# Surface Area of Solids Questions and Answers - Form 2 Topical Mathematics

## Questions

1. 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;
1. The area of the curved surface of the lampshade
2. 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.
2. 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.
3. 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.
4. 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 :
1. the area of the hemispherical surface
2. the slant height of cone from which the frustrum was cut
3. the surface area of frustrum
4. the area of the base
5. the total surface area of the model
5. 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.
1. Find the total area of the four walls
2. Find the area of the walls to be painted
3. 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:
1. Paint A
2. Paint B
4. 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.
6. 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:
1. the total surface area of the frustrum
2. the volume of the solid frustrum
3. the angle between the planes BCHG and the base EFGH.

1.
1.

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
2. 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
2. 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
3. Surface area = 2(0.6 x 2.8)m2 + 2(0.6 x 3.2)m2
= (3.36 + 3.84)m2
= 7.2m2
4.
1. Area of hemispherical part
= ½ X 4 UR2
= 2 X 22/7 x 35 X 35
= 7700cm2
2. Slant height for original cone
= 35/14
L – 60
L = 100cm
3. Surface area of frustrum
= URL – url
= 22/7 X 35 x 100 – 22/7 x 14 X 40
= 11000 – 1760 = 9240 cm2
4. Area of base
22/7 X 142 = 616 cm2
5. Total surface
= 7700 + 9240 + 616 = 17556cm2
5.
1. TA = 2 X 6.8 X 3.5 + 2 X 4.2 X 3.5m2
= 47.6 +29.4 m2 = 77m2
2. 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
3.
1. Cost of paint A
= 68.25 X 0.8 X 80 = Kshs.43681
2. Cost of paint B
68.25 X 35
0.5
= Kshs.4777.5
4. No of tins
= 54.6 X 1000
400
= 136.5 = 137 tins
No. of tins
= 136.5
1.25
= 109.2 = 110 tins
6. 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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