Questions
- 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:
- Calculate :-
- The area of one end of the roof
- The area of the curved surface of the roof
- 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
- Calculate :-
- The diagram below, not drawn to scale, is a regular pengtagon circumscribed in a circle of radius 10cm at centre O
Find;- The side of the pentagon
- The area of the shaded region
- Triangle PQR is inscribed in he circle PQ= 7.8cm, PR = 6.6cm and QR = 5.9cm.
Find:- The radius of the circle, correct to one decimal place
- The angles of the triangle
- The area of shaded region
- 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
Answers
- A = 120/360 x πx 102 – ½ x 100 x10sin 12
= 104. 72 – 43.30 = 61.42m2 - 120/360 x 2 x 10 x 20
= 418.9m2
- A = 120/360 x πx 102 – ½ x 100 x10sin 12
- Total area = 61.42 + 61.42 + 418.9
= 541.74m2
Cost = 541.74 x 310 = 167,939
- 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 - Area of circle = 22/7 x 10 x 10 = 314.8
Shaded region = 3/5 (3.143 – 237.8)
= 45.9cm2
- Cos 54° = x/10
- 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 - 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 - Area of shaded region
= 3.142 x 42 – ½ x 6.6 x 5.9sin 77
= 50.27 – 18.97 = 31.30
- 7.82 = 6.62 + 5.92 – 2 x 6.6 x 5.9 cos R
- (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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