Chapter 7 Mensuration Class 10 Maharashtra Board TextBook Solution
Practice set 7.1 | Q 5 | Page 145
The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Answer:-
The volume of the cuboid can be found by multiplying its three dimensions:
Volume of cuboid = 44 cm x 21 cm x 12 cm Volume of cuboid = 11,088 cm^3
Since the cuboid is melted to form a cone, the volume of the cone will be equal to the volume of the cuboid. The volume of a cone is given by:
Volume of cone = (1/3) x π x r^2 x h
where r is the radius of the base of the cone, and h is its height. We are given h = 24 cm.
Substituting the values of the volume of the cuboid and the height of the cone, we get:
11,088 = (1/3) x π x r^2 x 24
Simplifying, we get:
r^2 = (3 x 11,088) / (24 x π)
r^2 = 115.5
Taking the square root of both sides, we get:
r = √115.5 ≈ 10.75 cm
Therefore, the radius of the base of the cone is approximately 10.75 cm.
Solution
The volume of the cuboid is given by:
$$text{Volume of cuboid} = l times b times h = 44 text{ cm} times 21 text{ cm} times 12 text{ cm} = 11,088 text{ cm}^3$$
Since the cuboid is melted to form a cone, the volume of the cone will be equal to the volume of the cuboid. The volume of a cone is given by:
$$text{Volume of cone} = frac{1}{3} times pi times r^2 times h$$
where $r$ is the radius of the base of the cone, and $h$ is its height. We are given $h = 24$ cm.
Substituting the values of the volume of the cuboid and the height of the cone, we get:
$$11,088 = frac{1}{3} times pi times r^2 times 24$$
Simplifying, we get:
$$r^2 = frac{3 times 11,088}{24 times pi}$$
$$r^2 = 115.5$$
Taking the square root of both sides, we get:
$$r = sqrt{115.5} approx 10.75 text{ cm}$$
Therefore, the radius of the base of the cone is approximately 10.75 cm.
Chapter 7 Mensuration Textbook Solution 7.1
Practice set 7.1
Click Here for all Textbook Answers of Chapter 7 Mensuration