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Surface Areas and Volumes Exercise - 12.1

Hi Friends and Champs!
As you already learn the surface areas of different bodies in the previous class, so we will not go in deep to find out the total and curved surface areas i.e. TSA and CSA of different bodies. In this chapter, we will learn to find the resulting TSA and CSA by joining or scooping of different shape bodies. I will explain it, how to find the related surface areas question by question. Let's start.
 Exercise 12.1
1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. 
Solution:
volume of cube = 64 cm³
let side of the cube = a cm
then a³ = 64
⇒ a = 4 cm
resultant cuboid length = 2a = 2×4 = 8 cm
and resultant cuboid width = a = 4 cm
so, the surface area of the resultant cuboid 
= 4 × area of one rectangle face + 2×area of square side face of resultant cuboid
= 4× (8×4) + 2×4²
= 128 + 32
=160cm² Ans.
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. 
Solution:
Let radius "r" of the hemisphere and hollow cylinder = 14/2 = 7 cm
Since total height of the vessel is 13 cm, it's implies that height "h" of the hollow cylinder
= Total height of the vessel - radius of the hemisphere
= 13 - 7
= 6 cm
Inner surface area of the vessel 
= inner surface area of the hemisphere + inner surface area of the cylinder 
= 1/2 × ( 4πr² ) + 2πrh
=1/2  × 4 × 22/7 × 7 ×7 + 2× 22/7 × 7 × 6
= 572 cm² Ans.
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.
Solution:
Height of the cone h = Total height - radius of the hemisphere
h = 15.5 - 3.5 = 12 cm
slant height of cone 
l = √[12² + (3.5)²]
=√[144 + 49/4] 
= √(625/4)
=25/2
Total surface area = curved surface area of cone + curved surface area of hemisphere
=πrl + 2πr²
=πr( l + 2r )
=22/7 × 7/2 × (25/2 + 2 × 7/2 )
=11×(25/2 + 7 )
=11×39/2 
=214.5 cm² Ans.     
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. 
Solution:
Cubical block side = 7 cm, therefore the half of the diagonal of cubical box
Greatest diameter of the hemisphere = 7 cm  Ans.
surface area of the solid 
= curved surface area of the hemisphere + 5 side surface area of the cubical box + ( 1 side surface area of the cubical box - circle surface area of the hemisphere)
= 2πr² + 5×side² + (side² - πr²)
=  πr² + 6×side² 
= 22/7 × (7/2)² + 6×7² 
= 22/7 ×49/4 + 6×49
=332.5 Ans.
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. 
Solution:
Since the diameter is equal to the edge  of the cube , therefore radius of the hemisphere
r = l/2
surface area of the cubical box = 6×side² =6l²
surface area of the curved hemisphere = 2πr² =2πl²/4=πl²/2
surface area of the remaining side = 5 side surface area of the cubical box + ( 1 side surface area of the cubical box - hemisphere circle surface area + curved surface area of the hemisphere 
=5×side² +  ( side² - πr² + 2πr² )
=5×l² + l² + πl²/4
=6l²  +πl²/4
=l²/4 (24 + π) Ans.
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.
Solution:
Total length of capsule = 14 mm
diameter of the capsule = 5 mm
so the radius of the capsule = 5/2 mm
height "h" of the cylinder = total length of the capsule - 2×raidus of the hemisphere
h=14 - 2 × 5/2 = 9 mm
Surface area of the capsule 
= curved surface area of the cylinder + 2×curved surface area of the hemisphere
= 2πrh + 2 × 2πr²
=2πr(h+2r)
=2× 22/7 × 5/2 × ( 9 +  2 × 5/2 )
=22/7 × 5 × 14
=220 mm² Ans.
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.) 
Solution:
height of the cylinder h =2.1 m
radius of the cylinder and cone = 4/2 = 2m
slant height of the cone l = 2.8 m
area of the canvas 
= curved area of the cylinder + curved area of the cone
= 2πrh + πrl
=πr(2h + l)
=22/7 × 2 × (2×2.1 + 2.8)
=44 m² Ans.
cost of the canvas 
= area of the canvas × 500
=44 × 500
= 22000 Rs. Ans.
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. 
Solution:
Cylinder height h = 2.4 cm
Cylinder radius r = 1.4/2 =0.7=7/10 cm
conical cavity height  = cylinder height
conical cavity radius  = cylinder radius 
Total surface area of the remaining solid
=Total surface area of the cylinder - circle surface area of the conical cavity + curved surface area of the conical cavity
=2πr(h + r) - πr² + πrl 
=πr(2h +  2r - r + l)
=πr(2h + r + l)
=22/7 × 0.7 × [2 × 2.4 + 0.7 + √{(2.4)² + (0.7)²}] 
=2.2 × [ 5.5 + 2.5 ]  
=2.2 × 8.0
=17.6 ≈ 18 cm² ( nearest value of 17.6 ) Ans.
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.
Solution:
Height of the cylinder h = 10 cm
Radius of the cylinder r = 3.5 cm
Total surface area of the article 
=Total surface area of the cylinder - 2×circle surface area of the hemisphere + 2× curved surface area of the hemisphere 
=2πr(h + r) - 2 × πr² + 2 × 2πr²
=2πr(h + r) + 2πr²
=2πr(h + 2r)
=2 × 22/7 × 3.5 × [10 + 2 × 3.5]   
=22 × 17
=374 cm²  Ans.


I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏
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