NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes on Padhaiwise

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1

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

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1#

Exercise 12.1 covers the surface area of combinations of solids — cubes, cuboids, cylinders, cones, hemispheres. The total surface area excludes the joined faces.

Key Formulas:

Cylinder: Curved SA = $2\pi rh$, Total SA = $2\pi r(r+h)$
Cone: Curved SA = $\pi rl$, Total SA = $\pi r(r+l)$, where $l = \sqrt{r^2+h^2}$
Sphere: SA = $4\pi r^2$
Hemisphere: Curved SA = $2\pi r^2$, Total SA = $3\pi r^2$
Cube: SA = $6a^2$; Cuboid: SA = $2(lb+bh+hl)$

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid. #

Answer

Let the side of each cube be $a$ cm. The volume of each cube is 64 cm³, so:

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2

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

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2#

Exercise 12.2 covers the volume of combinations of solids — finding total volume when solids are combined together.

Key Volume Formulas:

Cylinder: $V = \pi r^2 h$
Cone: $V = \dfrac{1}{3}\pi r^2 h$
Sphere: $V = \dfrac{4}{3}\pi r^3$
Hemisphere: $V = \dfrac{2}{3}\pi r^3$
Cuboid: $V = l \times b \times h$; Cube: $V = a^3$

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. #

Answer

Radius $r$ = 1 cm
Height of cone $h$ = 1 cm