NCERT Solution Class 9th Maths Chapter – 11 Surface Areas and Volumes Examples

Solution – Curved surface area = πrl
=22/7 × 7 × 10 cm2
= 220 cm2

Solution – Here, h = 16 cm and r = 12 cm.
So, from l² = h² + r² , we have

l = √16² 12² + cm = 20 cm

So, curved surface area = πrl
= 3.14 × 12 × 20 cm2
= 753.6 cm

Further, total surface area = πrl + πr²
= (753.6 + 3.14 × 12 × 12) cm2
= (753.6 + 452.16) cm2
= 1205.76 cm2

Solution – Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height

Here, l = √r² h² + = √(2.1)² 20 + cm
= √404.41 cm = 20.11 cm

Therefore, the curved surface area of the corn cob = πrl
= 22/7 × 2.1 × 20.11 cm² = 132.726 cm² = 132.73 cm² (approx.)
Number of grains of corn on 1 cm² of the surface of the corn cob = 4
Therefore, number of grains on the entire curved surface of the cob
= 132.73 × 4 = 530.92 = 531 (approx.)
So, there would be approximately 531 grains of corn on the cob.

Solution – The surface area of a sphere of radius 7 cm would be 4πr²
= 4πr² = 4 × 22/7 × 7×7 cm² = 616 cm²

Solution – The curved surface area of a hemisphere of radius 21 cm would be
= 2πr² = 2 × 22/7 × 21 × 21 cm² = 2772 cm²

Solution – Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding
space available for the motorcyclist is the surface area of the ‘sphere’ which is given by

4πr² = 4 × 22/7 × 3.5 × 3.5 m²
= 154 m²

Solution – Since only the rounded surface of the dome is to be painted, we would need
to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2πr
So, the radius of the dome = 17.6 × 7/2×22 m = 2.8 m

The curved surface area of the dome = 2πr²
= 2 × 22/7 × 2.8 × 2.8 m²
= 49.28 m²
Now, cost of painting 100 cm² is ₹ 5.
So, cost of painting 1 m² = ₹500
Therefore, cost of painting the whole dome
= ₹500 × 49.28
= ₹24640

So, volume of the cone = 1/3 πr² h = 1/3 ×22/7 √7× 7 √7× √7× 21 cm³
= 7546 cm³

Solution – Since the area of the canvas = 551 m² and area of the canvas lost in wastage is 1 m², therefore the area of canvas available for making the tent is (551 – 1) m² = 550 m².Now, the surface area of the tent = 550 m² and the required base radius of the conical tent = 7m
Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!)

Solution – Required volume =4/3 πr³
= 4/3 × 22/7 × 11.2 11.2 11.2 cm³ = 5887.32 cm

Solution – Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.

Now, volume of the sphere = 4/3 πr³
= 4/3 × 22/7 × 4.9 × 4.9 × 4.9 cm³
= 493 cm³ (nearly)

Further, mass of 1 cm3 of metal is 7.8 g.
Therefore, mass of the shot-putt = 7.8 × 493 g
= 3845.44 g = 3.85 kg (nearly)

Solution – The volume of water the bowl can contain
= 2/3 πr 3
=2/3 × 22/7 × 3.5 × 3.5 × 3.5 = 89.8 cm