BCD and two semicircles with BC and CD as their diameters respectively. The two semicircles and the line BD meet at E as shown. BC = 10 m. CD = 24 m and BD = 26 m. (Take π = 3.14)
- Find the perimeter of the shaded region. Correct the answer to 2 decimal places.
- Find the area of the shaded region.
(a)
Circumference of Semicircle BEC
=
12 x 3.14 x 10
= 15.7 m
Circumference of Semicircle DEC
=
12 x 3.14 x 24
= 37.68 m
Perimeter of shaded region
= Circumference of semicircles + BD
= 15.7 m + 37.68 m + 26 m
= 79.38 m
(b)
Area of Triangle BCD
=
12 x 24 x 10
= 120 m
2 Area of Semicircle BEC
=
12 x 3.14 x 5 x 5
= 39.25 m
2 Area of Semicircle DEC
=
12 x 3.14 x 12 x 12
= 226.08 m
2 Area of the shaded region
= 39.25 + 226.08 - 120
= 145.33 m
2 Answer(s): (a) 79.38 m; (b) 145.33 m
2