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Area of part of a circle Questions

 

1.  The figure below shows a circle of radius 9cm and centre O. Chord AB is 7cm long. Calculate the area of the shaded region. (4mks)

 

Image From EcoleBooks.comImage From EcoleBooks.com

Image From EcoleBooks.com

 

2.  The figure below shows two intersecting circles with centres P and Q of radius 8cm and 10cm respectively. Length AB = 12cm

Image From EcoleBooks.comImage From EcoleBooks.com

Calculate:

a)  Image From EcoleBooks.com APB  (2mks)

b)  Image From EcoleBooks.comAQB  (2mks)

c)  Area of the shaded region  (6mks)

3.

 

 

ecolebooks.com

 

 

 

 

 

The diagram above represents a circle centre o of radius 5cm. The minor arc AB subtends an angle of 1200 at the centre. Find the area of the shaded part. (3mks)

4.  The figure below shows a regular pentagon inscribed in a circle of radius 12cm, centre O.  

Image From EcoleBooks.com

 Calculate the area of the shaded part. (3mks)

5.  Two circles of radii 13cm and 16cm intersect such that they share a common chord of length  20cm. calculate the area of the shaded part. Image From EcoleBooks.com (10mks)

Image From EcoleBooks.com

6.  Find the perimeter of the figure below, given AB,BC and AC are diameters. (4mks)

 

Image From EcoleBooks.com

Image From EcoleBooks.com7.  The figure below shows two intersecting circles. The radius of a circle A is 12cm and that of circle B is 8 cm.

 

 

 

 

 

 

 

 

 

If the angle MBN = 72o, calculate

  1. The size of the angle MAN

b)  The length of MN

c)  The area of the shaded region.

8.

Image From EcoleBooks.com

In the diagram above, two circles, centres A and C and radii 7cm and 24cm respectively intersect at B and D. AC = 25cm.

a) Show that angel ABC = 900

b) Calculate

i) the size of obtuse angel BAD

ii) the area of the shaded part  (10 Mks)

9.  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:

 

 

 

 

 

 

 (a) Calculate :

  (i) The area of one end of the roof

  (ii) The area of the curved surface of the roof

 (b) 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  

 

10.  The diagram below, not drawn to scale, is a regular pengtagon circumscribed in a circle

of radius 10cm at centre O

Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.com

Find;

Image From EcoleBooks.com  (a) The side of the pentagon

 (b) The area of the shaded region

 

 

 

11.  Triangle PQR is inscribed in he circle PQ= 7.8cm, PR = 6.6cm and QR = 5.9cm. Find:

 

 

 

 

 

 

 

 

 

 

 

 (a) The radius of the circle, correct to one decimal place

 (b) The angles of the triangle

 (c) The area of shaded region

 

12.  The figure below represents sector OAC and OBD with radius OA and OB respectively.

Image From EcoleBooks.comGiven that OB is twice OA and angle AOC = 60o. Calculate the area of the shaded region

Image From EcoleBooks.com in m2, given that OA = 12cm

Image From EcoleBooks.com

 

 

Image From EcoleBooks.comImage From EcoleBooks.com

 

 

Image From EcoleBooks.com

 

Image From EcoleBooks.com

 

Area of part of a circle Answers

1.  (a) A = 120 x πx 102 – ½ x 100 x10sin 12

 360

= 104. 72 – 43.30 = 61.42m2

 

(b) (ii) 120 x 2 x 10 x 20

  360

= 418.9m2

 

(b) Total area = 61.42 + 61.42 + 418.9

= 541.74m2

  Cost = 541.74 x 310  = 167,939

 

 

2.  a) Cos 54° = x/10

   X = 5.878

 size = 2 x 5.878 = 11.756

Area of  = ½ x 102 sin 72° = 47.55

Total area of s = 47.55 x 5 = 237.8cm2

 

b) Area of circle = 22/7 x 10 x 10 = 314.8

 

 Shaded region = 3/5 (3.143 – 237.8)

= 45.9cm2

 

 

3.  (a) 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 = 2r

  Sin 77

 

r = 7.8

  2 sin 77

  = 4 cm

 

(b) 5.9 = 7.8

   Sin p Sin 77

Sin P = 5.9 sin 77

  7.8

   = 0.7370

 ∠ P= 47.5o

∠ Q = 180 – (77 + 47.5)  = 55.5o

 

(c) Area of shaded region

= 3.142 x 42 – ½ x 6.6 x 5.9sin 77

= 50.27 – 18.97 = 31.30

 

4.  (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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