Chapter 12 Surface Areas and Volumes Exercise-01

EXERCISE 12.1

Unless stated otherwise, take π=227.

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

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2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

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Solution

It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7cm ).

Height of hemispherical part = Radius =7cm

Height of cylindrical part (h)=137=6cm

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part

=2πrh+2πr2

Inner surface area of vessel =2×227×7×6+2×227×7×7

=44(6+7)=44×13=572cm2

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

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Solution

It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5cm ).

Height of hemispherical part = Radius (r)=3.5=72cm

Height of conical part (h)=15.53.5=12cm

Slant height (l) of conical part =r2+h2

=(72)2+(12)2=494+144=49+5764=6254=252

Total surface area of toy = CSA of conical part + CSA of hemispherical part

=πrl+2πr2=227×72×252+2×227×72×72=137.5+77=214.5cm2

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

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Solution

From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm.

Radius (r. ) of hemispherical part =72=3.5cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

  • Area of base of hemispherical part

=6( Edge )2+2πr2πr2=6( Edge )2+πr2

Total surface area of solid =6(7)2+227×72×72

=294+38.5=332.5cm2

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

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Solution

Diameter of hemisphere = Edge of cube =I

Radius of hemisphere =l2

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

  • Area of base of hemispherical part

=6( Edge )2+2πr2πr2=6( Edge )2+πr2

Total surface area of solid =6l2+π×(l2)2

=6l2+πl24=14(24+π)l2 unit 2

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Fig. 12.10

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Solution

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical

part

= Diameter of the capsule 2=52

Length of cylindrical part (h)= Length of the entire capsule 2×r

=145=9cm

Surface area of capsule =2× CSA of hemispherical part + CSA of cylindrical part

=2×2πr2+2πrh=4π(52)2+2π(52)(9)=25π+45π=70πmm2=70×227=220mm2

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)

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Solution

Given that,

Height (h) of the cylindrical part =2.1m

Diameter of the cylindrical part =4m

Radius of the cylindrical part =2m

Slant height (I) of conical part =2.8m

Area of canvas used = CSA of conical part + CSA of cylindrical part =πrl+2πrh

=π×2×2.8+2π×2×2.1

=2π[2.8+2×2.1]=2π[2.8+4.2]=2×227×7

=44m2

Cost of 1m2 canvas = Rs 500

Cost of 44m2 canvas =44×500=22000

Therefore, it will cost Rs 22000 for making such a tent.

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

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Solution

Given that,

Height (h) of the conical part = Height (h) of the cylindrical part =2.4cm

Diameter of the cylindrical part =1.4cm

Therefore, radius (r) of the cylindrical part =0.7cm

Slant height (l) of conical part =r2+h2

=(0.7)2+(2.4)2=0.49+5.76=6.25=2.5

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

=2πrh+πrl+πr2

=2×227×0.7×2.4+227×0.7×2.5+227×0.7×0.7

=4.4×2.4+2.2×2.5+2.2×0.7=10.56+5.50+1.54=17.60cm2

The total surface area of the remaining solid to the nearest cm2 is 18cm2.

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Fig. 12.11

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Solution

Radius (r) of cylindrical part = Radius (r) of hemispherical part =3.5cm

Height of cylindrical part (h)=10cm

Surface area of article = CSA of cylindrical part +2× CSA of hemispherical part

=2πrh+2×2πr2

=2π×3.5×10+2×2π×3.5×3.5

=70π+49π

=119π

=17×22=374cm2



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