Class 10 - Chapter 12 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-12/ Mathematics is like music - Enjoy it!!! Thu, 12 Jun 2025 12:50:56 +0000 en-US hourly 1 https://wordpress.org/?v=7.0.2 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Class 10 - Chapter 12 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-12/ 32 32 Exercise 12.1 (NCERT) – Surface Areas and Volumes, Class 10 – All Solutions https://mathemafia.com/exercise-12-1-ncert-surface-areas-and-volumes-class-10-all-solutions/ Mon, 24 Feb 2025 12:02:03 +0000 https://mathemafia.com/?p=12962 Complete solution of Exercise 12.1, Chapter 12, Surface Areas and Volumes of NCERT book of Mathematics for Class 10. Animated gifs and colourful interactive images are used to explain the objects explained in the questions. Easily readable sections and embedded YouTube tutorials will help students understand the solutions quickly.

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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.

Click a link to view the solution of the corresponding question.

Below are the solutions of every question of exercise 12.1Β 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.Β Β 

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 cubeβ‡’Volume of cube=64π‘π‘šΒ³  β€¦(1)=π‘Ž=π‘ŽΒ³  β€¦(2)

From (1) and (2)Β 

π‘ŽΒ³β‡’π‘Ž=64=4π‘π‘š

=> Each side of both cubes is 4cm because the volumes of both cubes is 64 cmΒ³.

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=2Γ—(𝑙𝑏+π‘β„Ž+β„Žπ‘™)=2Γ—(4Γ—4+4Γ—8+8Γ—4)=2Γ—(16+32+32)=2Γ—80=160π‘π‘š2

SA – Surface Area

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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) = 14cmΒ 

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

Height = Radius = 7cm

Cylinder

Radius of Cylinder = Radius of Hemisphere = 7cmΒ 

BecauseΒ the cylinder is mounted on the hemisphere.Β 

Let height of cylinder = hΒ 

ht. of vesselβ‡’13β‡’β„Ž=ht. of Cylinder+ht. of Hemisphere=β„Ž+7=13–7=6π‘π‘š

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=2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ2=2πœ‹π‘Ÿ(β„Ž+π‘Ÿ)=2Γ—227Γ—7Γ—(6+7)=44Γ—13=572π‘π‘š2

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: ConeΒ & Hemisphere

Hemisphere

Radius = r = 3.5 cm

Height = Radius = 3.5 cm

Why ht. of hemisphere = radius of hemisphere?

Watch the video embedded below.Β 

Cone

Radius = r = 3.5 cmΒ 

Let ht. of Cone = h

ht. of Toyβ‡’15.5β‡’β„Ž=ht. of Cone+ht. of Hemisphere=β„Ž+3.5=15.5–3.5=12π‘π‘š

Slant ht. of Cone

Slant ht. of Cone (𝑙)=β„Ž2+π‘Ÿ2β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=122+3.52β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=144+12.25β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=156.25β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=12.5π‘π‘š

Surface Area of Toy

SA of toy=SA of Cone+SA of Hemisphere=πœ‹π‘Ÿπ‘™+2πœ‹π‘Ÿ2=227Γ—3.5Γ—12.5+2Γ—227Γ—3.52=137.5+77=214.5π‘π‘š2

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.=6π‘Ž2+2πœ‹π‘Ÿ2β€“πœ‹π‘Ÿ2=6π‘Ž2+πœ‹π‘Ÿ2=6Γ—72+227Γ—3.52=294+38.5=332.5π‘π‘š2

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 Ο€r2.Β 

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œ‹(𝑙2)2β€“πœ‹(𝑙2)2=6𝑙2+πœ‹(𝑙2)2=6𝑙2+πœ‹π‘™24=𝑙24(24+πœ‹)

H.S. – Hemisphere, SA – Surface Area

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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 Capsule1414β‡’β„Ž=Length of H.S. 1+Length of Cylinder+Length of H.S. 2=2.5+β„Ž+2.5=β„Ž+5=14–5=9π‘šπ‘š

Surface Area of Capsule

SA of Capsule=SA of H.S. 1+CSA of Cylinder+SA of H.S. 2=2πœ‹π‘Ÿ2+2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ2=4πœ‹π‘Ÿ2+2πœ‹π‘Ÿβ„Ž=2πœ‹π‘Ÿ(2π‘Ÿ+β„Ž)=2Γ—227Γ—2.5Γ—(2Γ—2.5+9)=1107Γ—(5+9)=1107Γ—14=110Γ—2=220π‘šπ‘š2

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 mΒ 

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=2πœ‹π‘Ÿβ„Ž+πœ‹π‘Ÿπ‘™=πœ‹π‘Ÿ(2β„Ž+𝑙)=227Γ—2Γ—(2Γ—2.1+2.8)=447Γ—(4.2+2.8)=447Γ—7=44π‘š2

Cost of Canvas

Cost of 1π‘š2β‡’Cost of 44π‘š2=500=44Γ—500=22,000  π‘…𝑠.

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=2πœ‹π‘Ÿβ„Ž+πœ‹π‘Ÿ2+πœ‹π‘Ÿπ‘™=πœ‹π‘Ÿ(2β„Ž+π‘Ÿ+𝑙)=227Γ—0.7Γ—(2Γ—2.4+0.7+2.5)=227Γ—0.7Γ—(4.8+3.2)=227Γ—0.7Γ—8=17.6π‘π‘š2

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=2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ2+2πœ‹π‘Ÿ2=2πœ‹π‘Ÿβ„Ž+4πœ‹π‘Ÿ2=2πœ‹π‘Ÿ(β„Ž+2π‘Ÿ)=2Γ—227Γ—3.5Γ—(10+2Γ—3.5)=22Γ—(10+7)=22Γ—17=374π‘π‘š2

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

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The post Exercise 12.1 (NCERT) – Surface Areas and Volumes, Class 10 – All Solutions appeared first on Mathemafia.

]]> Class 10 | Chapter 12 | All Solutions | Surface Areas and Volumes https://mathemafia.com/class-10-chapter-12-all-solutions-surface-areas-and-volumes/ Fri, 17 Dec 2021 07:40:14 +0000 http://sh016.global.temp.domains/~mathemaf/?p=7352 This playlist covers all the questions, concepts and examples of chapter 13 (Surface Areas and Volumes) of class 10, NCERT book. Click a thumbnail to watch the tutorial.Β  Watch complete chapter on YouTube

The post Class 10 | Chapter 12 | All Solutions | Surface Areas and Volumes appeared first on Mathemafia.

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This playlist covers all the questions, concepts and examples of chapter 13 (Surface Areas and Volumes) of class 10, NCERT book. Click a thumbnail to watch the tutorial.Β 

Watch complete chapter on YouTube

The post Class 10 | Chapter 12 | All Solutions | Surface Areas and Volumes appeared first on Mathemafia.

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