The figure is made up of a right-angled triangle LNP and two semicircles with LN and NP as their diameters respectively. The two semicircles and the line LP meet at R as shown. LN = 18 m. NP = 24 m and LP = 30 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 LRN
=
12 x 3.14 x 18
= 28.26 m
Circumference of Semicircle PRN
=
12 x 3.14 x 24
= 37.68 m
Perimeter of shaded region
= Circumference of semicircles + LP
= 28.26 m + 37.68 m + 30 m
= 95.94 m
(b)
Area of Triangle LNP
=
12 x 24 x 18
= 216 m
2 Area of Semicircle LRN
=
12 x 3.14 x 9 x 9
= 127.17 m
2 Area of Semicircle PRN
=
12 x 3.14 x 12 x 12
= 226.08 m
2 Area of the shaded region
= 127.17 + 226.08 - 216
= 137.25 m
2 Answer(s): (a) 95.94 m; (b) 137.25 m
2