Questions
- A lampshade is in the form of a frustrum of a cone. Its bottom and top diameters are
12cm and 8cm respectively. Its height is 6cm. Find;- The area of the curved surface of the lampshade
- The material used for making the lampshade is sold at Kshs.800 per square metre.
Find the cost of ten lampshades if a lampshade is sold at twice the cost of the material.
- A cylindrical piece of wood of radius 4.2cm and length 150cm is cut lengthwise into two
equal pieces. Calculate the surface area of one piece. - The base of an open rectangular tank is 3.2m by 2.8m. Its height is 2.4m. It contains water to a depth of 1.8m. Calculate the surface area inside the tank that is not in contact with water.
- The figure below represents a model of a solid structure in the shape of frustrum of a cone with a hemisphere top. The diameter of the hemispherical part is 70cm and is equal to the diameter of the top of the frustrum. The frustrum has a base diameter of 28cm and slant height of 60cm.
Calculate :- the area of the hemispherical surface
- the slant height of cone from which the frustrum was cut
- the surface area of frustrum
- the area of the base
- the total surface area of the model
- A room is 6.8m long, 4.2m wide and 3.5m high. The room has two glass doors each measuring 75cm by 2.5m and a glass window measuring 400cm by 1.25m. The walls are to be painted except the window and doors.
- Find the total area of the four walls
- Find the area of the walls to be painted
- Paint A costs Shs.80 per litre and paint B costs Shs.35 per litre. 0.8 litres of A covers an area of 1m2 while 0.5m2 uses 1 litre of paint B. If two coats of each paint are to be applied. Find the cost of painting the walls using:
- Paint A
- Paint B
- If paint A is packed in 400ml tins and paint B in 1.25litres tins, find the least number of tins of each type of paint that must be bought.
- The figure below shows a solid frustrum of pyramid with a square top of side 8cm and
a square base of side 12cm. The slant edge of the frustrum is 9cm.
Calculate:- the total surface area of the frustrum
- the volume of the solid frustrum
- the angle between the planes BCHG and the base EFGH.
Answers
-
-
x = 4/6
x+6
6x = 4x + 24
x = 12 cm
L = √(122 + 42)
= √160
= 12.65 (2 d.p)
L = √(182 + 62)
√360
= 18.97
SA = Π(RL – rL)
= 3.142 (6 x 18.97 – 4 x 12.65)
= 3.142 x 63.22 = 198.64 cm2 - Cost of material for one lamp shape
= 198.64 x 800
10000
= Sh15.90
Cost of 10 lamp shape = 2 x 10 x 15.90 = sh 318
-
- Area of the remaining cross-section
= 4.22 x Π
= (17.64Π)cm2
Area of the curved surface
= (8.4Π x 150
= 1260Π cm2
2
Area of the flat surface
= (150 x 8.4)cm2
=1260cm2
Total area = (1260 + 630Π + 17.64Π
= (1260 + 647.64Π)cm2
= 3295cm2/ 3295.44cm2 - Surface area = 2(0.6 x 2.8)m2 + 2(0.6 x 3.2)m2
= (3.36 + 3.84)m2
= 7.2m2 -
- Area of hemispherical part
= ½ X 4 UR2
= 2 X 22/7 x 35 X 35
= 7700cm2 - Slant height for original cone
L = 35/14
L – 60
L = 100cm - Surface area of frustrum
= URL – url
= 22/7 X 35 x 100 – 22/7 x 14 X 40
= 11000 – 1760 = 9240 cm2 - Area of base
22/7 X 142 = 616 cm2 - Total surface
= 7700 + 9240 + 616 = 17556cm2
- Area of hemispherical part
-
- TA = 2 X 6.8 X 3.5 + 2 X 4.2 X 3.5m2
= 47.6 +29.4 m2 = 77m2 - 77 – (75/100 X 2.5 X 2 + 400/100 X 1.25)m2
77 – (3.75 + 5) m2
77 – 68.25 m2 = 8.75m2 -
- Cost of paint A
= 68.25 X 0.8 X 80 = Kshs.43681 - Cost of paint B
68.25 X 35
0.5
= Kshs.4777.5
- Cost of paint A
- No of tins
= 54.6 X 1000
400
= 136.5 = 137 tins
No. of tins
= 136.5
1.25
= 109.2 = 110 tins
- TA = 2 X 6.8 X 3.5 + 2 X 4.2 X 3.5m2
- Top surface area = 8x8 =64cm2
Bottom surface area = 12x12=144cm2
Height of slanting faces
H = 92 – 22 = 8.775cm
Area of slanting face = ½ (12 + 8) x 8.775 x 4
= 351cm2
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