Chapter 13: Surface Areas and Volumes

EXERCISE 13.1

1. A plastic box 1.5 m long , 1.25 m wide and 65 cm deep is to be made . It is opened at the top . Ignoring the thickness of the plastic sheet , determine :

(i) The area of the sheet required for making the box .

(ii) The cost of sheet for it, if a sheet measuring 1 m2costs Rs 20 .

Solution : Here ,

l=1.5=150100m

,

b=1.25=125100 m

And

h=65 cm=65100 m

 (i) The area of the sheet required for making the box =2lb+lh+bh-lb

=2150100×125100+150100×65100+125100×65100-150100×125100

=21875010000+975010000+812510000-1875010000

=2×3662510000-1875010000

=7325010000-1875010000

=73250-1875010000

=5450010000=5.45 m²

 

(ii) The cost of sheet

=Rs 20m2×5.45 m2

=Rs 20×5.45=Rs 109 

2. The length , breadth and height of a room are 5 m , 4 m and 3 m respectively . Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m2 .

Solution : Here , l=5 m , b=4 m and h=3 m 

 The area of the room = the lateral surface area + Area of ceiling

 =2l+bh+lb

 =25+4×3+5×4

 =2×9×3+20

 =54+20=74 m² 

The cost of white washing the walls of the room and the ceiling

=Rs 7.50m2×74 m2

=Rs 555

 

3.The floor of a rectangular hall has a perimeter 250 m . If the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000 , find the height of the hall . [Hint : Area of the four walls = Lateral surface area]

Solution :  Let l , b and h be the length, breadth and height of the rectangular hall respectively .

  ∴    2l+b=250

⇒l+b=2502

⇒l+b=125 m

Area of the four walls =2hl+b

                                        =2h×125=250h m² 

The cost of painting the four walls

=Rs10m²×250 h m²=Rs 2500 h

A/Q , 2500 h=15000

⇒h=150002500

⇒h=8 m

Therefore , the height of the hall 8 m .

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container ?

Solution :  Here , l=22.5 cm , b=10 cm and h=7.5 cm

  The area of the brick =2lb+lh+hb

=222.5×10+22.5×7.5+7.5×10

=2225+168.75+75

=2×46.875

=937.5 cm2

=937.510000=9375100000 m²

The number of brick

=9.3759375100000

=9.375×1000009735

=9735009735=100

                                                                                                       

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long , 10 cm wide and 8 cm high .

(i) Which box has the greater lateral surface area and by how much ?

(ii) Which box has the smaller total surface area and by how much ?

Solution :  (i)  Cubical box :  Here , a=10 cm  

     The lateral surface area of cubical box =4×a²=4×10×10=400 cm²

For cuboidal box : Here , l=12.5 cm m , b=10 cm m and h=8 cm    

The lateral surface area of cuboidal box =2hl+b

=2×8×12.5+10 cm²

=16×22.5 cm²

=360 cm²

Therefore , the lateral surface area of cubical box is greater by =400-360=40 cm²

(ii)   Cubical box :  Here , a=10 cm 

The surface area of cubical box =6a2

 =6×102 cm²

=6×100 cm2 

=600 cm²  

For cuboidal box : Here , l=12.5 cm  , b=10 cm and h=8 cm  

    The surface area of cuboidal box =2lb+lh+hb

=212.5×10+12.5×8+8×10cm2

=2125+100+80cm2

=2×305 cm2

=610 cm²

 Total surface area of cuboidal box is greater by =610-600cm2

=10 cm²

 

 

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape . It is 30 cm long , 25 cm wide and 25 cm high .

(i) What is the area of the glass ?

(ii) How much of tape is needed for all the 12 edges ?

Solution : (i) Here , l=30 cm  , b=25 cm and h=25 cm  

  The surface area of the small indoor greenhouse =2lh+lb+bh

=230×25+30×25+25×25cm2

=2750+750+625cm2

=2×2125 cm2

=4250 cm² 

(ii)  The sum of all edges of the glass =4l+b+h

=430+25+25 cm

=4×80 cm

=320 cm

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets . Two sizes of boxes were required . The bigger of dimensions 25 cm×20 cm×5 cm and the smaller of dimensions 15 cm×12 cm×5 cm . For all the overlaps, 5% of the total surface area is required extra . If the cost of the cardboard is Rs 4 for 1000 cm2 , find the cost of cardboard required for supplying 250 boxes of each kind .

Solution :  For bigger box : Here , l=25 cm  , b=20 cm and h=5 cm

   Area of bigger box S1=2lh+lb+bh

=225×5+25×20+20×5 cm2

=2125+500+100cm2

=2×725 cm2

=1450 cm² 

For smaller box : Here , l=15 cm  , b=12 cm and h=5 cm

   Area of bigger box S2=2lh+lb+bh

=215×5+15×12+12×5cm2

=275+180+60cm2

=2×315 cm2

=630 cm²

  Area of two box =S1+S2

=1450+630 cm2

=2080 cm² 

Area of overlap parts

=5100×2080 cm2

=10400100 cm2

=104 cm²

The total surface area of the box =2080+104cm2

                                                           =2184 cm² 

 The cost of the one box (each kind of boxes)

=Rs41000 cm2×2184 cm2

=Rs 87361000

The cost of  250 boxes

=Rs 87361000×250

=Rs 21840001000

=Rs 2184

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up) . Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m , with base dimensions 4 m×3 m ?

Solution : Here , l=4 m , b=3 m and  h=2.5 m 

The area of shelter for car by making a box-like structure with tarpaulin =2lb+lh+bh-lb

={ 24×3+4×2.5+3×2.5-4×3 } m²

= 212+10+7.5-12  m²

=2×29.5-12 m²

=59-12 m²

=47 m²

 

EXERCISE 13.2
Assume =227 , unless stated otherwise.
1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm² . Find the diameter of the base of the cylinder.

Solution : let r be the radius of the right circular cylinder . Here , h=14 cm 

A/Q , 2πrh=88

⇒2×227×r×14=88

⇒44×r×2=88

⇒88×r=88

⇒r=8888

⇒r=1 cm

And  d=2×1=2 cm 

Therefore , the diameter of the base of the cylinder is 2 cm .
2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Solution : Here , h=1 m , d=140 cm 

And

r=1402 cm

 =70 cm

=70100 m=0.7 m

The total curve surface area of the cylindrical tank =2πrr+h

=2×227×0.7×0.7+1 m2

=44×0.1×1.7 m2

=7.48 m² 

Required the metal sheet is 7.48 m² .
 

 

 

 

 

 

 

3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Solution : Here , h=77 cm  ,

r=42=2 cm

and

 R=4.42=2.2 cm

(i) The inner curved surface area =2πrh

=2×227×2×77 cm2

=88×11 cm2

=968 cm2
(ii) The outer curved surface area =2πRh

=2×227×2.2×77 cm2

=44×2.2×11 cm2

=1064.8 cm²
(iii)The total surface area = Inner CSA + Outer CSA + Area of two base

  =968+1064.8+2πR2-r2

=2032.8+2×227×2.22-22

=2032.8+4474.84-4 cm2

=2032.8+447×0.84 cm2

=2032.8+44×0.12 cm2 

=2032.8+5.28cm2 

=2038.08 cm2 

4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m² .

Solution : Here, d=84 cm ,

r=842=42 cm

and  h=120 cm 

The curve surface area of one revolution by the roller =2πrh

=2×227×42×120 cm2

=44×6×120 cm2

=31680 cm2

=3168010000 m2

=3.168 m² 

The area of 500 complete revolutions =500×3.168 m2

=1584 m² 
5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m² .

Solution: Here, d=50 cm,

r=502=25100=0.25 m

 and

h=3.5=3510=72 cm

The curve surface area of cylindrical pillar =2πrh

=2×227×0.25×72

=22×0.25

=5.50 m²

Therefore, the cost of painting the curved surface of the pillar =Rs 12.50×550

=Rs 68.75
6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Solution: let h be the height of cylinder .

Here , r=0.7 m 

A/Q, 2πrh=4.4

⇒2×227×0.7×h=4.4

⇒44×0.1×h=4.4

⇒4.4h=4.4

⇒h=4.44.4

⇒h=1 m

Therefore, the height of cylinder is 1 m .
7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m2 .
Solution:  Here, d=3.5 m ,

r=3.52=352×10=3520=74 m

and h=10 m 

(i) The inner curved surface area of well =2πrh

=2×227×74×10 m2

=22×5 m2

=110 m²

(ii)  The cost of plastering of the well =Rs 40×110

                                            =Rs 4400

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:  Here, h=28 m ,

d=5 cm=5100 m=0.05 m

And

 r=0.052 m

The curve surface area of the cylindrical pipe =2πrh

=2×227×0.052×28 m2

=22×0.05×4 m2

=22×0.20 m2

=4.4 m²

Therefore, the total radiating surface in the system is 4.4 m² .

9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 112 of the steel actually used was wasted in making the tank.

Solution: Here, d=4.2 m , r=4.22=2.1 m and  h=4.5 m 

(i) The lateral surface area of cylindrical petrol storage =2πrh

=2×227×2.1×4.5 m2

=44×0.3×4.5 m2

=59.4 m² 

(ii) let the actual area of steel used be x m² .

The area of the wasted steel

=x-112×x

=x-x12  m²

=12x-x12 m²

=11x12 m²

A/Q,

11x12=59.4

⇒x=1211×59.4

⇒x=712.811

⇒x=64.8 m²

Therefore, the actual area of steel used is 64.8 m² .
10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution: Here, =20 cm ,

 r=202=10 cm

and h=30+2.5+2.5cm=35 cm

The curve surface area of a lampshade =2πrh

=2×227×10×35 cm2

=44×10×5 cm2

=2200 cm²

Therefore, the cloth is required for covering the lampshade is 2200 cm² .
11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution:  Here, r=3 cm  and  h=10.5 cm

The curve surface area of one the penholder =2πrh+πr2  

=πr(2h+r)

=227×3(2×10.5+3) cm2

=667×21+3cm2

=667×24 cm²

=66×247 cm²

So, the curve surface area of 35 the penholder

=66×247×35 cm2

=66×24×5 cm2

=7920 cm²

Required  the cardboard for the competition is 7920 cm² .

EXERCISE 13.3
Assume =227 , unless stated otherwise.
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution: Here, d=10.5 cm ,

 r=10.52=1052×10=10520=214 cm

and  l=10 cm 

The curve surface area of a cone =πrl

=227×214×10 cm2

=11×3×5 cm2

=165 cm²

 

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution: Here, l=21 m , d=24 m  and

r=242 m=12 m

The total surface area of cone =πrl+πr2

=πrl+r

=227×1221+12m2

=227×12×33 m2

=22×12×337 m2

=87127 m2

=1244.57 m² 
3. Curved surface area of a cone is 308 cm²  and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.

Solution:  let r be the radius of a cone .

Here, l=14 cm 

A/Q , πrl=308

⇒227×r×14=308

⇒r=308×722×14

⇒r=30822×2=30844=7 cm

⇒r=7 cm 

Therefore, the radius of a cone is 7 cm .

(ii)  Total surface area of the cone =πrl+πr2

=308+227×7×7 cm²

=308+22×7cm2

=308+154cm2

=462 cm²
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70 .

Solution:  Here, h=10 m and  r=24 m 

(i) The slant height of the tent l=r2+h2

=242+102

=576+100

=676

=26²=26 m 

(ii)  Here, h=10 m , l=26 m and  r=24 m

The curve surface area of a canvas =πrl

=227×24×26 m² 

The cost of the canvas required to make the tent

=Rs 70×227×24×26

=Rs 10×22×24×26

=Rs 137280 

 
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

Solution: Here, h=8 m and  r=6 m 

The slant height l=r2+h2

=62+82  m

=36+64 m

=100=10 m

The curve surface area of the conical tent =πrl

=3.14×6×10 m² 

The length of tarpaulin

=3.14×6×10 m23 m

=3.14×2×10 m

=62.8 m 

Therefore, the total length of tarpaulin =62.8 m+20 cm

=62.8+20100m

=62.8+0.2 m

=63 m
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m² .

Solution: Here, =25 m  ,  d=14 m and

r=142=7 m

The curve surface area of a conical tomb =πrl

=227×7×25 m2

=22×25 m2

=550 m²

Therefore , the cost of white-washing of a conical tomb

=Rs210100×550

= Rs 21×55

=Rs 1155 
7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:  Here, r=7 cm  and h=24 cm 

The slant height l=r2+h2

=72+242

=49+576

=625

=25²=25 cm 

The curve surface area of a joker’s caps =πrl

=227×7×25 cm2

=22×25 cm2

=550 cm² 

Therefore, the area of the sheet required to make 10 joker’s caps =550×10 cm2

=5500 cm²

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m² , what will be the cost of painting all these cones? (Use π=3.14 and take  1.04=1.02 ) .

Solution: Here, d=40 cm ,

r=402cm=20 cm=20100 cm=0.2 m

and h=1 m 

The slant height l=0.22+12

=0.04+1

=1.04=1.02 m

The curve surface area of a cone =πrl

=3.14×0.2×1.02 m2

=0.64056 m2

The curve surface area of 50 cones

=50×0.64056 m2

=32.028 m2

Therefore, the cost of painting of 50 cones

=Rs 12m2×32.028 m2

=Rs 12×32.028

=Rs 384.34  (approx.)