NCERT Solutions Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume on Padhaiwise

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.1#

Cube of side $a$: surface area $6a^2$, volume $a^3$.

Cuboid $l \times w \times h$: surface area $2(lw + wh + hl)$, volume $lwh$.

1. The volume of a cube is 64 cm³. What is its total surface area? #

Solution

$$a^3 = 64$$

$$\Rightarrow a = 4 \text{ cm}$$$$\text{TSA} = 6a^2$$

$$\text{TSA} = 6 \times 16$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.2#

Cylinder of radius $r$ and height $h$:
$$\text{CSA} = 2\pi rh, \qquad \text{Volume} = \pi r^2 h$$

1. Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B. #

Solution

Let A have radius $r$ and height $h$. Then B has radius $2r$ and height $\dfrac{h}{2}$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.3#

Cone of radius $r$, height $h$ and slant height $l$:
$$l^2 = h^2 + r^2$$

$$\text{CSA} = \pi rl, \qquad \text{TSA} = \pi r(l + r), \qquad V = \dfrac{1}{3}\pi r^2 h$$

For $\pi$, use $\dfrac{22}{7}$ or $3.14$.

1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. #

Solution

$r = 12$ m and $l = 21$ m.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.4#

Sphere: $\text{SA} = 4\pi r^2$, $V = \dfrac{4}{3}\pi r^3$

Hemisphere: $\text{CSA} = 2\pi r^2$, $\text{TSA} = 3\pi r^2$, $V = \dfrac{2}{3}\pi r^3$

1. A ball bearing has a radius of 0.7 cm. Find its surface area. #

Solution

$$\text{SA} = 4 \times \dfrac{22}{7} \times 0.7^2$$

$$\text{SA} = 4 \times \dfrac{22}{7} \times 0.49$$

$$\text{SA} = 6.16 \text{ cm}^2$$

*2. Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm. #

Solution

Same metal, so weight is proportional to volume, which is proportional to $r^3$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — End-of-Chapter Exercises#

Guesstimates (marked below) have no single exact answer. Make a guess first, state your assumptions clearly, and then calculate. Different sensible assumptions give different — but reasonable — answers.

1. Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm. #

Guesstimate

$$V_{\text{container}} = 10 \times 15 \times 20$$