Solids Class 10 Extra Questions Kerala State Syllabus Maths Chapter 12

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SSLC Maths Chapter 12 Solids Important Questions and Answers

Solids Class 10 Extra Questions Kerala Syllabus

Solids Class 10 Kerala Syllabus Extra Questions

Question 1.
The figure shows one lateral face of a square pyramid. Its sides are 5 centimetres, 5 centimetres and 6 centimetres. What would be the slant height of a square pyramid in centimetres?

(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Answer:
(b) 4 cm
In the figure, the height of the triangle is the slant height of the pyramid.
∴ Slant height = \(\sqrt{5^2-3^2}\) = √16 = 4 cm

Question 2.
The radii of two hemispheres are in the ratio 1 : 2. What is the ratio of their surface areas?
(a) 1 : 2
(b) 1 : 8
(c) 1 : 4
(d) 1 : 16
Answer:
(c) 1 : 4
Ratio of radii is 1 : 2
Surface area of hemisphere = 3πr²
In surface area, radius is in the second degree.
So, ratio of their surface area = 1² : 2²= 1 : 4

Question 3.
Consider the following statements.
Statement (A): A long wire of length a is cut equally, and the ends are joined to make a square pyramid. The pyramid will have slant \(\frac{a}{2} \sqrt{3}\).
Statement (B): The lateral faces are equilateral triangles and have a height √3 times half of their side.
(a) Statement A is true, Statement B is false.
(b) Statement B is true, Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A.
Answer:
(a) Statement B is true, Statement A is false.
Since a square pyramid has 8 edges (4 base edges and 4 lateral edges),
The wire of length a is cut into 8 equal pieces.
Each edge = \(\frac {a}{8}\)
So, lateral faces are equilateral triangles.
Thus, the first part of statement (A) is true.
Height of the equilateral triangle = \(\frac{\sqrt{3}}{2} \times \frac{a}{8}=\sqrt{3} \times \frac{1}{2} \times \frac{a}{8}\)
So, the height of the equilateral triangle is √3 times half of its side.
Hence, statement (B) is true.
Slant height = Height of the equilateral triangle
= \(\frac{\sqrt{3}}{2} \times \frac{a}{8}\)
= \(\frac{a \sqrt{3}}{16} \neq \frac{a}{2} \sqrt{3}\)
Hence, statement (A) is false.

Question 4.
A copper rod of length 96 cm is divided into 8 equal parts, ends joined to form a square pyramid.
(a) Find the slant height of the pyramid.
(b) Find the area of the paper required to cover the lateral face.
Answer:
(a) Lateral faces are equilateral triangles.
Each side = \(\frac {96}{8}\) = 12 cm
Slant height = 6√3 cm

(b) Area of one lateral face = \(\frac {1}{2}\) × 12 × 6√3 = 36√3 cm²
Lateral surface area = 4 × 36√3 = 144√3 cm²
The area of the paper required to cover the lateral face = 144√3 cm²

Question 5.
The length of the base edge of a square pyramid is twice its height. If the base perimeter is 40 cm.
(a) Find slant height
(b) Find the lateral surface area.
Answer:
(a) Height is half of the base edge.
Base perimeter = 40 cm
base edge =10 cm
height = 5 cm
Slant height = 5√2 cm
(b) Lateral surface area = 4 × \(\frac {1}{2}\) × 10 × 5√2 = 100√2 cm²

Question 6.
The base area of a square pyramid is 100 cm², slant height is 13 cm.
(a) What is the height of the pyramid?
(b) What is the volume of the pyramid?
Answer:
(a) Base area = 100 cm²
⇒ side = 10 cm
Height = \(\sqrt{13^2-5^2}\) = 12 cm
(b) Volume = \(\frac {1}{3}\) × 100 × 12 = 400 cm³

Question 7.
From a wooden square prism, a square pyramid of the same base area and height is carved out. Base area 400 sq cm, height 24 cm.
(a) What is the volume?
(b) What is the slant height?
(c) Find the surface area of the pyramid.
Answer:
(a) \(\frac {1}{3}\) × 400 × 24 = 3200 cm³
(b) Slant height = \(\sqrt{10^2+24^2}\) = √676 = 26 cm
(c) Surface area = 400 + 4 × \(\frac {1}{2}\) × 20 × 26 = 1440 cm²

Question 8.
The radii of two cones are in the ratio 1 : 2, and their heights are in the ratio 2 : 1.
(a) What is the ratio between their volume?
(b) If the volume of the small cone is 10, then what is the volume of the big cone?
Answer:
(a) 1² × 2 : 2² × 1 = 1 : 2
(b) 20 cm³

Question 9.
A circular plate of radius 20 cm is divided into 4 equal sectors. One of them rolled up to form a cone.
(a) What is the slant height?
(b) What is the radius of the cone?
Answer:
(a) 20 cm
(b) 5 cm

Question 10.
From a square prism of edge length 30 cm and height 20 cm, a cone of maximum size is carved out.
(a) What is the radius of the cone?
(b) What is the slant height of the cone?
(c) Find the curved surface area.
Answer:
(a) Diameter of the cone = Side of the square = 30 cm
So, radius = 15 cm
(b) Slant height = \(\sqrt{15^2+20^2}=\sqrt{625}\) = 25 cm
(c) Curved surface area = π × 15 × 25 = 375π cm²

Question 11.
Curved surface area of a cone is 135π sq.cm, radius 9 cm.
(a) What is the slant height?
(b) What is the height of the cone?
(c) What is the volume of the cone?
Answer:
(a) π × 9 × l = 135π
⇒ l = 15 cm
(b) h = \(\sqrt{15^2-9^2}\) = 12 cm
(c) Volume = \(\frac {1}{3}\) × π × 9² × 12 = 324π cm³

Question 12.
A sector of central angle 300° and radius 36 cm rolled up to form a cone.
(a) What is the slant height of the cone?
(b) What is the base radius?
(c) Base is formed using a circular plate. What is the surface area of the cone?
Answer:
(a) 36 cm
(b) \(\frac{300}{360}=\frac{5}{6}\)
Radius is \(\frac {5}{6}\) part of the slant height.
∴ Radius = \(\frac {5}{6}\) × 36 = 30 cm
(c) Surface area = π × 30² + π × 30 × 36
= 900π + 1080π
= 1980π cm²

Question 13.
A tent in the shape of a cone has a base perimeter of 24π meters and a height of 5 meters.
(a) What is the base radius?
(b) What is the slant height?
(c) What is the area of the canvas required to construct the tent?
Answer:
(a) 12 metre
(b) Slant height = \(\sqrt{12^2+5^2}\) = √169 = 13 metre
(c) If only the lateral surface is covered,
Area = π × 12 × 13 = 156π m²

Question 14.
A cone made of wax has a base radius of 15 cm and slant height of 25 cm.
(a) What is the height of the cone?
(b) What is the volume of the cone?
(c) If it is melted to form small cones of radius 1 cm and height 4 cm, what is the volume of the small cone?
(d) How many small cones can be made?
Answer:
(a) Height = \(\sqrt{25^2-15^2}\) = 20 cm
(b) Volume = \(\frac {1}{3}\) × π × 15² × 20 = 1500π cm3
(c) Volume of the small cone = \(\frac {1}{3}\) × π × 12 × 4 = \(\frac {4}{3}\)π cm³
(d) Number of small cones = \(\frac{\frac{1}{3} \times \pi \times 15^2 \times 20}{\frac{1}{3} \times \pi \times 1^2 \times 4}\)
= \(\frac{1500 \pi}{\frac{4}{3} \pi}\)
= 1125

Question 15.
The surface area of a solid sphere is 100π square centimetres.
(a) What is the radius of the sphere?
(b) What is the volume of the sphere?
Answer:
(a) 4πr² = 100π
⇒ r² = 25
⇒ r = 5 cm
(b) Volume = \(\frac{4}{3} \pi r^3\)
= \(\frac {4}{3}\) × π × 5³
= \(\frac{500 \pi}{3}\) cm³

Question 16.
The surface area of a sphere is 576π square centimetres.
(a) What is the radius of the sphere?
(b) Find the volume.
Answer:
(a) 4πr² = 576π
⇒ r² = 144
⇒ r = 12 cm
(b) Volume = \(\frac {4}{3}\) × π × 12³ = 2304π cm³

Question 17.
The radius of a solid sphere is 6 cm. It is melted to make small spheres of radius 1 cm.
(a) What is the volume of the small Sphere?
(b) How many such small spheres can be formed?
Answer:
(a) Volume = \(\frac{4}{3} \pi \times 1^3=\frac{4}{3} \pi \mathrm{~cm}^3\)
Number of Spheres = \(\frac{\frac{4}{3} \pi \times 6^3}{\frac{4}{3} \pi \times 1^3}\) = 6³ = 216

Question 18.
From a wooden cube of side 12 cm, a sphere of maximum size is carved out.
(a) What is the radius of the sphere?
(b) What is the surface area?
(c) Find the volume.
Answer:
(a) Diameter of the sphere = Side of the cube = 12 cm
So, radius = 6 cm
(b) Surface Area = 4πr²
= 4 × π × 6²
= 144π cm²
(c) Volume = \(\frac {4}{3}\) × π × 6³ = 288π cm³