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- 1 Surface Areas and Volumes CBSE NCERT Notes Class 10 Maths Chapter 13 PDF

# Surface Areas and Volumes CBSE NCERT Notes Class 10 Maths Chapter 13 PDF

## Cuboid

There are following formulas for cuboid.

1. Lateral Surface Area (LSA) = 2h(*l* + b)

2. Total Surface Area (TSA) = 2(*l*b + bh + h*l*)

3. Volume (V) = *l*bh

4. Diagonal (d) = √(*l*^{2} + b^{2} + h^{2})

### Class 10 Maths Chapter 13 Examples 1:

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 m^{2} is Rs 15000, find the height of the hall.

Area of Four walls = 15000/10 = 1500

⇒ 2h(*l* + b) = 1500

⇒ h × 250 = 1500

⇒ h = 1500/250 = 6 m

## Cube

There are following formulas for cube.

1. Lateral Surface Area (LSA) = 4*l*^{2}

2. Total Surface Area (TSA) = 6*l*^{2}

3. Volume (V) = *l*^{3}

4. Diagonal (d) = *l*√3

### Class 10 Maths Chapter 13 Examples 2:

A cubical box has each edge 10 cm. Find the length of its diagonal.

Since *l *= 10. Thus, the length of the diagonal is d = 10√3.

## Cylinder (Closed at Both Ends)

There are following formulas for cylinder, which are closed at both ends.

1. Curved Surface Area (CSA) = 2πrh

2. Total Surface Area (TSA) = 2πrh + πr^{2} + πr^{2 }= 2πr(h + r)

3. Volume (V) = πr^{2}h

## Cylinder (Open at Top)

There are following formulas for cylinder which are open at both ends.

1. Curved Surface Area (CSA) = 2πrh

2. Total Surface Area (TSA) = 2πrh + πr^{2 }= πr(2h + r)

3. Volume (V) = πr^{2}h

## Hollow Cylinder (Open at Both Ends)

There are following formulas for cylinder which are open at both ends (example: Pipe).

1. External Curved Surface Area (ECSA) = 2πRh

2. Internal Curved Surface Area (ICSA) = 2πrh

3. Circular Area of Top or Bottom = π(R^{2} – r^{2})

4. Total Surface Area (TSA) = 2πRh + 2πrh + 2π(R^{2} – r^{2})^{ }= πr(2h + r)

5. Volume (V) = πR^{2}h – πr^{2}h

### Class 10 Maths Chapter 13 Examples 3:

The curved surface area of a right circular cylinder of height 14 cm is 88 cm^{2}. Find the diameter of the base of the cylinder.

Since CSA = 88

⇒ 2πrh = 88

⇒ 2 × 22/7 × r × 14 = 88

⇒ r = 1

Thus, required diameter of the cylinder is d = 2r = 2 cm.

## Cone

There are following formulas for cone.

1. Slant height (*l*) = √(r^{2} + h^{2})

2. Curved Surface Area (CSA) = πr*l*

3. Total Surface Area (TSA) = πr*l* + πr^{2} = πr(*l* + r)

4. Volume (V) = (1/3)πr^{2}h

### Class 10 Maths Chapter 13 Examples 4:

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 28 m.

Since *l *= 21 and r = d/2 = 14

Thus, TSA = πr(*l* + r) = 22/7 × 12 × (21 + 14) = 22 × 12 × 5 = 1320 m^{2}.

## Frustum of Cone

There are following formulas for cone.

1. Slant height (*l*) = √(h^{2} + (R – r)^{2})

2. Curved Surface Area (CSA) = π(R + r)*l*

3. Total Surface Area (TSA) = π(R + r)*l* + πr^{2 }+ πR^{2}

4. Volume (V) = (1/3)π(R^{2} + r^{2} + rR)h

### Class 10 Maths Chapter 13 Examples 5:

Find the volume of the frustum of cone whose base and top radius is 5 mm and 2 mm respectively. The height of the frustum of cone is 15 mm.

Volume (V) = (1/3)π(R^{2} + r^{2} + rR)h = 1/3 × 22/7 × (25 + 4 + 10) × 15 = 612.3 mm^{2}.

## Sphere

There are following formulas for sphere.

1. Surface Area (SA) = 4πr^{2}

2. Volume (V) = (4/3)πr^{3}

### Class 10 Maths Chapter 13 Examples 6:

Find the surface area of a sphere of radius 7 cm

SA = 4πr^{2} = 4 × 22/7 × 7 × 7 = 616 cm^{2}.

## Hemi-Sphere

There are following formulas for hemisphere.

1. Curved Surface Area (CSA) = 2πr^{2}

2. Total Surface Area (TSA) = 3πr^{2}

3. Volume (V) = (2/3)πr^{3}

### Class 10 Maths Chapter 13 Examples 7:

Find the volume of a hemisphere of radius 7 cm

V = (2/3)πr^{3} = 4/3 × 22/7 × 7 × 7 × 7 = 718.66 cm^{3}.

## Combination of Solids

If two or more solids are combined. Here, you will get some examples of combination of solids.

## Surface Areas of Combination of Above Solids

For Solid Figure 1 : Cone on Cylinder

TSA of Solid 1 = CSA of Cone + CSA of Cylinder + Area of Circular Base = πrl + 2πrh_{2} + πr^{2}

For Solid Figure 2 : Cone on Hemisphere

TSA of Solid 2 = CSA of Cone + CSA of Hemisphere = πrl + 2πr^{2}

For Solid Figure 3 : Conical Cavity in Cylinder

TSA of Solid 3 = CSA of Cylinder + Area of Base of Cylinder + CSA of Cone = 2πrh + πr^{2} + πrl

For Solid Figure 4 : Cones on Both Ends of Cylinder

TSA of Solid 4 = CSA of Two Cones + CSA of Cylinder = 2πrl + 2πrh_{1}

For Solid Figure 5 : Hemispheres on Both Ends of Cylinder

TSA of Solid 5 = CSA of Two Hemispheres + CSA of Cylinder = 2πrh + 4πr^{2}

For Solid Figure 6 : Hemispheres on Cubes

TSA of Solid 6

= TSA of Cube – Area of Hemisphere Face + CSA of Hemisphere = 6a^{2} – πr^{2} + 2πr^{2} = 6a^{2} + πr^{2}

### Class 10 Maths Chapter 13 Examples 8:

Find the total surface area of the solids given as below.

Surface Area

= TSA of Cube – Base Area of Hemisphere + CSA of Hemisphere

= (6 × 5 × 5) – πr^{2} + 2πr^{2}

= (150 + πr^{2})

= 150 + 22/7 × 4.2/2 × 4.2/2

= 163.86 cm^{2}.

## Volumes of Combination of Above Solids

For Solid Figure 1 : Cone on Cylinder

Volume of Solid 1 = Volume of Cone + Volume of Cylinder = (1/3)πr^{2}h_{1} + πr^{2}h_{2}

For Solid Figure 2 : Cone on Hemisphere

Volume of Solid 2 = Volume of Cone + Volume of Hemisphere = (1/3)πr^{2}h + (4/3)πr^{3}

For Solid Figure 3 : Conical Cavity in Cylinder

Volume of Solid 3 = Volume of Cylinder – Volume of Cone = πr^{2}h – (1/3)πr^{2}h = (2/3)πr^{2}h

For Solid Figure 4 : Cones on Both Ends of Cylinder

Volume of Solid 4 = Volume of Two Cones + Volume of Cylinder = (2/3)πr^{2}h_{2} + πr^{2}h_{1}

For Solid Figure 5 : Hemispheres on Both Ends of Cylinder

Volume of Solid 5 = Volume of Two Hemisphere + Volume of Cylinder = (4/3)πr^{3} + πr^{2}h

For Solid Figure 6 : Hemispheres on Cubes

Volume of Solid 6 = Volume of Cube + Volume of Hemisphere = a^{3} + (4/3)πr^{3}

### Class 10 Maths Chapter 13 Examples 9:

A solid is composed of a cylinder with hemispherical ends. If the total height of the solid is 105 cm and the radius of the hemispherical ends is 14 cm, find the volume of the solid.

Radius of the hemispherical ends = radius of the cylinder = 14 cm.

Volume = Volume of Two Hemisphere + Volume of Cylinder

= (4/3)πr^{3} + πr^{2}h

= (4/3) × 22/7 × 14^{3} + 22/7 × 14^{2 }× 77

= 22/7 × 14 × 14 × [4/3 × 14 + 77]

= 44 × 14 × 287/3

= 58930.67 cm^{3}.

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