The figure is made up of a right-angled triangle CDE and two semicircles with CD and DE as their diameters respectively. The two semicircles and the line CE meet at F as shown. CD = 12 m. DE = 16 m and CE = 20 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 CFD
=
12 x 3.14 x 12
= 18.84 m
Circumference of Semicircle EFD
=
12 x 3.14 x 16
= 25.12 m
Perimeter of shaded region
= Circumference of semicircles + CE
= 18.84 m + 25.12 m + 20 m
= 63.96 m
(b)
Area of Triangle CDE
=
12 x 16 x 12
= 96 m
2 Area of Semicircle CFD
=
12 x 3.14 x 6 x 6
= 56.52 m
2 Area of Semicircle EFD
=
12 x 3.14 x 8 x 8
= 100.48 m
2 Area of the shaded region
= 56.52 + 100.48 - 96
= 61 m
2 Answer(s): (a) 63.96 m; (b) 61 m
2