# Area of Part of a Circle Questions and Answers - Form 2 Topical Mathematics

## Questions

1. The ends of the roof of a workshop are segments of a circle of radius 10m. The roof is 20m long. The angle at the centre of the circle is 120o as shown in the figure below:

1. Calculate :-
1. The area of one end of the roof
2. The area of the curved surface of the roof
2. What would be the cost to the nearest shilling of covering the two ends and the curved surface with galvanized iron sheets costing shs.310 per square metre
2. The diagram below, not drawn to scale, is a regular pengtagon circumscribed in a circle of radius 10cm at centre O

Find;
1. The side of the pentagon
2. The area of the shaded region
3. Triangle PQR is inscribed in he circle PQ= 7.8cm, PR = 6.6cm and QR = 5.9cm.

Find:
1. The radius of the circle, correct to one decimal place
2. The angles of the triangle
3. The area of shaded region
4. The figure below represents sector OAC and OBD with radius OA and OB respectively. Given that OB is twice OA and angle AOC = 60o. Calculate the area of the shaded region in m2, given that OA = 12cm

1. A = 120/360 x πx 102 – ½ x 100 x10sin 12
= 104. 72 – 43.30 = 61.42m
2
2. 120/360 x 2 x 10 x 20
= 418.9m2
1.  Total area = 61.42 + 61.42 + 418.9
= 541.74m2
Cost = 541.74 x 310 = 167,939
1. Cos 54° = x/10
X = 5.878
size = 2 x 5.878 = 11.756
Area of Δ= ½ x 102 sin 72° = 47.55
Total area of Δs = 47.55 x 5 = 237.8cm2
2. Area of circle = 22/7 x 10 x 10 = 314.8
Shaded region = 3/5 (3.143 – 237.8)
= 45.9cm2
1. 7.82 = 6.62 + 5.92 – 2 x 6.6 x 5.9 cos R

Cos R = 6.62 + 5.92 – 7.82/(2 x 6.6 x 5.9)
= (78.37 – 60.84)/77.88
= 0.2251
R = 77o
7.8/Sin 77 = 2r
r = 7.8/2 sin 77
= 4 cm
2. 5.9/Sin p = 7.8/Sin 77

Sin P = 5.9 sin 77/7.8
= 0.7370
P= 47.5o
Q = 180 – (77 + 47.5) = 55.5o
= 3.142 x 42 – ½ x 6.6 x 5.9sin 77
= 50.27 – 18.97 = 31.30
1. (60/360 x 22/7 x 24 x 24) – (60/360 x 22/7 x12 x 12)
301.71 – 75.43 = 226.26

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