VOLUME OF 3D SHAPES WHEN BASE AREA IS GIVEN

Calculate, to 3 significant figures, the volume of :

Problem 1 :

Solution :

Given, base area = 142.3 cm²

Height = 24.9 cm

Volume of pyramid = 1/3 × base area × height

= 1/3 × 142.3 × 24.9

= 3543.27/3

V = 1181 cm³

Problem 2 :

Solution :

Given, base area = 56.8 m²

Height = 11.2 m

Volume of pyramid = 1/3 × base area × height

= 1/3 × 56.8 × 11.2

= 636.27/3

V = 212 m³

Problem 3 :

Solution :

Given, base area = 14.2 cm²

Height = 7.8 cm

Volume of cone = 1/3 × base area × height

= 1/3 × 14.2 × 7.8

= 110.76/3

V = 36.9 cm³

Problem 4:

Solution :

Radius of hemisphere = 4.8 cm

Volume of hemisphere = 2/3 × πr³

= 2/3 × 22/7 × (4.8)³

= 4866/21

V = 230 cm³

Problem 5 :

Solution :

Given, diameter of sphere = 3.7 cm

Radius r = 3.7/2 = 1.85 cm

Volume of sphere = 4/3 × πr³

= 4/3 × 22/7 × (1.85)³

= 557/21

V = 26.5 cm³

Problem 6 :

Solution :

Given, base area = 4.2 m²

Height = 1.87 m

Volume of cone = 1/3 × base area × height

= 1/3 × 4.2 × 1.87

= 7.85/3

V = 2.62 cm³

Problem 7 :

Solution :

Given, length = 2.8 cm

Width = 1.7 cm

Height = 2.9 cm

Volume of rectangular based pyramid = 1/3 × length × width × height

= 1/3 × 2.8 × 1.7 × 2.9

= 13.80/3

V = 4.60 cm³

Problem 8 :

Solution :

Given, base = 8.9 cm

Height = 6.8 cm

Area of base triangle = 1/2 × base × height

= 1/2 × 8.9 × 6.8

= 60.52/2

Area of base = 30.26 cm

Volume of triangular prism = 1/3 × base area × height

= 1/3 × 30.26 × 7.2

= 217.87/3

= 72.6 cm³

Problem 9 :

Solution :

diameter = 18.2 m

Radius = 18.2/2 = 9.1 m

Height = 21.6 m

Base area = Area of circle

= πr²

= 22/7 × 9.1 × 9.1

= 260.26

Volume of cone = 1/3 × πr²h

= 1/3 × 260.26 × 21.6

V = 1873 m³

Problem 10 :

A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto an height of 12 cm, find how many litres of milk is needed to serve 1600 students.

Solution :

Quantity of milk in one cylindrical glass = base area of cylinder x height

= πr² x h

Radius = 7/2 ==> 3.5 cm

Height = 12 cm

= 3.14 x 3.5² x 12

= 461.58 cm³

Quantity of milk in 1600 glasses = 1600 (461.58)

= 738528 cm³

1000 cm³ = 1 liter

= 738528/1000

= 738.528 liter

Approximately 739 liter.

Problem 11 :

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m². How many revolutions did it make?

Solution :

Area covered by the cylinder in one roll = 2πr x h

radius = 1.75 m and height = 2.5 m

= 2 x 3.14 x 1.75 x 2.5

= 27.47 m²

Number of revolutions = 5500/27.47

= 200

So, number of revolutions is 200.

Problem 12 :

A small village, having a population of 5000, requires 75 litres of water per head per day. The village has got an overhead tank of measurement 40 m × 25 m × 15 m. For how many days will the water of this tank last?

Solution :

Quantity of water in the tank = 40 x 25 x 15

= 15000 m³

1 m³ = 1000 liter

15000 = 15000000 liter

Quantity of water required per head = 75 liters

= 15000000/75

= 200000

= 200000/5000

= 40

So, water in this tank will be enough for 40 days.

Problem 13 :

A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made?

Solution :

Number of laddoo's required = Volume of spherical shape laddoo with the radius of 5 cm / Volume of spherical shape laddoo with the radius of 2.5 cm

= 4/3 πR³ / 4/3 πr³

= 5³ / 2.5³

= 125/15.625

= 8

So, the required number of laddoos we may make is 8.

Problem 14 :

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume of the solid so formed.

Solution :

Volume of cone = 1/3 πr² h

When 8 cm is the height of the cone, radius will be 6 cm and slant height is 10 cm.

= (1/3) x 3.14 x 6² x  8

= 301.44

So, the required volume is 301 cm³