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 = 6 m. DE = 8 m and CE = 10 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 6
= 9.42 m
Circumference of Semicircle EFD
=
12 x 3.14 x 8
= 12.56 m
Perimeter of shaded region
= Circumference of semicircles + CE
= 9.42 m + 12.56 m + 10 m
= 31.98 m
(b)
Area of Triangle CDE
=
12 x 8 x 6
= 24 m
2 Area of Semicircle CFD
=
12 x 3.14 x 3 x 3
= 14.13 m
2 Area of Semicircle EFD
=
12 x 3.14 x 4 x 4
= 25.12 m
2 Area of the shaded region
= 14.13 + 25.12 - 24
= 15.25 m
2 Answer(s): (a) 31.98 m; (b) 15.25 m
2