##### Class 10^{th} Mathematics Gujarat Board Solution

**Exercise 14.1**- A toy is made by mounting a cone onto a hemisphere. The radius of the cone and…
- A show - piece shown in figure 14.10 is made of two solids - a cube and a…
- A vessel is in the form of a hemisphere mounted on a hollow cylinder. The…
- Chirag made a bird - bath for his garden in the shape of a cylinder with a…
- A solid is composed of a cylinder with hemispherical ends on both the sides.…
- The radius of a conical tent is 4 m and slant height is 5 m. How many meters of…
- If the radius of a cone is 60 cm and its curved surface area is 23.55 m^2 ,…
- The cost of painting the surface of sphere is Rs. 1526 at the rate of Rs. 6 per…

**Exercise 14.2**- The curved surface area of a cone is 550 cm^2 . If its diameter is 14 cm, find…
- A solid is in the form of cone with hemispherical base. The radius of the cone…
- How many litres of milk can be stored in a cylindrical tank with radius 1.4 m…
- The spherical balloon with radius 21 cm is filled with air. Find the volume of…
- A solid has hemi - spherical base with diameter 8.5 cm and it is surmounted by…
- A playing top is made up of steel. The top is shaped like a cone surmounted by…
- How many litres of petrol will be contained in a closed cylindrical tank with…
- The capacity of a cylindrical tank at a petrol pump is 57750 litres. If its…
- A hemispherical pond is filled with 523.908 m^3 of water. Find the maximum…
- A gulab - jamun contain 40% sugar syrup in it. Find how much syrup would be…
- The height and the slant height of a cone are 12 cm and 20 cm respectively.…
- Find the total volume of a cone having a hemispherical base. If the radius of…
- If the slant height of a cone is 18.7 cm and the curved surface area is 602.8…
- If the surface area of a spherical ball is 1256 cm^2 , then find the volume of…

**Exercise 14.3**- A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid…
- A cylindrical container having diameter 16 cm and height 40 cm is full of ice -…
- A cylindrical tank of diameter 3 m and height 7 m is completely filled with…
- A cylinder of radius 2 cm and height 10 cm is melted into small spherical balls…
- A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is…
- There are 45 conical heaps of wheat, each of them having diameter 80 cm and…
- A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled…

**Exercise 14.4**- A metal bucket is in the shape of a frustum of a cone, mounted on a hollow…
- A container, open from the top and made up of a metal sheet is the form of…

**Exercise 14**- A tent is in the shape of cylinder surmounted by a conical top. If the height…
- A metallic sphere of radius 5.6 cm is melted and recast into the shape of a…
- How many spherical balls of radius 2 cm can be made out of a solid cube of lead…
- A hemispherical bowl of internal radius 18 cm contains an edible oil to be…
- A hemispherical tank of radius 2.4 m is full of water. It is connected with a…
- A shuttle cock used for playing badminton has the shape of a frustum of a cone…
- A fez, the headgear cap used by the trucks is shaped like the frustum of a…
- A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm^3 of…
- The volume of sphere with diameter 1 cm is …….. cm^3 .A. 2/3 Ï€ B. 1/6 Ï€ C. Ï€…
- The volume of hemisphere with radius 1.2 cm is ……….. cm^3 .A. 1.152Ï€ B. 0.96Ï€…
- The volume of sphere is in 4/3 Ï€ cm^3 . Then its diameter is ………. cm.A. 0.5 B.…
- The volume of cone with radius 2 cm and height 6 cm is ……….. cm^3 .A. 8Ï€ B.…
- The diameter of the base of cone is 10 cm and its slant height is 17 cm. Then…
- The diameter and the height of the cylinder are 14 cm and 10 cm respectively.…
- The ratio of the radii of two cones having equal height is 2 : 3. Then, the…
- If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm,…
- The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then…

**Exercise 14.1**

- A toy is made by mounting a cone onto a hemisphere. The radius of the cone and…
- A show - piece shown in figure 14.10 is made of two solids - a cube and a…
- A vessel is in the form of a hemisphere mounted on a hollow cylinder. The…
- Chirag made a bird - bath for his garden in the shape of a cylinder with a…
- A solid is composed of a cylinder with hemispherical ends on both the sides.…
- The radius of a conical tent is 4 m and slant height is 5 m. How many meters of…
- If the radius of a cone is 60 cm and its curved surface area is 23.55 m^2 ,…
- The cost of painting the surface of sphere is Rs. 1526 at the rate of Rs. 6 per…

**Exercise 14.2**

- The curved surface area of a cone is 550 cm^2 . If its diameter is 14 cm, find…
- A solid is in the form of cone with hemispherical base. The radius of the cone…
- How many litres of milk can be stored in a cylindrical tank with radius 1.4 m…
- The spherical balloon with radius 21 cm is filled with air. Find the volume of…
- A solid has hemi - spherical base with diameter 8.5 cm and it is surmounted by…
- A playing top is made up of steel. The top is shaped like a cone surmounted by…
- How many litres of petrol will be contained in a closed cylindrical tank with…
- The capacity of a cylindrical tank at a petrol pump is 57750 litres. If its…
- A hemispherical pond is filled with 523.908 m^3 of water. Find the maximum…
- A gulab - jamun contain 40% sugar syrup in it. Find how much syrup would be…
- The height and the slant height of a cone are 12 cm and 20 cm respectively.…
- Find the total volume of a cone having a hemispherical base. If the radius of…
- If the slant height of a cone is 18.7 cm and the curved surface area is 602.8…
- If the surface area of a spherical ball is 1256 cm^2 , then find the volume of…

**Exercise 14.3**

- A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid…
- A cylindrical container having diameter 16 cm and height 40 cm is full of ice -…
- A cylindrical tank of diameter 3 m and height 7 m is completely filled with…
- A cylinder of radius 2 cm and height 10 cm is melted into small spherical balls…
- A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is…
- There are 45 conical heaps of wheat, each of them having diameter 80 cm and…
- A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled…

**Exercise 14.4**

- A metal bucket is in the shape of a frustum of a cone, mounted on a hollow…
- A container, open from the top and made up of a metal sheet is the form of…

**Exercise 14**

- A tent is in the shape of cylinder surmounted by a conical top. If the height…
- A metallic sphere of radius 5.6 cm is melted and recast into the shape of a…
- How many spherical balls of radius 2 cm can be made out of a solid cube of lead…
- A hemispherical bowl of internal radius 18 cm contains an edible oil to be…
- A hemispherical tank of radius 2.4 m is full of water. It is connected with a…
- A shuttle cock used for playing badminton has the shape of a frustum of a cone…
- A fez, the headgear cap used by the trucks is shaped like the frustum of a…
- A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm^3 of…
- The volume of sphere with diameter 1 cm is …….. cm^3 .A. 2/3 Ï€ B. 1/6 Ï€ C. Ï€…
- The volume of hemisphere with radius 1.2 cm is ……….. cm^3 .A. 1.152Ï€ B. 0.96Ï€…
- The volume of sphere is in 4/3 Ï€ cm^3 . Then its diameter is ………. cm.A. 0.5 B.…
- The volume of cone with radius 2 cm and height 6 cm is ……….. cm^3 .A. 8Ï€ B.…
- The diameter of the base of cone is 10 cm and its slant height is 17 cm. Then…
- The diameter and the height of the cylinder are 14 cm and 10 cm respectively.…
- The ratio of the radii of two cones having equal height is 2 : 3. Then, the…
- If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm,…
- The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then…

###### Exercise 14.1

**Question 1.**A toy is made by mounting a cone onto a hemisphere. The radius of the cone and a hemisphere is 5 cm. The total height of the toy is 17 cm. Find the total surface area of the toy.

**Answer:**Given.

A toy is made by mounting a cone onto a hemisphere

The radius of the cone and a hemisphere is 5 cm.

The total height of the toy is 17 cm

Formula used/Theory.

Curve surface area of Cone = Ï€rl

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ TSA of toy is sum of CSA of both cone and hemisphere

TSA of toy = CSA of cone + CSA of Hemisphere

=

=

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cone = Height of Toy – Radius

= 17cm – 5cm

= 12 cm

TSA of toy

**Question 2.**A show - piece shown in figure 14.10 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 7 cm and the hemisphere fixed on the top has diameter 5.2 cm. Find the total surface area of the piece.

**Answer:**Given.

A show - piece is made by mounting a hemisphere on cube

The Diameter of hemisphere is 5.2 cm.

Length of side of cube is 7 cm

Formula used/Theory.

Total surface area of Cube = 6 × side^{2}

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put hemisphere on Cube

The circle part of hemisphere will attach to cube 1 plane

∴ TSA of show - piece is sum of CSA of hemisphere and TSA of cube subtracted by Area of circle of hemisphere

TSA of show - piece =

CSA of Hemisphere + TSA of cube–Area of circle

= 2Ï€r^{2} + 6 × side^{2}–Ï€r^{2}

= Ï€r^{2} + 6 × side^{2}

Diameter = 5.2 cm

Radius = = 2.6

TSA of show - piece = 3.14 × 2.6 × 2.6 + 6 × 7 × 7

= 21.22 + 294

= 315.22 cm^{2}

**Question 3.**A vessel is in the form of a hemisphere mounted on a hollow cylinder. The diameter of the hemisphere is 21 cm and the height of vessel is 25 cm. If the vessel is to be painted at the rate of Rs. 3.5 per cm^{2}, then find the total cost to paint the vessel from outside.

**Answer:**Given.

A vessel is made by mounting hemisphere on cylinder

The radius of the hemisphere is cm = 10.5 cm.

The total height of the cylinder is 25 cm

Rates of painting = 3.5 per cm^{2}

Formula used/Theory.

Curve surface area of Cylinder = 2Ï€rh

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put hemisphere on hollow Cylinder

∴ Area of Vessel is sum of CSA of both hemisphere and cylinder

Area of vessel = CSA of cylinder + CSA of Hemisphere

= 2Ï€rh + 2Ï€r^{2}

= 2Ï€r[h + r]

As height of hemisphere is equal to radius of hemisphere

Then;

Area of vessel = 2 × × 10.5 × [25 + 10.5]

= 2 × × 10.5 × [35.5]

= 2343 cm^{2}

Rates for 1cm^{2} = Rs.3.5

Rates of 1648.5 cm^{2} = Rs.3.5 × 2343

= Rs. 8200.5

**Question 4.**Chirag made a bird - bath for his garden in the shape of a cylinder with a hemispherical depression at one end, (see the figure 14.11). The height of the cylinder is 1.5 m and its radius is 50 cm. Find the total area of the bird - bath. (Ï€ = 3.14)

**Answer:**Given.

A vessel is made by depressing hemisphere on cylinder

The radius of the hemisphere is 50 cm

Height of the cylinder is 150 cm

Formula used/Theory.

Curve surface area of Cylinder = 2Ï€rh

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put hemisphere in Cylinder

∴ Area of bird - bath is sum of CSA of both hemisphere and cylinder

Area of Bird - bath = CSA of cylinder + CSA of Hemisphere

= 2Ï€rh + 2Ï€r^{2}

= 2Ï€r[h + r]

TSA of toy = 2 × 3.14 × 50 × [150 + 50]

= 2 × 3.14 × 50 × [200]

= 62800 cm^{2}

**Question 5.**A solid is composed of a cylinder with hemispherical ends on both the sides. The radius and the height of the cylinder are 20 cm and 35 cm respectively. Find the total surface area of the solid.

**Answer:**Given.

A Solid is made by mounting 2 hemisphere on cylinder

The radius of the hemisphere is 20 cm.

The total height of the cylinder is 35 cm

Formula used/Theory.

Curve surface area of Cylinder = 2Ï€rh

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put 2 hemisphere on both sides of Cylinder

∴ Area of Solid is sum of CSA of both hemisphere and of cylinder

Area of Solid = CSA of cylinder + 2 × CSA of Hemisphere

= 2Ï€rh + 2 × (2Ï€r^{2})

= 2Ï€r[h + 2r]

Area of solid

**Question 6.**The radius of a conical tent is 4 m and slant height is 5 m. How many meters of canvas of width 125 cm will be used to prepare 12 tents? If the cost of canvas is Rs. 20 per meter, then what is total cost of 12 tents (Ï€ = 3.14)

**Answer:**Given.

Radius of conical tent is 4 m

Slant height of conical tent is 5m

Formula used/Theory.

CSA of cone = Ï€rl

⇒ As the conical tent is not on ground

∴ only CSA is required to construct the tent

CSA of conical tents = 3.14 × 5 × 4

= 62.8 m^{2}

Let length of cloth be x cm

∴ Area of Cloth = 1.25 × x m^{2}

If 12 tents are to be constructed

Then area of cloth required to make 12 conical tents is

12 × 62.8 m^{2}

= 753.6 m^{2}

1.25 m × x m = 753.6 m^{2}

x = = 602.88 m

If 1m length of canvas cost Rs. 20

Then 602.88 m length of canvas cost Rs. 20 × 602.88

= Rs. 12056.7

**Question 7.**If the radius of a cone is 60 cm and its curved surface area is 23.55 m^{2}, then find its slant height. (Ï€ = 3.14)

**Answer:**Given.

The radius of a cone is 60 cm

Curved surface area is 23.55 m^{2}

Formula used/Theory.

CSA of cone = Ï€rl

Radius is in cm but area is given in m^{2}

Then convert radius into m

Radius = m = 0.6m

23.55 m^{2} = 3.14 × 0.6 × l

L × 1.884 m = 23.55 m^{2}

L = = 12.5 m

∴ The slant height of cone is 12.5 m

**Question 8.**The cost of painting the surface of sphere is Rs. 1526 at the rate of Rs. 6 per m^{2}. Find the radius of sphere.

**Answer:**Given.

Cost of painting the surface of sphere is Rs. 1526

Rate of Rs. 6 per m^{2}

Formula used/Theory.

TSA of sphere = 4Ï€r^{2}

Let radius of sphere is r

Then area of sphere is 4Ï€r^{2} m^{2}

Rate of painting = Rs. 6 per m^{2}

Cost of painting of sphere =

4Ï€r^{2} m^{2} × Rs. 6 per m^{2}

Rs. 24Ï€r^{2}

Rs. 24Ï€r^{2} = Rs. 1526

24 × × r^{2} = 1526

r^{2} = = 20.23

r = √20.23 = 4.49 m

= 4.5 m

**Question 1.**

A toy is made by mounting a cone onto a hemisphere. The radius of the cone and a hemisphere is 5 cm. The total height of the toy is 17 cm. Find the total surface area of the toy.

**Answer:**

Given.

A toy is made by mounting a cone onto a hemisphere

The radius of the cone and a hemisphere is 5 cm.

The total height of the toy is 17 cm

Formula used/Theory.

Curve surface area of Cone = Ï€rl

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ TSA of toy is sum of CSA of both cone and hemisphere

TSA of toy = CSA of cone + CSA of Hemisphere

=

=

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cone = Height of Toy – Radius

= 17cm – 5cm

= 12 cm

TSA of toy

**Question 2.**

A show - piece shown in figure 14.10 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 7 cm and the hemisphere fixed on the top has diameter 5.2 cm. Find the total surface area of the piece.

**Answer:**

Given.

A show - piece is made by mounting a hemisphere on cube

The Diameter of hemisphere is 5.2 cm.

Length of side of cube is 7 cm

Formula used/Theory.

Total surface area of Cube = 6 × side^{2}

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put hemisphere on Cube

The circle part of hemisphere will attach to cube 1 plane

∴ TSA of show - piece is sum of CSA of hemisphere and TSA of cube subtracted by Area of circle of hemisphere

TSA of show - piece =

CSA of Hemisphere + TSA of cube–Area of circle

= 2Ï€r^{2} + 6 × side^{2}–Ï€r^{2}

= Ï€r^{2} + 6 × side^{2}

Diameter = 5.2 cm

Radius = = 2.6

TSA of show - piece = 3.14 × 2.6 × 2.6 + 6 × 7 × 7

= 21.22 + 294

= 315.22 cm^{2}

**Question 3.**

A vessel is in the form of a hemisphere mounted on a hollow cylinder. The diameter of the hemisphere is 21 cm and the height of vessel is 25 cm. If the vessel is to be painted at the rate of Rs. 3.5 per cm^{2}, then find the total cost to paint the vessel from outside.

**Answer:**

Given.

A vessel is made by mounting hemisphere on cylinder

The radius of the hemisphere is cm = 10.5 cm.

The total height of the cylinder is 25 cm

Rates of painting = 3.5 per cm^{2}

Formula used/Theory.

Curve surface area of Cylinder = 2Ï€rh

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put hemisphere on hollow Cylinder

∴ Area of Vessel is sum of CSA of both hemisphere and cylinder

Area of vessel = CSA of cylinder + CSA of Hemisphere

= 2Ï€rh + 2Ï€r^{2}

= 2Ï€r[h + r]

As height of hemisphere is equal to radius of hemisphere

Then;

Area of vessel = 2 × × 10.5 × [25 + 10.5]

= 2 × × 10.5 × [35.5]

= 2343 cm^{2}

Rates for 1cm^{2} = Rs.3.5

Rates of 1648.5 cm^{2} = Rs.3.5 × 2343

= Rs. 8200.5

**Question 4.**

Chirag made a bird - bath for his garden in the shape of a cylinder with a hemispherical depression at one end, (see the figure 14.11). The height of the cylinder is 1.5 m and its radius is 50 cm. Find the total area of the bird - bath. (Ï€ = 3.14)

**Answer:**

Given.

A vessel is made by depressing hemisphere on cylinder

The radius of the hemisphere is 50 cm

Height of the cylinder is 150 cm

Formula used/Theory.

Curve surface area of Cylinder = 2Ï€rh

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put hemisphere in Cylinder

∴ Area of bird - bath is sum of CSA of both hemisphere and cylinder

Area of Bird - bath = CSA of cylinder + CSA of Hemisphere

= 2Ï€rh + 2Ï€r^{2}

= 2Ï€r[h + r]

TSA of toy = 2 × 3.14 × 50 × [150 + 50]

= 2 × 3.14 × 50 × [200]

= 62800 cm^{2}

**Question 5.**

A solid is composed of a cylinder with hemispherical ends on both the sides. The radius and the height of the cylinder are 20 cm and 35 cm respectively. Find the total surface area of the solid.

**Answer:**

Given.

A Solid is made by mounting 2 hemisphere on cylinder

The radius of the hemisphere is 20 cm.

The total height of the cylinder is 35 cm

Formula used/Theory.

Curve surface area of Cylinder = 2Ï€rh

Curve surface area of Hemisphere = 2Ï€r^{2}

⇒ As we put 2 hemisphere on both sides of Cylinder

∴ Area of Solid is sum of CSA of both hemisphere and of cylinder

Area of Solid = CSA of cylinder + 2 × CSA of Hemisphere

= 2Ï€rh + 2 × (2Ï€r^{2})

= 2Ï€r[h + 2r]

Area of solid

**Question 6.**

The radius of a conical tent is 4 m and slant height is 5 m. How many meters of canvas of width 125 cm will be used to prepare 12 tents? If the cost of canvas is Rs. 20 per meter, then what is total cost of 12 tents (Ï€ = 3.14)

**Answer:**

Given.

Radius of conical tent is 4 m

Slant height of conical tent is 5m

Formula used/Theory.

CSA of cone = Ï€rl

⇒ As the conical tent is not on ground

∴ only CSA is required to construct the tent

CSA of conical tents = 3.14 × 5 × 4

= 62.8 m^{2}

Let length of cloth be x cm

∴ Area of Cloth = 1.25 × x m^{2}

If 12 tents are to be constructed

Then area of cloth required to make 12 conical tents is

12 × 62.8 m^{2}

= 753.6 m^{2}

1.25 m × x m = 753.6 m^{2}

x = = 602.88 m

If 1m length of canvas cost Rs. 20

Then 602.88 m length of canvas cost Rs. 20 × 602.88

= Rs. 12056.7

**Question 7.**

If the radius of a cone is 60 cm and its curved surface area is 23.55 m^{2}, then find its slant height. (Ï€ = 3.14)

**Answer:**

Given.

The radius of a cone is 60 cm

Curved surface area is 23.55 m^{2}

Formula used/Theory.

CSA of cone = Ï€rl

Radius is in cm but area is given in m^{2}

Then convert radius into m

Radius = m = 0.6m

23.55 m^{2} = 3.14 × 0.6 × l

L × 1.884 m = 23.55 m^{2}

L = = 12.5 m

∴ The slant height of cone is 12.5 m

**Question 8.**

The cost of painting the surface of sphere is Rs. 1526 at the rate of Rs. 6 per m^{2}. Find the radius of sphere.

**Answer:**

Given.

Cost of painting the surface of sphere is Rs. 1526

Rate of Rs. 6 per m^{2}

Formula used/Theory.

TSA of sphere = 4Ï€r^{2}

Let radius of sphere is r

Then area of sphere is 4Ï€r^{2} m^{2}

Rate of painting = Rs. 6 per m^{2}

Cost of painting of sphere =

4Ï€r^{2} m^{2} × Rs. 6 per m^{2}

Rs. 24Ï€r^{2}

Rs. 24Ï€r^{2} = Rs. 1526

24 × × r^{2} = 1526

r^{2} = = 20.23

r = √20.23 = 4.49 m

= 4.5 m

###### Exercise 14.2

**Question 1.**The curved surface area of a cone is 550 cm^{2}. If its diameter is 14 cm, find its volume.

**Answer:**Given.

CSA of cone = 550 cm^{2}

Diameter is 14 cm

Formula used/Theory.

CSA of cone = Ï€r]

Volume of cone = Ï€r^{2}h

If diameter is 14 cm

Then radius is half of diameter

Radius = = 7 cm

CSA of cone = Ï€r[]

= × 7 × []

= 22 ×

=

Squaring both sides

h^{2} + 49 = 25^{2}

h^{2} = 625 – 49

h^{2} = 576

h = √576 = 24 cm

Volume of cone = Ï€r^{2}h

= × 7 × 7 × 24

= 8 × 7 × 22

= 1232 cm^{2}

**Question 2.**A solid is in the form of cone with hemispherical base. The radius of the cone is 15 cm and the total height of the solid is 55 cm. Find the volume of the solid. (Ï€ = 3.14)

**Answer:**Given.

A solid is made by mounting a cone onto a hemisphere

The radius of the cone and a hemisphere is 15 cm.

The total height of the solid is 55 cm

Formula used/Theory.

Volume of Cone = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

⇒ as we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ Volume of Solid is sum of volume of both cone and hemisphere

Volume of solid = Volume of cone + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + 2r]

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cone = Height of Solid – Radius

= 55cm – 15cm

= 40 cm

Volume of solid = × 3.14 × 15 × 15 × [40 + 2 × 15]

= × 15 × 15 × [40 + 30]

= × 5 × 15 × 70

= 22 × 5 × 15 × 10

= 16500 cm^{3}

**Question 3.**How many litres of milk can be stored in a cylindrical tank with radius 1.4 m and height 3 m?

**Answer:**Given.

Radius of cylindrical tank = 1.4 m

Height of cylindrical tank = 3 m

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

1m^{3} = 1000 Litres

Volume of Cylinder = Ï€r^{2}h

= × 1.4 × 1.4 × 3

= 22 × 0.2 × 1.4 × 3

= 18.48 m^{3}

1m^{3} = 1000 Litres

18.48m^{3} = 18.48 × 1000 Litres

= 18480 Litres

**Question 4.**The spherical balloon with radius 21 cm is filled with air. Find the volume of air contained in it.

**Answer:**Given.

Radius of spherical balloon = 21 cm

Formula used/Theory.

Volume of Sphere = Ï€r^{3}

1cm^{3} = Litres

Volume of Balloon = Ï€r^{3}

= × 21 × 21 × 21

= 22 × 4 × 21 × 21

= 38808 cm^{3}

1cm^{3} = Litres

38808 cm^{3} = 38808 × Litres

= 38.808 Litres

**Question 5.**A solid has hemi - spherical base with diameter 8.5 cm and it is surmounted by a cylinder with height 8 cm and diameter of cylinder is 2 cm. Find the volume of this solid. (Ï€ = 3.14)

**Answer:**Given

Diameter of hemisphere = 8.5

Diameter of cylinder = 2cm

Height of cylinder = 8cm

Formula used

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Radius of cylinder =

= = 1 cm

⇒ Volume of Cylinder = Ï€r^{2}h

= × (1)^{2} × 8

=

= 25.14 cm^{3}

Radius of hemisphere =

= = 4.25 cm

⇒ Volume of hemisphere = Ï€r^{3}

= × (4.25)^{3}

= 160.84 cm^{3}

Volume of solid = Volume of cylinder + volume of hemisphere

= 25.14 cm^{3} + 160.84 cm^{3}

= 185.98 cm^{3}

**Question 6.**A playing top is made up of steel. The top is shaped like a cone surmounted by a hemisphere. The total height of top is 5 cm and the diameter of the top is 3.5 cm. Find the volume of the top.

**Answer:**Given.

Diameter of cone and hemisphere = 3.5cm

Total height of top = 5cm

Formula used/Theory.

Volume of Cone = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

⇒ As we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ Volume of top is sum of volume of both cone and hemisphere

Volume of top = Volume of cone + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + 2r]

Radius of hemisphere =

= = 1.75

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cone = Height of top – Radius

= 5cm – 1.75cm

= 3.25 cm

Volume of top = × 1.75 × 1.75 × [3.25 + 2 × 1.75]

= × 1.75 × 1.75 × [3.25 + 3.5]

= × 1.75 × 1.75 × [6.75]

= 22 × 0.25 × 1.75 × 2.25

= 21.65 cm^{3}

**Question 7.**How many litres of petrol will be contained in a closed cylindrical tank with hemisphere at one end having radius 4.2 cm and total height 27.5 cm?

**Answer:**Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

1 cm^{3} = litres

⇒ As we put Cylinder on hemisphere

The circle part of both cylinder and hemisphere will attach

∴ Volume of container is sum of volume of both cylinder and hemisphere

Volume of container = Volume of cylinder + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + r]

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cylinder = Height of Container – Radius

= 27.5cm – 4.2cm

= 23.3 cm

Volume of container = × 4.2 × 4.2 × [23.3 + × 4.2]

= 22 × 0.6 × 4.2 × [23.3 + 2.8]

= 22 × 0.6 × 4.2 × [26.1]

= 1446.98 cm^{3}

For Volume in litres

1 cm^{3} = litres

1446.98 cm^{3} = 1446.98 × litres

1.44698 litres

1.45 litres (approx.)

**Question 8.**The capacity of a cylindrical tank at a petrol pump is 57750 litres. If its diameter is 3.5 m, find the height of cylinder.

**Answer:**Given.

Capacity of petrol tank = 57750 litres

Diameter of tank = 3.5 m

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

1m^{3} = 1000 litres

Radius of tank =

= = 1.75

Let’s take height of tank to be h

Volume of cylinder = Ï€r^{2}h

= × 1.75 × 1.75 × h

9.625 × h m^{3}

Volume in litres = 57750 litres

1 m^{3} = 1000 litres

⇒ Volume = 9.625 × h × (1 m^{3})

= 9.625 × h × 1000 litres

57750 = 9625 × h

h = = 6 m

**Question 9.**A hemispherical pond is filled with 523.908 m^{3} of water. Find the maximum depth of pond.

**Answer:**Given.

Volume of hemisphere = 523.908 m^{3}

Formula used/Theory.

Volume of hemisphere = Ï€r^{3}

As maximum depth is height of pound

But in case of hemisphere radius is height of pound

Ï€r^{3} = × r^{3} = volume

Volume = 523.908 m^{3}

Equating both

× r^{3} = 523.908 m^{3}

× r^{3} = 523.908 m^{3}

r^{3} =

r^{3} = 250.047 m^{3}

r = ∛(250.047 m^{3})

r = 6.3 m

**Question 10.**A gulab - jamun contain 40% sugar syrup in it. Find how much syrup would be there in 50 gulab - jamuns, each shaped like a cylinder with two hemispherical ends with total length 5 cm and diameter 2.8 cm.

**Answer:**Given.

40% of sugar syrup in 1 gulab - jamun

Having length is 5 cm

And diameter is 2.8 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

If gulab - jamun shape like cylinder between 2 hemisphere

Volume of gulab - jamun = volume of cylinder

+ 2 × Volume of hemisphere

Diameter of hemisphere = 2.8 cm

Radius of hemisphere = = 1.4 cm

Height of cylinder = length of gulab - jamun–2 × radius

= 5 cm–2 × 1.4

= 5 cm – 2.8 cm

= 2.2 cm

Volume of gulab - jamun = volume of cylinder

+ 2 × Volume of hemisphere

= Ï€r^{2}h + 2 × Ï€r^{3}

= Ï€r^{2}[h + r]

= × 1.4 × 1.4 × [2.2 + × 1.4]

= 22 × 0.2 × 1.4 × [2.2 + 1.86]

= 22 × 0.2 × 1.4 × [4.06]

= 25.0096 cm^{3}

Volume of sugar syrup = × 25.0096 cm^{3}

Volume of sugar syrup = 10.0038 cm^{3}

Volume of sugar syrup in 50 gulab - jamun = 50 × 10.0038 cm^{3}

= 500.192 cm^{3}

1 cm^{3} = litres

500.192 cm^{3} = litres

= 0.500192 litres

= 0.5 litres (approx.)

**Question 11.**The height and the slant height of a cone are 12 cm and 20 cm respectively. Find its volume. (Ï€ = 3.14)

**Answer:**Given.

Slant height of cone = 20 cm

Height of cone = 12 cm

Formula used/Theory.

Volume of cone = Ï€r^{2}h

In cone,

The Radius, height and slant height makes a right angled triangle

With hypotenuse as slant height of triangle

∴ By Pythagoras Theorem

Radius^{2} + Height^{2} = (slant height) ^{2}

Radius^{2} + (12 cm) ^{2} = (20 cm) ^{2}

Radius^{2} + 144 cm^{2} = 400 cm^{2}

Radius^{2} = 400 cm^{2} – 144 cm^{2}

Radius = √(256 cm^{2}) = 16 cm

Volume of Cone = Ï€r^{2}h

= × 3.14 × 16cm × 16cm × 12cm

= 3.14 × 16cm × 16cm × 4cm

= 3215.36 cm^{3}

**Question 12.**Find the total volume of a cone having a hemispherical base. If the radius of the base is 21 cm and height 60 cm.

**Answer:**Given.

Radius of cone and hemisphere = 21 cm

Total height of cone = 60 cm

Formula used/Theory.

Volume of Cone = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

⇒ As we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ Volume of solid is sum of volume of both cone and hemisphere

Volume of solid = Volume of cone + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + 2r]

As height of hemisphere is equal to radius of hemisphere

Then;

Volume of top = × 21 × 21 × [60 + 2 × 21]

= × 21 × 21 × [60 + 42]

= × 21 × 21 × [102]

= 22 × 21 × 102

= 47124 cm^{3}

**Question 13.**If the slant height of a cone is 18.7 cm and the curved surface area is 602.8 cm^{2}, find the volume of cone. (Ï€ = 3.14)

**Answer:**Given.

Slant height of cone = 18.7 cm

CSA of cone = 602.8 cm^{2}

Formula used/Theory.

Volume of cone = Ï€r^{2}h

CSA of cone = Ï€rl

In cone,

CSA of cone = 602.8 cm^{2}

Ï€rl = 602.8 cm^{2}

3.14 × r × 18.7cm = 602.8 cm^{2}

r = = 10.26 cm

In cone,

The Radius, height and slant height makes a right angled triangle

With hypotenuse as slant height of triangle

∴ By Pythagoras Theorem

Radius^{2} + Height^{2} = (slant height) ^{2}

Height^{2} + (10.26 cm) ^{2} = (18.7 cm) ^{2}

Height^{2} + 105.26 cm^{2} = 349.69 cm^{2}

Height^{2} = 349.69 cm^{2} – 105.26 cm^{2}

Height = √(244.43 cm^{2}) = 15.63 cm

Volume of Cone = Ï€r^{2}h

= × 3.14 × 10.26cm × 10.26cm × 15.63cm

= 3.14 × 10.26cm × 10.26cm × 5.21cm

= 1722.11 cm^{3}

**Question 14.**If the surface area of a spherical ball is 1256 cm^{2}, then find the volume of sphere. (Take Ï€ = 3.14)

**Answer:**Given.

The surface area of a spherical ball is 1256 cm^{2}

Formula used/Theory.

Area of sphere = 4Ï€r^{2}

Volume of sphere = Ï€r^{3}

Volume of sphere = Ï€r^{3}

= × (4Ï€r^{2}) × r

Area of sphere = 4Ï€r^{2}

= 1256 cm^{2}

Radius of sphere is

r^{2} = = 100 cm^{2}

r = √100 cm^{2}

r = 10 cm

∴ Volume of sphere = × 4Ï€r^{2} × r

= × 1256 cm^{2} × 10 cm

= 418.66 cm^{2} × 10 cm

= 4186.67 cm^{3}

**Question 1.**

The curved surface area of a cone is 550 cm^{2}. If its diameter is 14 cm, find its volume.

**Answer:**

Given.

CSA of cone = 550 cm^{2}

Diameter is 14 cm

Formula used/Theory.

CSA of cone = Ï€r]

Volume of cone = Ï€r^{2}h

If diameter is 14 cm

Then radius is half of diameter

Radius = = 7 cm

CSA of cone = Ï€r[]

= × 7 × []

= 22 ×

=

Squaring both sides

h^{2} + 49 = 25^{2}

h^{2} = 625 – 49

h^{2} = 576

h = √576 = 24 cm

Volume of cone = Ï€r^{2}h

= × 7 × 7 × 24

= 8 × 7 × 22

= 1232 cm^{2}

**Question 2.**

A solid is in the form of cone with hemispherical base. The radius of the cone is 15 cm and the total height of the solid is 55 cm. Find the volume of the solid. (Ï€ = 3.14)

**Answer:**

Given.

A solid is made by mounting a cone onto a hemisphere

The radius of the cone and a hemisphere is 15 cm.

The total height of the solid is 55 cm

Formula used/Theory.

Volume of Cone = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

⇒ as we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ Volume of Solid is sum of volume of both cone and hemisphere

Volume of solid = Volume of cone + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + 2r]

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cone = Height of Solid – Radius

= 55cm – 15cm

= 40 cm

Volume of solid = × 3.14 × 15 × 15 × [40 + 2 × 15]

= × 15 × 15 × [40 + 30]

= × 5 × 15 × 70

= 22 × 5 × 15 × 10

= 16500 cm^{3}

**Question 3.**

How many litres of milk can be stored in a cylindrical tank with radius 1.4 m and height 3 m?

**Answer:**

Given.

Radius of cylindrical tank = 1.4 m

Height of cylindrical tank = 3 m

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

1m^{3} = 1000 Litres

Volume of Cylinder = Ï€r^{2}h

= × 1.4 × 1.4 × 3

= 22 × 0.2 × 1.4 × 3

= 18.48 m^{3}

1m^{3} = 1000 Litres

18.48m^{3} = 18.48 × 1000 Litres

= 18480 Litres

**Question 4.**

The spherical balloon with radius 21 cm is filled with air. Find the volume of air contained in it.

**Answer:**

Given.

Radius of spherical balloon = 21 cm

Formula used/Theory.

Volume of Sphere = Ï€r^{3}

1cm^{3} = Litres

Volume of Balloon = Ï€r^{3}

= × 21 × 21 × 21

= 22 × 4 × 21 × 21

= 38808 cm^{3}

1cm^{3} = Litres

38808 cm^{3} = 38808 × Litres

= 38.808 Litres

**Question 5.**

A solid has hemi - spherical base with diameter 8.5 cm and it is surmounted by a cylinder with height 8 cm and diameter of cylinder is 2 cm. Find the volume of this solid. (Ï€ = 3.14)

**Answer:**

Given

Diameter of hemisphere = 8.5

Diameter of cylinder = 2cm

Height of cylinder = 8cm

Formula used

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Radius of cylinder =

= = 1 cm

⇒ Volume of Cylinder = Ï€r^{2}h

= × (1)^{2} × 8

=

= 25.14 cm^{3}

Radius of hemisphere =

= = 4.25 cm

⇒ Volume of hemisphere = Ï€r^{3}

= × (4.25)^{3}

= 160.84 cm^{3}

Volume of solid = Volume of cylinder + volume of hemisphere

= 25.14 cm^{3} + 160.84 cm^{3}

= 185.98 cm^{3}

**Question 6.**

A playing top is made up of steel. The top is shaped like a cone surmounted by a hemisphere. The total height of top is 5 cm and the diameter of the top is 3.5 cm. Find the volume of the top.

**Answer:**

Given.

Diameter of cone and hemisphere = 3.5cm

Total height of top = 5cm

Formula used/Theory.

Volume of Cone = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

⇒ As we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ Volume of top is sum of volume of both cone and hemisphere

Volume of top = Volume of cone + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + 2r]

Radius of hemisphere =

= = 1.75

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cone = Height of top – Radius

= 5cm – 1.75cm

= 3.25 cm

Volume of top = × 1.75 × 1.75 × [3.25 + 2 × 1.75]

= × 1.75 × 1.75 × [3.25 + 3.5]

= × 1.75 × 1.75 × [6.75]

= 22 × 0.25 × 1.75 × 2.25

= 21.65 cm^{3}

**Question 7.**

How many litres of petrol will be contained in a closed cylindrical tank with hemisphere at one end having radius 4.2 cm and total height 27.5 cm?

**Answer:**

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

1 cm^{3} = litres

⇒ As we put Cylinder on hemisphere

The circle part of both cylinder and hemisphere will attach

∴ Volume of container is sum of volume of both cylinder and hemisphere

Volume of container = Volume of cylinder + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + r]

As height of hemisphere is equal to radius of hemisphere

Then;

Height of cylinder = Height of Container – Radius

= 27.5cm – 4.2cm

= 23.3 cm

Volume of container = × 4.2 × 4.2 × [23.3 + × 4.2]

= 22 × 0.6 × 4.2 × [23.3 + 2.8]

= 22 × 0.6 × 4.2 × [26.1]

= 1446.98 cm^{3}

For Volume in litres

1 cm^{3} = litres

1446.98 cm^{3} = 1446.98 × litres

1.44698 litres

1.45 litres (approx.)

**Question 8.**

The capacity of a cylindrical tank at a petrol pump is 57750 litres. If its diameter is 3.5 m, find the height of cylinder.

**Answer:**

Given.

Capacity of petrol tank = 57750 litres

Diameter of tank = 3.5 m

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

1m^{3} = 1000 litres

Radius of tank =

= = 1.75

Let’s take height of tank to be h

Volume of cylinder = Ï€r^{2}h

= × 1.75 × 1.75 × h

9.625 × h m^{3}

Volume in litres = 57750 litres

1 m^{3} = 1000 litres

⇒ Volume = 9.625 × h × (1 m^{3})

= 9.625 × h × 1000 litres

57750 = 9625 × h

h = = 6 m

**Question 9.**

A hemispherical pond is filled with 523.908 m^{3} of water. Find the maximum depth of pond.

**Answer:**

Given.

Volume of hemisphere = 523.908 m^{3}

Formula used/Theory.

Volume of hemisphere = Ï€r^{3}

As maximum depth is height of pound

But in case of hemisphere radius is height of pound

Ï€r^{3} = × r^{3} = volume

Volume = 523.908 m^{3}

Equating both

× r^{3} = 523.908 m^{3}

× r^{3} = 523.908 m^{3}

r^{3} =

r^{3} = 250.047 m^{3}

r = ∛(250.047 m^{3})

r = 6.3 m

**Question 10.**

A gulab - jamun contain 40% sugar syrup in it. Find how much syrup would be there in 50 gulab - jamuns, each shaped like a cylinder with two hemispherical ends with total length 5 cm and diameter 2.8 cm.

**Answer:**

Given.

40% of sugar syrup in 1 gulab - jamun

Having length is 5 cm

And diameter is 2.8 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

If gulab - jamun shape like cylinder between 2 hemisphere

Volume of gulab - jamun = volume of cylinder

+ 2 × Volume of hemisphere

Diameter of hemisphere = 2.8 cm

Radius of hemisphere = = 1.4 cm

Height of cylinder = length of gulab - jamun–2 × radius

= 5 cm–2 × 1.4

= 5 cm – 2.8 cm

= 2.2 cm

Volume of gulab - jamun = volume of cylinder

+ 2 × Volume of hemisphere

= Ï€r^{2}h + 2 × Ï€r^{3}

= Ï€r^{2}[h + r]

= × 1.4 × 1.4 × [2.2 + × 1.4]

= 22 × 0.2 × 1.4 × [2.2 + 1.86]

= 22 × 0.2 × 1.4 × [4.06]

= 25.0096 cm^{3}

Volume of sugar syrup = × 25.0096 cm^{3}

Volume of sugar syrup = 10.0038 cm^{3}

Volume of sugar syrup in 50 gulab - jamun = 50 × 10.0038 cm^{3}

= 500.192 cm^{3}

1 cm^{3} = litres

500.192 cm^{3} = litres

= 0.500192 litres

= 0.5 litres (approx.)

**Question 11.**

The height and the slant height of a cone are 12 cm and 20 cm respectively. Find its volume. (Ï€ = 3.14)

**Answer:**

Given.

Slant height of cone = 20 cm

Height of cone = 12 cm

Formula used/Theory.

Volume of cone = Ï€r^{2}h

In cone,

The Radius, height and slant height makes a right angled triangle

With hypotenuse as slant height of triangle

∴ By Pythagoras Theorem

Radius^{2} + Height^{2} = (slant height) ^{2}

Radius^{2} + (12 cm) ^{2} = (20 cm) ^{2}

Radius^{2} + 144 cm^{2} = 400 cm^{2}

Radius^{2} = 400 cm^{2} – 144 cm^{2}

Radius = √(256 cm^{2}) = 16 cm

Volume of Cone = Ï€r^{2}h

= × 3.14 × 16cm × 16cm × 12cm

= 3.14 × 16cm × 16cm × 4cm

= 3215.36 cm^{3}

**Question 12.**

Find the total volume of a cone having a hemispherical base. If the radius of the base is 21 cm and height 60 cm.

**Answer:**

Given.

Radius of cone and hemisphere = 21 cm

Total height of cone = 60 cm

Formula used/Theory.

Volume of Cone = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

⇒ As we put Cone on hemisphere

The circle part of both cone and hemisphere will attach

∴ Volume of solid is sum of volume of both cone and hemisphere

Volume of solid = Volume of cone + Volume of Hemisphere

= Ï€r^{2}h + Ï€r^{3}

= Ï€r^{2}[h + 2r]

As height of hemisphere is equal to radius of hemisphere

Then;

Volume of top = × 21 × 21 × [60 + 2 × 21]

= × 21 × 21 × [60 + 42]

= × 21 × 21 × [102]

= 22 × 21 × 102

= 47124 cm^{3}

**Question 13.**

If the slant height of a cone is 18.7 cm and the curved surface area is 602.8 cm^{2}, find the volume of cone. (Ï€ = 3.14)

**Answer:**

Given.

Slant height of cone = 18.7 cm

CSA of cone = 602.8 cm^{2}

Formula used/Theory.

Volume of cone = Ï€r^{2}h

CSA of cone = Ï€rl

In cone,

CSA of cone = 602.8 cm^{2}

Ï€rl = 602.8 cm^{2}

3.14 × r × 18.7cm = 602.8 cm^{2}

r = = 10.26 cm

In cone,

The Radius, height and slant height makes a right angled triangle

With hypotenuse as slant height of triangle

∴ By Pythagoras Theorem

Radius^{2} + Height^{2} = (slant height) ^{2}

Height^{2} + (10.26 cm) ^{2} = (18.7 cm) ^{2}

Height^{2} + 105.26 cm^{2} = 349.69 cm^{2}

Height^{2} = 349.69 cm^{2} – 105.26 cm^{2}

Height = √(244.43 cm^{2}) = 15.63 cm

Volume of Cone = Ï€r^{2}h

= × 3.14 × 10.26cm × 10.26cm × 15.63cm

= 3.14 × 10.26cm × 10.26cm × 5.21cm

= 1722.11 cm^{3}

**Question 14.**

If the surface area of a spherical ball is 1256 cm^{2}, then find the volume of sphere. (Take Ï€ = 3.14)

**Answer:**

Given.

The surface area of a spherical ball is 1256 cm^{2}

Formula used/Theory.

Area of sphere = 4Ï€r^{2}

Volume of sphere = Ï€r^{3}

Volume of sphere = Ï€r^{3}

= × (4Ï€r^{2}) × r

Area of sphere = 4Ï€r^{2}

= 1256 cm^{2}

Radius of sphere is

r^{2} = = 100 cm^{2}

r = √100 cm^{2}

r = 10 cm

∴ Volume of sphere = × 4Ï€r^{2} × r

= × 1256 cm^{2} × 10 cm

= 418.66 cm^{2} × 10 cm

= 4186.67 cm^{3}

###### Exercise 14.3

**Question 1.**A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid is to be filled into cylindrical bottles of diameter 4 cm and height 6 cm. How many bottles can be filled with this liquid?

**Answer:**Given.

Hemispherical bowl of internal radius 12 cm

Cylindrical bottles of diameter 4 cm and height 6 cm

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Volume of water will be equal to volume of hemisphere

Volume of hemisphere = Ï€r^{3}

= × Ï€ × (12 cm) ^{3}

= 1152 × Ï€ cm^{3}

Radius of bottle = = 2 cm

Volume of each bottle = Ï€r^{2}h

= Ï€ × (2cm) ^{2} × 6 cm

= 24Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

Number of bottles filled will be dividing the total volume of water by volume of water contained by 1 bottle

Number of bottles = = 48

**Question 2.**A cylindrical container having diameter 16 cm and height 40 cm is full of ice - cream. The ice - cream is to be filled into cones of height 12 cm and diameter 4 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with the ice - cream.

**Answer:**Given.

Radius of Hemispherical bowl and cone is = 2 cm

Cylindrical container of diameter 16 cm and height 40 cm

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Volume of Cone = Ï€r^{2}h

Volume of ice - cream will be equal to sum of volume of hemisphere bowl and volume of cone

Volume of hemisphere = Ï€r^{3}

= × Ï€ × (2 cm) ^{3}

= Ï€ cm^{3}

Volume of Cone = Ï€r^{2}h

= × Ï€ × (2 cm) ^{2} × 12 cm

16Ï€ cm^{3}

Volume of ice - cream = 16Ï€ cm^{3} + Ï€ cm^{3}

= × 16Ï€ cm^{3}

Radius of container = = 8 cm

Volume of container = Ï€r^{2}h

= Ï€ × (8cm) ^{2} × 40 cm

= 2560Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

Number of ice - cream will be dividing the total volume of ice - cream in container by volume of ice - cream contained by 1 cone

Number of ice - creams =

= 120

∴ 120 ice - cream can be filled by the container

**Question 3.**A cylindrical tank of diameter 3 m and height 7 m is completely filled with groundnut oil. It is to be emptied in 15 tins each of capacity 15 litres. Find the number of such tins required.

**Answer:**Given.

Cylindrical tank of diameter 3 m and height 7 m

Tins of capacity 15 litres

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

1 m^{3} = 1000 litres

Volume of cylinder tank = Ï€r^{2}h

Radius of tank = = 1.5 m

⇒ Volume of cylinder = × (1.5 m) ^{2} × 7 m

= 22 × 2.25 m^{3}

= 49.5 m^{3}

If 1 m^{3} = 1000 litres

Then 49.5 m^{3} = 49.5 × 1000 litres

= 49500 litres

Volume of 1 tin = 15 litres

Number of tins is dividing volume of cylindrical tank by volume of tins

⇒ Number of tins = = 3300 tins

∴ 3300 tins required to empty the tank

**Question 4.**A cylinder of radius 2 cm and height 10 cm is melted into small spherical balls of diameter 1 cm. Find the number of such balls.

**Answer:**Given.

Cylinder of radius 2 cm and height 10 cm

spherical balls of diameter 1 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of sphere = Ï€r^{3}

Volume of cylinder = Ï€r^{2}h

= Ï€ × (2 cm) ^{2} × 10 cm

= 40Ï€ cm^{3}

Radius of sphere = = 0.5 cm

Volume of sphere = Ï€r^{3}

= Ï€ (0.5 cm) ^{3}

= Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

Number of spherical balls will be dividing the total volume of cylinder by volume of 1 spherical ball

Number of spherical balls = = 240

∴ 240 Spherical balls can be made by melting cylinder

**Question 5.**A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is drawn from it. Find the length of the wire.

**Answer:**Given.

Cylindrical wire of diameter 1 cm

Metallic spherical of radius 15 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of sphere = Ï€r^{3}

Let the length of wire be x

Volume of cylinder = Ï€r^{2}h

= Ï€ × ( cm) ^{2} × x

= Ï€ cm^{2}

Volume of sphere = Ï€r^{3}

= Ï€ (15 cm) ^{3}

= 4500Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

As we melt the metallic sphere in cylindrical wire the volume of both will be equal

∴ equating both we will get the length of wire

Volume of wire = Volume of metallic sphere

Ï€ cm^{2} = 4500Ï€ cm^{3}

x = 4500 × 4

x = 18000 cm = 180 m

∴ length of wire is 180 m

**Question 6.**There are 45 conical heaps of wheat, each of them having diameter 80 cm and height 30 cm. To store the wheat in a cylindrical container of the same radius, what will be the height of cylinder?

**Answer:**Given.

Height of cone is 30 cm

Cone and cylinder of diameter 80 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of cone = Ï€r^{2}h

Let the Height of container be x

Radius of cylinder and cone is = 40cm

Volume of container = Ï€r^{2}h

= Ï€ × (40 cm)^{2} × x

= 1600Ï€x cm^{2}

Volume of conical heaps = Ï€r^{2}h

= Ï€(40 cm)^{2} × 30 cm

= 16000Ï€ cm^{3}

Volume of 45 conical heaps = 45 × 16000Ï€ cm^{3}

= 720000Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

As we put 45 conical heaps of wheat in cylindrical container volume of 45 conical heaps will be equal to volume of container

∴ equating both we will get the Height of container

Volume of container = Volume of 45 conical heaps

1600Ï€ × x cm^{2} = 720000Ï€ cm^{3}

x =

x = 450 cm

∴ Height of container is 450 cm

**Question 7.**A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 33 cm, find the radius and the slant height of the heap.

**Answer:**Given.

Height of bucket is 44 cm

Radius of bucket is 21 cm

Height of conical heap is 33 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of cone = Ï€r^{2}h

Let the Radius of conical heap be x

Volume of bucket = Ï€r^{2}h

= Ï€ × (21 cm)^{2} × 44 cm

= 19404Ï€ cm^{3}

Volume of conical heaps = Ï€r^{2}h

= Ï€ × x^{2} × 33 cm

= 11Ï€ × x^{2} cm

**Note we will not put value of Ï€ as it will be divided in next step

As we put bucket of sand on ground it will form a conical heap volume of conical heaps will be equal to volume of bucket

∴ equating both we will get the Radius of conical heap

Volume of bucket = Volume of conical heaps

11Ï€ × x^{2} cm = 19404Ï€ cm^{3}

x^{2} =

x^{2} = 1764 cm^{2}

x = √ (1764 cm^{2})

x = 42 cm

∴ Radius of conical heap is 42 cm

In cone;

As the radius , height and slant height makes Right angled triangle where hypotenuse is slant height

Then by Pythagoras theorem

(Slant height)^{2} = (height)^{2} + (radius)^{2}

(Slant height)^{2} = (33 cm)^{2} + (42 cm)^{2}

(Slant height)^{2} = 1089 cm^{2} + 1764 cm^{2}

(Slant height)^{2} = 2853 cm^{2}

Slant height = √(2853 cm^{2})

= 53.41 cm

**Question 1.**

A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid is to be filled into cylindrical bottles of diameter 4 cm and height 6 cm. How many bottles can be filled with this liquid?

**Answer:**

Given.

Hemispherical bowl of internal radius 12 cm

Cylindrical bottles of diameter 4 cm and height 6 cm

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Volume of water will be equal to volume of hemisphere

Volume of hemisphere = Ï€r^{3}

= × Ï€ × (12 cm) ^{3}

= 1152 × Ï€ cm^{3}

Radius of bottle = = 2 cm

Volume of each bottle = Ï€r^{2}h

= Ï€ × (2cm) ^{2} × 6 cm

= 24Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

Number of bottles filled will be dividing the total volume of water by volume of water contained by 1 bottle

Number of bottles = = 48

**Question 2.**

A cylindrical container having diameter 16 cm and height 40 cm is full of ice - cream. The ice - cream is to be filled into cones of height 12 cm and diameter 4 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with the ice - cream.

**Answer:**

Given.

Radius of Hemispherical bowl and cone is = 2 cm

Cylindrical container of diameter 16 cm and height 40 cm

Formula used/Theory.

Volume of Cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Volume of Cone = Ï€r^{2}h

Volume of ice - cream will be equal to sum of volume of hemisphere bowl and volume of cone

Volume of hemisphere = Ï€r^{3}

= × Ï€ × (2 cm) ^{3}

= Ï€ cm^{3}

Volume of Cone = Ï€r^{2}h

= × Ï€ × (2 cm) ^{2} × 12 cm

16Ï€ cm^{3}

Volume of ice - cream = 16Ï€ cm^{3} + Ï€ cm^{3}

= × 16Ï€ cm^{3}

Radius of container = = 8 cm

Volume of container = Ï€r^{2}h

= Ï€ × (8cm) ^{2} × 40 cm

= 2560Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

Number of ice - cream will be dividing the total volume of ice - cream in container by volume of ice - cream contained by 1 cone

Number of ice - creams =

= 120

∴ 120 ice - cream can be filled by the container

**Question 3.**

A cylindrical tank of diameter 3 m and height 7 m is completely filled with groundnut oil. It is to be emptied in 15 tins each of capacity 15 litres. Find the number of such tins required.

**Answer:**

Given.

Cylindrical tank of diameter 3 m and height 7 m

Tins of capacity 15 litres

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

1 m^{3} = 1000 litres

Volume of cylinder tank = Ï€r^{2}h

Radius of tank = = 1.5 m

⇒ Volume of cylinder = × (1.5 m) ^{2} × 7 m

= 22 × 2.25 m^{3}

= 49.5 m^{3}

If 1 m^{3} = 1000 litres

Then 49.5 m^{3} = 49.5 × 1000 litres

= 49500 litres

Volume of 1 tin = 15 litres

Number of tins is dividing volume of cylindrical tank by volume of tins

⇒ Number of tins = = 3300 tins

∴ 3300 tins required to empty the tank

**Question 4.**

A cylinder of radius 2 cm and height 10 cm is melted into small spherical balls of diameter 1 cm. Find the number of such balls.

**Answer:**

Given.

Cylinder of radius 2 cm and height 10 cm

spherical balls of diameter 1 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of sphere = Ï€r^{3}

Volume of cylinder = Ï€r^{2}h

= Ï€ × (2 cm) ^{2} × 10 cm

= 40Ï€ cm^{3}

Radius of sphere = = 0.5 cm

Volume of sphere = Ï€r^{3}

= Ï€ (0.5 cm) ^{3}

= Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

Number of spherical balls will be dividing the total volume of cylinder by volume of 1 spherical ball

Number of spherical balls = = 240

∴ 240 Spherical balls can be made by melting cylinder

**Question 5.**

A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is drawn from it. Find the length of the wire.

**Answer:**

Given.

Cylindrical wire of diameter 1 cm

Metallic spherical of radius 15 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of sphere = Ï€r^{3}

Let the length of wire be x

Volume of cylinder = Ï€r^{2}h

= Ï€ × ( cm) ^{2} × x

= Ï€ cm^{2}

Volume of sphere = Ï€r^{3}

= Ï€ (15 cm) ^{3}

= 4500Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

As we melt the metallic sphere in cylindrical wire the volume of both will be equal

∴ equating both we will get the length of wire

Volume of wire = Volume of metallic sphere

Ï€ cm^{2} = 4500Ï€ cm^{3}

x = 4500 × 4

x = 18000 cm = 180 m

∴ length of wire is 180 m

**Question 6.**

There are 45 conical heaps of wheat, each of them having diameter 80 cm and height 30 cm. To store the wheat in a cylindrical container of the same radius, what will be the height of cylinder?

**Answer:**

Given.

Height of cone is 30 cm

Cone and cylinder of diameter 80 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of cone = Ï€r^{2}h

Let the Height of container be x

Radius of cylinder and cone is = 40cm

Volume of container = Ï€r^{2}h

= Ï€ × (40 cm)^{2} × x

= 1600Ï€x cm^{2}

Volume of conical heaps = Ï€r^{2}h

= Ï€(40 cm)^{2} × 30 cm

= 16000Ï€ cm^{3}

Volume of 45 conical heaps = 45 × 16000Ï€ cm^{3}

= 720000Ï€ cm^{3}

**Note we will not put value of Ï€ as it will be divided in next step

As we put 45 conical heaps of wheat in cylindrical container volume of 45 conical heaps will be equal to volume of container

∴ equating both we will get the Height of container

Volume of container = Volume of 45 conical heaps

1600Ï€ × x cm^{2} = 720000Ï€ cm^{3}

x =

x = 450 cm

∴ Height of container is 450 cm

**Question 7.**

A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 33 cm, find the radius and the slant height of the heap.

**Answer:**

Given.

Height of bucket is 44 cm

Radius of bucket is 21 cm

Height of conical heap is 33 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of cone = Ï€r^{2}h

Let the Radius of conical heap be x

Volume of bucket = Ï€r^{2}h

= Ï€ × (21 cm)^{2} × 44 cm

= 19404Ï€ cm^{3}

Volume of conical heaps = Ï€r^{2}h

= Ï€ × x^{2} × 33 cm

= 11Ï€ × x^{2} cm

**Note we will not put value of Ï€ as it will be divided in next step

As we put bucket of sand on ground it will form a conical heap volume of conical heaps will be equal to volume of bucket

∴ equating both we will get the Radius of conical heap

Volume of bucket = Volume of conical heaps

11Ï€ × x^{2} cm = 19404Ï€ cm^{3}

x^{2} =

x^{2} = 1764 cm^{2}

x = √ (1764 cm^{2})

x = 42 cm

∴ Radius of conical heap is 42 cm

In cone;

As the radius , height and slant height makes Right angled triangle where hypotenuse is slant height

Then by Pythagoras theorem

(Slant height)^{2} = (height)^{2} + (radius)^{2}

(Slant height)^{2} = (33 cm)^{2} + (42 cm)^{2}

(Slant height)^{2} = 1089 cm^{2} + 1764 cm^{2}

(Slant height)^{2} = 2853 cm^{2}

Slant height = √(2853 cm^{2})

= 53.41 cm

###### Exercise 14.4

**Question 1.**A metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The total vertical height of the bucket is 40 cm and that of cylindrical base is 10 cm, radii of two circular ends are 60 cm and 20 cm. Find the area of the metallic sheet used. Also find the volume of water the bucket can hold. (Ï€ = 3.14)

**Answer:**Given.

Height of bucket is 40 cm

Height of cylinder is 10 cm

Radii of both circular ends of frustum are 60 cm, 20 cm

Formula used/Theory.

CSA of cylinder = 2Ï€rh

CSA of frustum = Ï€(r + R) × l

Volume of cylinder = Ï€r^{2}h

Volume of frustum = h[R^{2} + r^{2} + Rr]

As the bucket is always open from mouth

Metallic sheet require will be sum of CSA of Frustum and CSA of cylinder and Area of base circle

In frustum

L^{2} = height^{2} +

Height of frustum = height of bucket – height of cylinder

= 40 – 10 = 30 cm

L^{2} = 30^{2} + [60 – 20]^{2}

L^{2} = 900 cm^{2} + 1600 cm^{2} = 2500 cm^{2}

L = √(2500 cm^{2}) = 50cm

⇒ CSA of Frustum = Ï€(60 + 20)50 cm

= 3.14 × 80cm × 50cm

= 12560cm^{2}

⇒ CSA of cylinder = 2Ï€rh

= 2 × 3.14 × 20cm × 10cm

= 1256 cm^{2}

⇒ Area of base = Ï€r^{2}

= 3.14 × 20 × 20

= 1256 cm^{2}

Area of metallic sheet = 12560 cm^{2} + 1256 cm^{2} + 1256 cm^{2}

= 15072 cm^{2}

⇒ Volume of Frustum = h[R^{2} + r^{2} + Rr]

= × 30cm × [(60 cm)^{2} + (20 cm)^{2} + 60 cm × 20 cm]

= 31.4 cm × [3600cm^{2} + 400cm^{2} + 1200cm^{2}]

= 31.4 cm × 5200 cm^{2}

= 163280 cm^{3}

1 cm^{3} = litres

163280 cm^{3} = litres

= 163.280 litres

∴ Area of metallic sheet used in making bucket is 15072 cm^{2}

Volume of bucket is 163.280 litres

**Question 2.**A container, open from the top and made up of a metal sheet is the form of frustum of a cone of height 30 cm with radii 30 cm and 10 cm. Find the cost of the milk which can completely fill container at the rate of Rs. 30 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 50 per 100 cm^{2}. (Ï€ = 3.14)

**Answer:**Given.

Radii of frustum is 10 cm and 30 cm

Height of frustum is 30 cm

Rate of milk is Rs. 30 per litre

Rate metallic sheet is Rs . 50 per 100 cm^{2}

Formula used/Theory.

CSA of frustum = Ï€(r + R) × l

Volume of frustum = h[R^{2} + r^{2} + Rr]

⇒ Volume of Frustum = h[R^{2} + r^{2} + Rr]

= × 30cm × [(30 cm)^{2} + (10 cm)^{2} + 30 cm × 10 cm]

= 31.4 cm × [900cm^{2} + 100cm^{2} + 300cm^{2}]

= 31.4 cm × 1300 cm^{2}

= 40820 cm^{3}

1 cm^{3} = litres

40820 cm^{3} = litres

= 40.820 litres

As milk is Rs. 30 per litre

For 40.82 litres

Rs. 30 × 40.82

= Rs. 1224.6

In frustum

L^{2} = height^{2} +

L^{2} = 30^{2} + [30 – 10]^{2}

L^{2} = 900 cm^{2} + 400 cm^{2} = 1300 cm^{2}

L = √(1300 cm^{2}) = 10√13 cm

⇒ CSA of Frustum = Ï€(30 + 10)10√13 cm

= 3.14 × 40cm × 10√13cm

= 1256√13cm^{2}

⇒ Area of base = Ï€r^{2}

= 3.14 × 10 × 10

= 314 cm^{2}

Area of container = 1256√13 + 314

= 314[4√13 + 1]

= 314[4 × 3.6 + 1]

= 314 × 15.4

= 4835.6 cm^{2}

Cost of 100 cm^{2} metal sheet = Rs. 50

Cost of 1 cm^{2} metal sheet = Rs. 0.5

Cost of 4835.6 cm^{2} metal sheet = Rs. 4835.6 × 0.5

= Rs. 2417.8

**Question 1.**

A metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The total vertical height of the bucket is 40 cm and that of cylindrical base is 10 cm, radii of two circular ends are 60 cm and 20 cm. Find the area of the metallic sheet used. Also find the volume of water the bucket can hold. (Ï€ = 3.14)

**Answer:**

Given.

Height of bucket is 40 cm

Height of cylinder is 10 cm

Radii of both circular ends of frustum are 60 cm, 20 cm

Formula used/Theory.

CSA of cylinder = 2Ï€rh

CSA of frustum = Ï€(r + R) × l

Volume of cylinder = Ï€r^{2}h

Volume of frustum = h[R^{2} + r^{2} + Rr]

As the bucket is always open from mouth

Metallic sheet require will be sum of CSA of Frustum and CSA of cylinder and Area of base circle

In frustum

L^{2} = height^{2} +

Height of frustum = height of bucket – height of cylinder

= 40 – 10 = 30 cm

L^{2} = 30^{2} + [60 – 20]^{2}

L^{2} = 900 cm^{2} + 1600 cm^{2} = 2500 cm^{2}

L = √(2500 cm^{2}) = 50cm

⇒ CSA of Frustum = Ï€(60 + 20)50 cm

= 3.14 × 80cm × 50cm

= 12560cm^{2}

⇒ CSA of cylinder = 2Ï€rh

= 2 × 3.14 × 20cm × 10cm

= 1256 cm^{2}

⇒ Area of base = Ï€r^{2}

= 3.14 × 20 × 20

= 1256 cm^{2}

Area of metallic sheet = 12560 cm^{2} + 1256 cm^{2} + 1256 cm^{2}

= 15072 cm^{2}

⇒ Volume of Frustum = h[R^{2} + r^{2} + Rr]

= × 30cm × [(60 cm)^{2} + (20 cm)^{2} + 60 cm × 20 cm]

= 31.4 cm × [3600cm^{2} + 400cm^{2} + 1200cm^{2}]

= 31.4 cm × 5200 cm^{2}

= 163280 cm^{3}

1 cm^{3} = litres

163280 cm^{3} = litres

= 163.280 litres

∴ Area of metallic sheet used in making bucket is 15072 cm^{2}

Volume of bucket is 163.280 litres

**Question 2.**

A container, open from the top and made up of a metal sheet is the form of frustum of a cone of height 30 cm with radii 30 cm and 10 cm. Find the cost of the milk which can completely fill container at the rate of Rs. 30 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 50 per 100 cm^{2}. (Ï€ = 3.14)

**Answer:**

Given.

Radii of frustum is 10 cm and 30 cm

Height of frustum is 30 cm

Rate of milk is Rs. 30 per litre

Rate metallic sheet is Rs . 50 per 100 cm^{2}

Formula used/Theory.

CSA of frustum = Ï€(r + R) × l

Volume of frustum = h[R^{2} + r^{2} + Rr]

⇒ Volume of Frustum = h[R^{2} + r^{2} + Rr]

= × 30cm × [(30 cm)^{2} + (10 cm)^{2} + 30 cm × 10 cm]

= 31.4 cm × [900cm^{2} + 100cm^{2} + 300cm^{2}]

= 31.4 cm × 1300 cm^{2}

= 40820 cm^{3}

1 cm^{3} = litres

40820 cm^{3} = litres

= 40.820 litres

As milk is Rs. 30 per litre

For 40.82 litres

Rs. 30 × 40.82

= Rs. 1224.6

In frustum

L^{2} = height^{2} +

L^{2} = 30^{2} + [30 – 10]^{2}

L^{2} = 900 cm^{2} + 400 cm^{2} = 1300 cm^{2}

L = √(1300 cm^{2}) = 10√13 cm

⇒ CSA of Frustum = Ï€(30 + 10)10√13 cm

= 3.14 × 40cm × 10√13cm

= 1256√13cm^{2}

⇒ Area of base = Ï€r^{2}

= 3.14 × 10 × 10

= 314 cm^{2}

Area of container = 1256√13 + 314

= 314[4√13 + 1]

= 314[4 × 3.6 + 1]

= 314 × 15.4

= 4835.6 cm^{2}

Cost of 100 cm^{2} metal sheet = Rs. 50

Cost of 1 cm^{2} metal sheet = Rs. 0.5

Cost of 4835.6 cm^{2} metal sheet = Rs. 4835.6 × 0.5

= Rs. 2417.8

###### Exercise 14

**Question 1.**A tent is in the shape of cylinder surmounted by a conical top. If the height and the radius of the cylindrical part are 3.5 m and 2 m respectively and the slant height of the top is 3.5 m, find the area of the canvas used for making the tent. Also find the cost of canvas of the tent at the rate of Rs. 1000 per m^{2}.

**Answer:**

Given.

Radius of cylinder and cone is 2 m

Height of cylinder is 3.5 m

Slant height of cone is 3.5 m

Formula used/Theory.

CSA of cylinder = 2Ï€rh

CSA of cone = Ï€rl

CSA of cylinder = 2Ï€rh

= 2 × × 3.5 m × 2 m

= 44 m^{2}

CSA of cone = Ï€rl

= × 2 m × 3.5 m

= 22 m^{2}

Total Area of cloth = 44 m^{2} + 22 m^{2}

= 66 m^{2}

Rate of cloth for 1 m^{2} = Rs. 1000

Rate of cloth for 66 m^{2} = Rs. 1000 × 66

= Rs. 66000

Cost of canvas is Rs. 66000

**Question 2.**A metallic sphere of radius 5.6 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

**Answer:**Given.

Radius of sphere is 5.6 cm

Radius of cylinder is 6 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of sphere = Ï€r^{3}

Volume of sphere = Ï€r^{3}

= × Ï€ × (5.6 cm)^{3}

Volume of cylinder = Ï€r^{2}h

= Ï€ × (6 cm)^{2} × h

When a metallic sphere is recast in shape of cylinder

Volume of both sphere and cylinder will be equal

∴ comparing both we get height of cylinder

Ï€ × (6 cm)^{2} × h = × Ï€ × (5.6 cm)^{3}

h =

h = 6.5 cm

∴ Height of cylinder is 6.5 cm

**Question 3.**How many spherical balls of radius 2 cm can be made out of a solid cube of lead whose side measures 44 cm?

**Answer:**

Given.

Radius of sphere is 2 cm

Side of cube is 44 cm

Formula used/Theory.

Volume of sphere = Ï€r^{3}

Volume of cube = side^{3}

Volume of sphere = Ï€r^{3}

= × (2 cm)^{3}

Volume of cube = side^{3}

= (44 cm)^{3}

Number of spherical balls will be get by dividing Volume of cube by Volume of each Sphere

Number of spheres =

=

=

= 2541

**Question 4.**A hemispherical bowl of internal radius 18 cm contains an edible oil to be filled in cylindrical bottles of radius 3 cm and height 9 cm. How many bottles are required to empty the bowl?

**Answer:**Given.

Radius of hemisphere is 18 cm

Radius of cylinder is 3 cm

Height of cylinder is 9 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Volume of cylindrical bottle = Ï€r^{2}h

= Ï€ × (3cm)^{2} × 9 cm

= 81Ï€ cm^{3}

Volume of hemispherical bowl = Ï€r^{3}

= × Ï€ × (18 cm)^{3}

= 3888 Ï€ cm^{3}

Number of cylindrical bottle will be get by dividing Volume of hemispherical bowl by Volume of each Cylindrical bottle

Number of cylindrical bottle =

=

= 48 bottles

**Question 5.**A hemispherical tank of radius 2.4 m is full of water. It is connected with a pipe which empties it at the rate of 7 litres per second. How much time will it take to empty the tank completely?

**Answer:**Given.

A hemispherical tank of radius 2.4 m is full of water

Rate of water flow is 7 litres per second

Formula used/Theory.

Volume of hemisphere = Ï€r^{3}

Volume of hemispherical bowl = Ï€r^{3}

= × (2.4 m)^{3}

= 28.96m^{3}

1m^{3} = 1000 litres

28.96m^{3} = 28.964 × 1000 litres

= 28964 litres

Rate of flow is 7 litres in 1 second

Rate of flow is 1 litres in second

Rate of flow is 28964 litres in second

Is 4137.8 seconds

In minutes = = 68.96 minutes

**Question 6.**A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.

**Answer:**Given.

The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm

Formula used/Theory.

CSA of frustum = Ï€(R + r) × L

CSA of hemisphere = 2Ï€r^{2}

Radii of frustum is and = 2.5cm and 1cm

As in hemisphere radius and height are same

Height of frustum = height of cock – Radius

= 7 cm – 1 cm

= 6 cm

In frustum

L^{2} = height^{2} +

L^{2} = (6 cm)^{2} + [2.5 cm – 1 cm]^{2}

L^{2} = 36 cm^{2} + 2.25 cm^{2} = 38.25 cm^{2}

L = √(38.25 cm^{2}) = 6.18 cm

CSA of hemisphere = 2 × 3.14 × (1 cm)^{2}

= 6.28 cm^{2}

CSA of frustum = 3.14 × [2.5cm + 1cm] × 6.18cm

= 3.14 × 3.5cm × 6.18cm

= 67.92 cm^{2}

External surface area = CSA of hemisphere + CSA of frustum

= 6.28 cm^{2} + 67.92 cm^{2}

= 74.2 cm^{2}

**Question 7.**A fez, the headgear cap used by the trucks is shaped like the frustum of a cone. If its radius on the open side is 12 cm and radius at the upper base is 5 cm and its slant height is 15 cm, find the area of material used for making it. (Ï€ = 3.14)

**Answer:**Given.

Radii of frustum are 12 cm and 5 cm

Slant height is 15 cm

Formula used/Theory.

CSA of frustum is Ï€[R + r] × l

Area of material = CSA of frustum

= Ï€[R + r] × l

= 3.14[12cm + 5cm] × 15cm

= 3.14 × 17cm × 15cm

= 800.7 cm^{2}

**Question 8.**A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm^{3} of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of bucket and the cost of making it at the rate of Rs. 10 per cm^{2}.

**Answer:**

Given.

Volume of frustum = 12308.8 cm^{3}

Radii of frustum are 20cm and 12cm

Formula used/Theory.

CSA of frustum = Ï€[R + r]l

Volume of frustum = h[R^{2} + r^{2} + Rr]

Volume of frustum = h[R^{2} + r^{2} + Rr]

= × h[(20cm)^{2} + (12cm)^{2} + 20cm × 12cm]

= × h[400cm^{2} + 144cm^{2} + 240cm^{2}]

= × h × 784cm^{2}

= × h × 22 × 112

= × 22 × 112 × h = 12308.8 cm^{3}

h = = 15 cm

In frustum

(Slant height)^{2} = (difference in radii )^{2} + (height)^{2}

(Slant height)^{2} = (R - r)^{2} + (height)^{2}

(Slant height)^{2} = (20cm – 12cm)^{2} + (15 cm)^{2}

(Slant height)^{2} = 64 cm^{2} + 225 cm^{2}

(Slant height)^{2} = 289 cm^{2}

Slant height = √ (289 cm^{2})

Slant height = 17 cm

CSA of cone = Ï€[R + r]l

= [12cm + 20cm] × 17cm

= × 32cm × 17cm

= 1709.71 cm^{2}

Rate of 1 cm^{2} = Rs. 10

Rate of 1709.71 cm^{2} = Rs. 10 × 1709.71

= Rs. 17097.1

**Question 9.**The volume of sphere with diameter 1 cm is …….. cm^{3}.

A. Ï€

B. Ï€

C. Ï€

D. Ï€

**Answer:**Volume of sphere = Ï€r^{3}

When diameter of sphere is 1cm

Radius of sphere is cm

= Ï€( cm)^{3}

= Ï€ cm^{3}

= Ï€ cm^{3}

**Question 10.**The volume of hemisphere with radius 1.2 cm is ……….. cm^{3}.

A. 1.152Ï€

B. 0.96Ï€

C. 2.152Ï€

D. 3.456Ï€

**Answer:**Volume of hemisphere = Ï€r^{3}

When radius is 1.2 cm

= Ï€(1.2 cm)^{3}

= 1.152Ï€ cm^{3}

**Question 11.**The volume of sphere is in 4/3 Ï€ cm^{3}. Then its diameter is ………. cm.

A. 0.5

B. 1

C. 2

D. 2.5

**Answer:**Volume of sphere = Ï€r^{3}

= Ï€ cm^{3}

Ï€ × r^{3}cm^{3} = Ï€ cm^{3}

(r cm)^{3} = 1 cm^{3}

r = ∛1 cm^{3}

r = 1

if radius = 1 cm then diameter = 2 × radius

= 2 cm

**Question 12.**The volume of cone with radius 2 cm and height 6 cm is ……….. cm^{3}.

A. 8Ï€

B. 12Ï€

C. 14Ï€

D. 16Ï€

**Answer:**Volume of cone = Ï€r^{2}h

= Ï€ × (2cm)^{2} × 6 cm

= 8 Ï€

**Question 13.**The diameter of the base of cone is 10 cm and its slant height is 17 cm. Then the curved surface area of the cone is ………. cm^{2}.

A. 85Ï€

B. 170Ï€

C. 95Ï€

D. 88Ï€

**Answer:**CSA of cone = Ï€rl

= Ï€ × cm × 17cm

= Ï€ × 5cm × 17cm

= 85 Ï€ cm

**Question 14.**The diameter and the height of the cylinder are 14 cm and 10 cm respectively. Then the total surface area is ……….. cm^{2}.

A. 44

B. 308

C. 748

D. 1010

**Answer:**Radius of cylinder = = = 7cm

TSA of cylinder = 2Ï€r[r + h]

= 2 × × 7cm × [7cm + 10cm]

= 44cm × 17cm

= 748 cm^{2}

**Question 15.**The ratio of the radii of two cones having equal height is 2 : 3. Then, the ratio of their volumes is ……..

A. 4 : 6

B. 8 : 27

C. 3 : 2

D. 4 : 9

**Answer:**If ratio of radii are 2:3

Then;

Let the radii of both cones be 2x and 3x

Volume of cone = Ï€r^{2}h

Volume of 1^{st} cone = Ï€(2r)^{2}h

= Ï€r^{2}h

Volume of 2^{nd} cone = Ï€(3r)^{2}h

= 3Ï€r^{2}h

The ratio of their volumes is

Ï€r^{2}h: 3Ï€r^{2}h

4:9

**Question 16.**If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm, then the curved surface area is …………cm^{2}.

A. 50Ï€

B. 25Ï€

C. 35Ï€

D. 63Ï€

**Answer:**In frustum

(Slant height)^{2} = (difference in radii )^{2} + (height)^{2}

(Slant height)^{2} = (R - r)^{2} + (height)^{2}

(Slant height)^{2} = (7cm – 3cm)^{2} + (3 cm)^{2}

(Slant height)^{2} = 16 cm^{2} + 9 cm^{2}

(Slant height)^{2} = 25 cm^{2}

Slant height = √ (25 cm^{2})

Slant height = 5 cm

CSA of frustum = Ï€[R + r]l

= Ï€[7cm + 3cm] × 5cm

= 50 Ï€ cm^{2}

**Question 17.**The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then the volume is ………….. cm^{3}.

A. 320Ï€

B. 151Ï€

C. 302Ï€

D. 98Ï€

**Answer:**Volume of frustum = h[R^{2} + r^{2} + Rr]

= × 6 cm[(9cm)^{2} + (5cm)^{2} + 9cm × 5cm]

= 2Ï€ [81cm^{2} + 25cm^{2} + 45cm^{2}]

= 2Ï€ × 151cm^{2}

= 302 cm^{2}

^{}

**Question 1.**

A tent is in the shape of cylinder surmounted by a conical top. If the height and the radius of the cylindrical part are 3.5 m and 2 m respectively and the slant height of the top is 3.5 m, find the area of the canvas used for making the tent. Also find the cost of canvas of the tent at the rate of Rs. 1000 per m^{2}.

**Answer:**

Given.

Radius of cylinder and cone is 2 m

Height of cylinder is 3.5 m

Slant height of cone is 3.5 m

Formula used/Theory.

CSA of cylinder = 2Ï€rh

CSA of cone = Ï€rl

CSA of cylinder = 2Ï€rh

= 2 × × 3.5 m × 2 m

= 44 m^{2}

CSA of cone = Ï€rl

= × 2 m × 3.5 m

= 22 m^{2}

Total Area of cloth = 44 m^{2} + 22 m^{2}

= 66 m^{2}

Rate of cloth for 1 m^{2} = Rs. 1000

Rate of cloth for 66 m^{2} = Rs. 1000 × 66

= Rs. 66000

Cost of canvas is Rs. 66000

**Question 2.**

A metallic sphere of radius 5.6 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

**Answer:**

Given.

Radius of sphere is 5.6 cm

Radius of cylinder is 6 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of sphere = Ï€r^{3}

Volume of sphere = Ï€r^{3}

= × Ï€ × (5.6 cm)^{3}

Volume of cylinder = Ï€r^{2}h

= Ï€ × (6 cm)^{2} × h

When a metallic sphere is recast in shape of cylinder

Volume of both sphere and cylinder will be equal

∴ comparing both we get height of cylinder

Ï€ × (6 cm)^{2} × h = × Ï€ × (5.6 cm)^{3}

h =

h = 6.5 cm

∴ Height of cylinder is 6.5 cm

**Question 3.**

How many spherical balls of radius 2 cm can be made out of a solid cube of lead whose side measures 44 cm?

**Answer:**

Given.

Radius of sphere is 2 cm

Side of cube is 44 cm

Formula used/Theory.

Volume of sphere = Ï€r^{3}

Volume of cube = side^{3}

Volume of sphere = Ï€r^{3}

= × (2 cm)^{3}

Volume of cube = side^{3}

= (44 cm)^{3}

Number of spherical balls will be get by dividing Volume of cube by Volume of each Sphere

Number of spheres =

=

=

= 2541

**Question 4.**

A hemispherical bowl of internal radius 18 cm contains an edible oil to be filled in cylindrical bottles of radius 3 cm and height 9 cm. How many bottles are required to empty the bowl?

**Answer:**

Given.

Radius of hemisphere is 18 cm

Radius of cylinder is 3 cm

Height of cylinder is 9 cm

Formula used/Theory.

Volume of cylinder = Ï€r^{2}h

Volume of hemisphere = Ï€r^{3}

Volume of cylindrical bottle = Ï€r^{2}h

= Ï€ × (3cm)^{2} × 9 cm

= 81Ï€ cm^{3}

Volume of hemispherical bowl = Ï€r^{3}

= × Ï€ × (18 cm)^{3}

= 3888 Ï€ cm^{3}

Number of cylindrical bottle will be get by dividing Volume of hemispherical bowl by Volume of each Cylindrical bottle

Number of cylindrical bottle =

=

= 48 bottles

**Question 5.**

A hemispherical tank of radius 2.4 m is full of water. It is connected with a pipe which empties it at the rate of 7 litres per second. How much time will it take to empty the tank completely?

**Answer:**

Given.

A hemispherical tank of radius 2.4 m is full of water

Rate of water flow is 7 litres per second

Formula used/Theory.

Volume of hemisphere = Ï€r^{3}

Volume of hemispherical bowl = Ï€r^{3}

= × (2.4 m)^{3}

= 28.96m^{3}

1m^{3} = 1000 litres

28.96m^{3} = 28.964 × 1000 litres

= 28964 litres

Rate of flow is 7 litres in 1 second

Rate of flow is 1 litres in second

Rate of flow is 28964 litres in second

Is 4137.8 seconds

In minutes = = 68.96 minutes

**Question 6.**

A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.

**Answer:**

Given.

The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm

Formula used/Theory.

CSA of frustum = Ï€(R + r) × L

CSA of hemisphere = 2Ï€r^{2}

Radii of frustum is and = 2.5cm and 1cm

As in hemisphere radius and height are same

Height of frustum = height of cock – Radius

= 7 cm – 1 cm

= 6 cm

In frustum

L^{2} = height^{2} +

L^{2} = (6 cm)^{2} + [2.5 cm – 1 cm]^{2}

L^{2} = 36 cm^{2} + 2.25 cm^{2} = 38.25 cm^{2}

L = √(38.25 cm^{2}) = 6.18 cm

CSA of hemisphere = 2 × 3.14 × (1 cm)^{2}

= 6.28 cm^{2}

CSA of frustum = 3.14 × [2.5cm + 1cm] × 6.18cm

= 3.14 × 3.5cm × 6.18cm

= 67.92 cm^{2}

External surface area = CSA of hemisphere + CSA of frustum

= 6.28 cm^{2} + 67.92 cm^{2}

= 74.2 cm^{2}

**Question 7.**

A fez, the headgear cap used by the trucks is shaped like the frustum of a cone. If its radius on the open side is 12 cm and radius at the upper base is 5 cm and its slant height is 15 cm, find the area of material used for making it. (Ï€ = 3.14)

**Answer:**

Given.

Radii of frustum are 12 cm and 5 cm

Slant height is 15 cm

Formula used/Theory.

CSA of frustum is Ï€[R + r] × l

Area of material = CSA of frustum

= Ï€[R + r] × l

= 3.14[12cm + 5cm] × 15cm

= 3.14 × 17cm × 15cm

= 800.7 cm^{2}

**Question 8.**

A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm^{3} of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of bucket and the cost of making it at the rate of Rs. 10 per cm^{2}.

**Answer:**

Given.

Volume of frustum = 12308.8 cm^{3}

Radii of frustum are 20cm and 12cm

Formula used/Theory.

CSA of frustum = Ï€[R + r]l

Volume of frustum = h[R^{2} + r^{2} + Rr]

Volume of frustum = h[R^{2} + r^{2} + Rr]

= × h[(20cm)^{2} + (12cm)^{2} + 20cm × 12cm]

= × h[400cm^{2} + 144cm^{2} + 240cm^{2}]

= × h × 784cm^{2}

= × h × 22 × 112

= × 22 × 112 × h = 12308.8 cm^{3}

h = = 15 cm

In frustum

(Slant height)^{2} = (difference in radii )^{2} + (height)^{2}

(Slant height)^{2} = (R - r)^{2} + (height)^{2}

(Slant height)^{2} = (20cm – 12cm)^{2} + (15 cm)^{2}

(Slant height)^{2} = 64 cm^{2} + 225 cm^{2}

(Slant height)^{2} = 289 cm^{2}

Slant height = √ (289 cm^{2})

Slant height = 17 cm

CSA of cone = Ï€[R + r]l

= [12cm + 20cm] × 17cm

= × 32cm × 17cm

= 1709.71 cm^{2}

Rate of 1 cm^{2} = Rs. 10

Rate of 1709.71 cm^{2} = Rs. 10 × 1709.71

= Rs. 17097.1

**Question 9.**

The volume of sphere with diameter 1 cm is …….. cm^{3}.

A. Ï€

B. Ï€

C. Ï€

D. Ï€

**Answer:**

Volume of sphere = Ï€r^{3}

When diameter of sphere is 1cm

Radius of sphere is cm

= Ï€( cm)^{3}

= Ï€ cm^{3}

= Ï€ cm^{3}

**Question 10.**

The volume of hemisphere with radius 1.2 cm is ……….. cm^{3}.

A. 1.152Ï€

B. 0.96Ï€

C. 2.152Ï€

D. 3.456Ï€

**Answer:**

Volume of hemisphere = Ï€r^{3}

When radius is 1.2 cm

= Ï€(1.2 cm)^{3}

= 1.152Ï€ cm^{3}

**Question 11.**

The volume of sphere is in 4/3 Ï€ cm^{3}. Then its diameter is ………. cm.

A. 0.5

B. 1

C. 2

D. 2.5

**Answer:**

Volume of sphere = Ï€r^{3}

= Ï€ cm^{3}

Ï€ × r^{3}cm^{3} = Ï€ cm^{3}

(r cm)^{3} = 1 cm^{3}

r = ∛1 cm^{3}

r = 1

if radius = 1 cm then diameter = 2 × radius

= 2 cm

**Question 12.**

The volume of cone with radius 2 cm and height 6 cm is ……….. cm^{3}.

A. 8Ï€

B. 12Ï€

C. 14Ï€

D. 16Ï€

**Answer:**

Volume of cone = Ï€r^{2}h

= Ï€ × (2cm)^{2} × 6 cm

= 8 Ï€

**Question 13.**

The diameter of the base of cone is 10 cm and its slant height is 17 cm. Then the curved surface area of the cone is ………. cm^{2}.

A. 85Ï€

B. 170Ï€

C. 95Ï€

D. 88Ï€

**Answer:**

CSA of cone = Ï€rl

= Ï€ × cm × 17cm

= Ï€ × 5cm × 17cm

= 85 Ï€ cm

**Question 14.**

The diameter and the height of the cylinder are 14 cm and 10 cm respectively. Then the total surface area is ……….. cm^{2}.

A. 44

B. 308

C. 748

D. 1010

**Answer:**

Radius of cylinder = = = 7cm

TSA of cylinder = 2Ï€r[r + h]

= 2 × × 7cm × [7cm + 10cm]

= 44cm × 17cm

= 748 cm^{2}

**Question 15.**

The ratio of the radii of two cones having equal height is 2 : 3. Then, the ratio of their volumes is ……..

A. 4 : 6

B. 8 : 27

C. 3 : 2

D. 4 : 9

**Answer:**

If ratio of radii are 2:3

Then;

Let the radii of both cones be 2x and 3x

Volume of cone = Ï€r^{2}h

Volume of 1^{st} cone = Ï€(2r)^{2}h

= Ï€r^{2}h

Volume of 2^{nd} cone = Ï€(3r)^{2}h

= 3Ï€r^{2}h

The ratio of their volumes is

Ï€r^{2}h: 3Ï€r^{2}h

4:9

**Question 16.**

If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm, then the curved surface area is …………cm^{2}.

A. 50Ï€

B. 25Ï€

C. 35Ï€

D. 63Ï€

**Answer:**

In frustum

(Slant height)^{2} = (difference in radii )^{2} + (height)^{2}

(Slant height)^{2} = (R - r)^{2} + (height)^{2}

(Slant height)^{2} = (7cm – 3cm)^{2} + (3 cm)^{2}

(Slant height)^{2} = 16 cm^{2} + 9 cm^{2}

(Slant height)^{2} = 25 cm^{2}

Slant height = √ (25 cm^{2})

Slant height = 5 cm

CSA of frustum = Ï€[R + r]l

= Ï€[7cm + 3cm] × 5cm

= 50 Ï€ cm^{2}

**Question 17.**

The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then the volume is ………….. cm^{3}.

A. 320Ï€

B. 151Ï€

C. 302Ï€

D. 98Ï€

**Answer:**

Volume of frustum = h[R^{2} + r^{2} + Rr]

= × 6 cm[(9cm)^{2} + (5cm)^{2} + 9cm × 5cm]

= 2Ï€ [81cm^{2} + 25cm^{2} + 45cm^{2}]

= 2Ï€ × 151cm^{2}

= 302 cm^{2}

^{}