Exercise 12.1 (NCERT) – Surface Areas and Volumes, Class 10 – All Solutions

24 Feb
Class 10, Class 10 - Chapter 12, Class 10 - Surface Areas and Volumes, , , , ,
Ex: 8.3, Class 10
Ex: 14.1, Class 10

Exercise 12.1 (NCERT) - Surface Areas and Volumes, Class 10

Chapter 12 Exercise 12.1 Class 10

Below are the Quick links for all questions of Exercise 12.1, Surface Areas and Volumes, Class 10.

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Below are the solutions of every question of exercise 12.1 of chapter 12, Surface Areas and Volumes, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

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

Volume of one cubeLet side of cube⇒Volume of cube=64𝑐𝑚³  …(1)=𝑎=𝑎³  …(2)

From (1) and (2) 

𝑎³⇒𝑎=64=4𝑐𝑚

=> Each side of both cubes is 4cm because the volumes of both cubes is 64 cm³.

This means joining 2 cubes will make them a cuboid of edges of lengths: 4cm, 4cm and 8cm.

Surface Area of Cuboid

Surface Area (SA) of cuboid=2×(𝑙𝑏+𝑏ℎ+ℎ𝑙)=2×(4×4+4×8+8×4)=2×(16+32+32)=2×80=160𝑐𝑚2

SA – Surface Area

Question 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.

The vessel is made of two objects: Hollow hemisphere & Hollow cylinder

Hemisphere

Diameter (d) = 14cm 

Radius (r) = d/2 = 14/2 = 7cm

Height = Radius = 7cm

Cylinder

Radius of Cylinder = Radius of Hemisphere = 7cm 

Because the cylinder is mounted on the hemisphere. 

Let height of cylinder = h 

ht. of vessel⇒13⇒ℎ=ht. of Cylinder+ht. of Hemisphere=ℎ+7=13–7=6𝑐𝑚

Inner Surface Area of Vessel

The cylinder is hollow and mounted on top of the hemisphere. Hence it has no top and base.

=> Inner SA of cylinder = 2𝝅rh

Inner SA of vessel=Inner CSA of Cylinder+Inner SA of Hemisphere=2𝜋𝑟ℎ+2𝜋𝑟2=2𝜋𝑟(ℎ+𝑟)=2×227×7×(6+7)=44×13=572𝑐𝑚2

SA – Surface Area, CSA – Curved Surface Area

Question 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.

The toy is made of two objects: Cone & Hemisphere

Hemisphere

Radius = r = 3.5 cm

Height = Radius = 3.5 cm

Why ht. of hemisphere = radius of hemisphere?

Watch the video embedded below. 

Cone

Radius = r = 3.5 cm 

Let ht. of Cone = h

ht. of Toy⇒15.5⇒ℎ=ht. of Cone+ht. of Hemisphere=ℎ+3.5=15.5–3.5=12𝑐𝑚

Slant ht. of Cone

Slant ht. of Cone (𝑙)=ℎ2+𝑟2‾‾‾‾‾‾‾√=122+3.52‾‾‾‾‾‾‾‾‾‾√=144+12.25‾‾‾‾‾‾‾‾‾‾‾√=156.25‾‾‾‾‾‾√=12.5𝑐𝑚

Surface Area of Toy

SA of toy=SA of Cone+SA of Hemisphere=𝜋𝑟𝑙+2𝜋𝑟2=227×3.5×12.5+2×227×3.52=137.5+77=214.5𝑐𝑚2

SA – Surface Area

Question 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.

Hemisphere with greatest diameter should touch all edges of the cube.

Largest Hemisphere

Diameter = Side of Cube = 7cm

Radius (r) = 7/2 = 3.5 cm

Surface Area of Solid

SA of the Solid=SA of Cube+SA of H.S.–Area of base of H.S.=6𝑎2+2𝜋𝑟2–𝜋𝑟2=6𝑎2+𝜋𝑟2=6×72+227×3.52=294+38.5=332.5𝑐𝑚2

H.S. – Hemisphere, SA – Surface Area

Base of the hemisphere hides the equal area from the top of the cube. That’s why we have to subtract πr2. 

NOTE: Next question is similar except the hemisphere is inside the cube.

Question 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.

Edge of CubeDiameter of HemisphereRadius of Hemisphere=𝑙=𝑙=𝑙2

Surface Area of Solid

SA of the Solid=SA of Cube+SA of H.S.–Area of base of H.S.=6𝑙2+2𝜋(𝑙2)2–𝜋(𝑙2)2=6𝑙2+𝜋(𝑙2)2=6𝑙2+𝜋𝑙24=𝑙24(24+𝜋)

H.S. – Hemisphere, SA – Surface Area

Question 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.

Shape of Capsule = Hemisphere 1 + Cylinder + Hemisphere 2

Diameter of Capsule = 5 mm

Radius of Capsule = 2.5 mm

Radius of Hemisphere 1 = 2.5 mm

Radius of Cylinder = 2.5 mm

Radius of Hemisphere 2 = 2.5 mm

Length of Capsule = 14 mm

Length of H.S.1 = Radius of H.S. 1 = 2.5 mm

Length of H.S.2 = Radius of H.S. 2 = 2.5 mm

Let length of Cylinder = h

Length of Capsule1414⇒ℎ=Length of H.S. 1+Length of Cylinder+Length of H.S. 2=2.5+ℎ+2.5=ℎ+5=14–5=9𝑚𝑚

Surface Area of Capsule

SA of Capsule=SA of H.S. 1+CSA of Cylinder+SA of H.S. 2=2𝜋𝑟2+2𝜋𝑟ℎ+2𝜋𝑟2=4𝜋𝑟2+2𝜋𝑟ℎ=2𝜋𝑟(2𝑟+ℎ)=2×227×2.5×(2×2.5+9)=1107×(5+9)=1107×14=110×2=220𝑚𝑚2

H.S. – Hemisphere, SA – Surface Area, CSA – Curved Surface Area

Question 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 Rs. 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Shape of Tent = Cylinder + Cone

Cylinder

Height, h = 2.1 m 

Diameter = 4 m

Radius, r = 2 m

Cone

Slant height, l = 2.8 m

Radius of Cone = Radius of Cylinder, r = 2 m

Because the Cone is surmounting the Cylinder.

Area of required Canvas = Surface Area of Tent

SA of Tent=CSA of Cylinder+CSA of Cone=2𝜋𝑟ℎ+𝜋𝑟𝑙=𝜋𝑟(2ℎ+𝑙)=227×2×(2×2.1+2.8)=447×(4.2+2.8)=447×7=44𝑚2

Cost of Canvas

Cost of 1𝑚2⇒Cost of 44𝑚2=500=44×500=22,000  𝑅𝑠.

SA – Surface Area, CSA – Curved Surface Area

Question 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 .

Cylinder

Height, h = 2.4 cm

Diameter = 1.4 cm

Radius, r = 1.4 cm

Cone

The Cone (Conical cavity) has same diameter and height as that of the Cylinder, hence

Height, h = 2.4 cm

Radius, r = 1.4 cm

Surface Area of Solid

SA of Solid=CSA of Cylinder+Area of Base of Cylinder+CSA of Cone=CSA of Cylinder+Area of Circle+CSA of Cone=2𝜋𝑟ℎ+𝜋𝑟2+𝜋𝑟𝑙=𝜋𝑟(2ℎ+𝑟+𝑙)=227×0.7×(2×2.4+0.7+2.5)=227×0.7×(4.8+3.2)=227×0.7×8=17.6𝑐𝑚2

SA – Surface Area, CSA – Curved Surface Area

Question 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.

Radius of all 3 shapes is 3.5 cm.

=> Radius r = 3.5 cm

(For both hemispheres and cylinder)

Height of Cylinder, h = 10 cm

Surface Area of Article

SA of the Article=CSA of Cylinder+Inner SA of H.S. 1+Inner SA of H.S. 2=2𝜋𝑟ℎ+2𝜋𝑟2+2𝜋𝑟2=2𝜋𝑟ℎ+4𝜋𝑟2=2𝜋𝑟(ℎ+2𝑟)=2×227×3.5×(10+2×3.5)=22×(10+7)=22×17=374𝑐𝑚2

SA – Surface Area, CSA – Curved Surface Area, H.S. – Hemisphere

Ex: 8.3, Class 10
Ex: 14.1, Class 10