Solids Questions and Answers Class 10 Maths Chapter 12 Kerala Syllabus Solutions

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 12 Solids Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 12 Solids Questions and Answers

Solids Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 12 Solids Solutions

Class 10 Maths Chapter 12 Kerala Syllabus – Pyramids and Area

(Textbook Page No. 259-260)

Question 1.
A square of side 5 centimetres, and four isosceles triangles of base 5 centimetres and height 8 centimetres, are to be put together to make a square pyramid. How many square centimetres of paper are needed?
Answer:
Area of one lateral face = \(\frac {1}{2}\) × 8 × 5 = 20 sq.cm
Lateral surface area = 4 × 20 = 80 sq.cm
Base area = 5² = 25 sq.cm
Area of the paper required to make the square pyramid = 80 + 25 = 105 sq.cm

Question 2.
A toy is in the shape of a square pyramid of base edge 16 centimetres and slant height 10 centimetres. What is the total cost of painting 500 such toys, at 80 rupees per square metre?
Answer:
Area of one lateral face = \(\frac {1}{2}\) × 16 × 10 = 80 sq.cm
Lateral surface area = 4 × 80 = 320 sq. cm
Base area = 16² = 256 sq. cm
Total surface area = 320 + 256 = 576 sq. cm
Area of 500 such toys = 500 × 576
= 288000 sq. cm
= 28.8 sq. cm
(1 metre = 100 cm, 1 sq.m = 100 × 100 = 10000 sq.cm)
Total cost of paining 500 toys = 80 × 28.8 = 2,304 rupees

Question 3.
The lateral faces of a square pyramid are equilateral triangles and the length of a base edge is 30 centimetres. What is its surface area?
Answer:
Length of all edges are 30 cm.
Lateral faces are equilateral triangles.
Height of the triangle = 15√3 cm
Area = \(\frac {1}{2}\) × 30 × 15√3 = 225√3 sq.cm
Lateral surface area = 4 × 225√3 = 900√3 sq.cm
Base area = 30² = 900 sq.cm
Total surface area = 900√3 + 900 sq.cm

Question 4.
The perimeter of the base of a square pyramid is 40 centimetres, and the total length of all its edges is 92 centimetres. Calculate its surface area.
Answer:
Length of base edge = \(\frac {40}{4}\) = 10 cm
Total length of all edges is the sum of the lengths of the 4 base edges and the 4 lateral edges.
40 + 4 × lateral edge = 92
⇒ 4 × lateral edge = 92 – 40 = 52 cm
⇒ Lateral edge = \(\frac {52}{4}\) = 13 cm

Slant height = \(\sqrt{13^2-5^2}\) = 12 cm
Total surface area of the pyramid = 10² + 4 × \(\frac {1}{2}\) × 10 × 12
= 100 + 240
= 340 sq. cm

Question 5.
Can we make a square pyramid with the lateral surface area equal to the base area?
Answer:
Assume that the lateral surface area is equal to the base area.
Then the area of each triangle formed by drawing diagonals of the base is equal to the area of each lateral face.
In such cases, no square pyramids can be formed.

SCERT Class 10 Maths Chapter 12 Solutions – Height and Slant Height

(Textbook Page No. 261-262)

Question 1.
Using a square and four triangles with dimensions as specified in the picture, a pyramid is made. What is the height of this pyramid?

What if the squares and triangles are like this?

Answer:
(a) Height = \(\sqrt{18^2-12^2}\)
= √180
= \(\sqrt{36 \times 5}\)
= 6√5 cm

(b) Base edge = 24 cm
So, base diagonal = 24√2 cm
Half of the base diagonal = 12√2 cm
Height = \(\sqrt{30^2-(12 \sqrt{2})^2}\) = √612 cm

Question 2.
A square pyramid of base edge 10 centimetres and height 12 centimetres is to be made of paper. What should be the dimensions of the triangles?
Answer:
Slant height = \(\sqrt{5^2+12^2}\) = 13
Lateral faces are isosceles triangles. There are 4 triangles.
The base side of each triangle is 10 cm, and the height to that side is 13 cm.
The other two sides are equal.
So, the other sides = \(\sqrt{13^2+\left(\frac{10}{2}\right)^2}\)
= \(\sqrt{13^2+5^2}\)
= √194 cm

Question 3.
Prove that in any square pyramid, the squares of the height, slant height, and lateral edge are in arithmetic sequence.
Answer:
If we subtract the square of height from the square of slant height, we get the square of half of the base edge.
slant height² – height² = \(\left(\frac{\text { base edge }}{2}\right)^2\)
Also, if we subtract the square of the height from the square of the lateral edge, we get the square of half of the base edge.
lateral edge² – height² = \(\left(\frac{\text { base edge }}{2}\right)^2\)
These square differences are the common difference of the arithmetic sequence.
∴ The squares of the height, slant height, and lateral edge are in an arithmetic sequence.

Question 4
A square pyramid is to be made with the triangle shown here as a lateral face. What would be its height? What if the base edge is 40 centimetres instead of 30 centimetres?

Answer:
Half of the base edge = 15 cm
Lateral edge = 25 cm
Slant height = \(\sqrt{25^2-15^2}\)
= √400
= 20 cm
Height of the pyramid = \(\sqrt{20^2-15^2}\)
= √175
= \(\sqrt{25 \times 7}\)
= 5√7 cm

Class 10 Maths Kerala Syllabus Chapter 12 Solutions – Volume of a Pyramid

(Textbook Page No. 263)

Question 1.
What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?
Answer:
Form a right triangle joining half of the base edge and the slant height.
Height = \(\sqrt{15^2-5^2}\)
= √200
= 10√2 cm
Volume = \(\frac {1}{3}\) × Base area × Height
= \(\frac {1}{3}\) × 10² × 10√2
= \(\frac{1000 \sqrt{2}}{3}\) cm³

Question 2.
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times is the height of the second pyramid the height of the first?
Answer:
Let the base edge of the first pyramid be a and its height be h.
Then, volume = \(\frac {1}{3}\) × Base area × height
= \(\frac {1}{3}\) × a² × h
Let the base edge of the second pyramid be \(\frac {a}{2}\) and its height be x.
Then, volume = \(\frac{1}{3} \times\left(\frac{a}{2}\right)^2 \times x\)
⇒ \(\frac{1}{3} \times a^2 \times h=\frac{1}{3} \times\left(\frac{a}{2}\right)^2 \times x\)
⇒ a²h = \(\frac{a^2 x}{4}\)
⇒ h = \(\frac {x}{4}\)
⇒ x = 4h
∴ The height of the second pyramid is 4 times the height of the first.
OR
One-third of the square of the base edge multiplied by the height gives the volume.
If the base edge becomes half, then its square becomes one-fourth.
Then, to get the same volume, the height must become 4 times.

Question 3.
The base edges of two square pyramids are in the ratio 1 : 2 and their heights in the ratio 1 : 3. The volume of the first is 180 cubic centimetres. What is the volume of the second?
Answer:
Let base edges be a, 2a, and heights be h, 3h
Volume of the first pyramid = \(\frac {1}{3}\) × a² × h = 180
⇒ a²h = 540
Volume of the second pyramid = \(\frac {1}{3}\) × (2a)² × 3h
= 4a²h
= 4 × 540
= 2160 cm³
OR
The ratio of base edges is 1 : 2.
Therefore, the ratio of the base areas is 1:4.
The ratio of heights is 1 : 3.
Therefore, the ratio between the product of base area and height is 1:12.
That is, the ratio between their volumes is 1 : 12.
The volume of the second pyramid is 12 times the volume of the first pyramid.
∴ Volume of the second pyramid = 12 × 180 = 2160 cm³

Question 4.
All edges of a square pyramid are 18 centimetres. What is its volume?
Answer:
Form a right triangle joining half of the base edge = 9 cm, lateral edge = 18 cm, and slant height.
It is a 30° – 60° – 90° triangle.
Slant height = 9√3 cm
Form a right triangle joining the slant height, half of the base edge, and the height.
Height = \(\sqrt{(9 \sqrt{3})^2-9^2}\) = 9√2 cm
Volume = \(\frac {1}{3}\) × 18² × 9√2 = 972√2 cm³

Question 5.
The slant height of a square pyramid is 25 centimetres, and its surface area is 896 square centimetres. What is its volume?
Answer:
If base edge be x, then x² + 4 × \(\frac {1}{2}\) × x × 25 = 896
⇒ x² + 50x – 896 = 0
x² + 50x – 896 = x² + (a + b)x + ab
⇒ a + b = 50, ab = -896
(a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 50² – 4 × -896 = 6084
⇒ a – b = 78
a + b = 50, a – b = 78
⇒ 2a = 128
⇒ a = 64, b = -14
x² + 50x – 896 = (x + 64)(x – 14) = 0
∴ x = 14 is the base edge.
Height = \(\sqrt{25^2-7^2}\) = 24 cm
Volume = \(\frac {1}{3}\) × 14² × 24 = 1568 cm³

Question 6.
All edges of a square pyramid are of the same length, and its height is 12 centimetres. What is its volume?
Answer:
Let a be the length of all edges.
Then, base edge = lateral edge = a
Form a right triangle joining half of the base edge, the lateral edge, and the slant height.
It is a 30° – 60° – 90° triangle.
Slant height = \(\frac{a \sqrt{3}}{2}\) cm
Form a right triangle joining the slant height, half of the base edge, and the height.
Height = \(\sqrt{\left(\frac{a \sqrt{3}}{2}\right)^2-\left(\frac{a}{2}\right)^2}\) = 12 cm
⇒ \(\sqrt{\frac{2 a^2}{4}}\) = 12
⇒ \(\frac{a}{\sqrt{2}}\) = 12
⇒ a = 12√2 cm
∴ Volume = \(\frac {1}{3}\) × (12√2)² × 12 = 1152 cm³

Question 7.
What is the surface area of a square pyramid of base perimeter 64 centimetres and volume 1280 cubic centimetres?
Answer:
Base edge = \(\frac {64}{4}\) = 16 cm
Base area =16² = 256 cm²
Let h be the height.
\(\frac {1}{3}\) × 256 × h = 1280
⇒ h = \(\frac{1280 \times 3}{256}\)
⇒ h = 15 cm
Slant height = \(\sqrt{15^2+8^2}\) = 17 cm
Total surface area = 16² + 4 × \(\frac {1}{2}\) × 16 × 17 = 800 cm²

SSLC Maths Chapter 12 Questions and Answers – Cone

(Textbook Page No. 265)

Question 1.
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60° cut out from a circle of radius 10 centimetres?
Answer:
\(\frac{60}{360}=\frac{1}{6}\)
Base radius = 10 × \(\frac {1}{6}\) = \(\frac {10}{6}\) = \(\frac {5}{3}\) cm
Slant height = 10 cm

Question 2.
What is the central angle of the sector to be used to make a cone of base radius 10 centimetres and slant height 25 centimetres?
Answer:
Slant height of the cone is the radius of the sector, which is the radius of the circle from which the sector is cut off.
So, radius of the cone = 10 cm
\(\frac{10}{25}=\frac{2}{5}\)
Central angle is \(\frac {2}{5}\) part of 360°.
So, central angle = 360 × \(\frac {2}{5}\) = 144°

Question 3.
What is the ratio of the base radius and slant height of a cone made by rolling up a semicircle?
Answer:
Central angle of the semicircle = 180°, which is \(\frac {1}{2}\) part of 360°.
The radius of the sector (semicircle) is the slant height.
Its \(\frac {1}{2}\) part is the radius of the cone.
The ratio of base radius and slant height is 1 : 2.

Kerala Syllabus Class 10 Maths Chapter 12 Solutions – Curved Surface Area

(Textbook Page No. 267)

Question 1.
What is the area of the curved surface of a cone of base radius 12 centimetres and slant height 25 centimetres?
Answer:
Curved surface area = πrl
= π × 12 × 25
= 300π cm²

Question 2.
What is the surface area of a cone of base diameter 30 centimetres and height 20 centimetres?
Answer:
Base radius = 15 cm
Height = 20 cm
Slant height = \(\sqrt{15^2+20^2}\)
= √625
= 25 cm
Surface area = Base area + Curved surface area
= π × 15² + π × 15 × 25
= 600π cm²

Question 3.
A conical firework is of base diameter 10 centimetres and height 12 centimetres. 10000 such fireworks are to be wrapped in coloured paper. The price of the coloured paper is 2 rupees per square metre. What is the total cost?
Answer:
Slant height = \(\sqrt{12^2+5^2}\) = 13 cm
Surface area of a cone = π × 5² + π × 5 × 13 = 90π cm²
Total surface area of 10000 fireworks = 90π × 10000 = 90π m²
Total cost = 90π × 2
= 180π
= 180 × 3.14
= 565.2 rupees

Question 4.
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area.
Answer:
If the radius of the semicircle is r, then the slant height of the cone is r.
Base radius = \(\frac {r}{2}\)
Curved surface area = π × r × \(\frac {r}{2}\) = \(\frac{\pi r^2}{2}\)
Base area = \(\pi \times\left(\frac{r}{2}\right)^2=\frac{\pi r^2}{4}\)
The curved surface area \(\frac{\pi r^2}{2}\) is twice the base area \(\frac{\pi r^2}{4}\).

Class 10 Maths Chapter 12 Kerala Syllabus – Volume of a Cone

(Textbook Page No. 268)

Question 1.
The base radius and height of a cylindrical block of wood are 15 centimetres and 40 centimetres. What is the volume of the largest cone that can be carved out of this?
Answer:
Volume = \(\frac {1}{3}\) × π × 15² × 40 = 3000π cm³

Question 2.
The base radius and height of a solid metal cylinder are 12 centimetres and 20 centimetres. By melting it and recasting, how many cones of base radius 4 centimetres and height 5 centimetres can be made?
Answer:
The number of cones that can be made = Volume of the melted solid metal cylinder ÷ Volume of the cone made by recasting
= \(\frac{\pi \times 12^2 \times 20}{\frac{1}{3} \times \pi \times 4^2 \times 5}\)
= 108

Question 3.
A sector of central angle 216° is cut out from a circle of radius 25 centimetres and is rolled up into a cone. What are the base radius and height of the cone? What is its volume?
Answer:
\(\frac{216}{360}=\frac{3}{5}\)
Central angle of the sector is \(\frac {3}{5}\) part of 360°.
The radius of the cone is \(\frac {3}{5}\) part of the radius of the sector.
Radius of the cone = \(\frac {3}{5}\) × 25 = 15 cm
Slant height = 25 cm
Height of the cone = \(\sqrt{25^2-15^2}\) = 20 cm
Volume = \(\frac {1}{3}\) × π × 15² × 20 = 1500π cm³

Question 4.
The base radii of two cones are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?
Answer:
Radii are 3r, 5r, and heights 2h, 3h
Ratio of their volume = \(\frac {1}{3}\) × π × (3r)² × 2h : \(\frac {1}{3}\) × π × (5r)² × 3h
= 18 : 75
= 6 : 25

Question 5.
Two cones have the same volume, and their base radii are in the ratio 4 : 5. What is the ratio of their heights?
Answer:
Let their heights be h₁, h₂
\(\frac{1}{3} \times(4 r)^2 \times h_1=\frac{1}{3} \times(5 r)^2 \times h_2\)
⇒ 16h1 = 25h2
⇒ \(\frac{h_1}{h_2}=\frac{25}{16}\)
∴ Ratio of height = 25 : 16

SCERT Class 10 Maths Chapter 12 Solutions – Sphere

(Textbook Page No. 270)

Question 1.
The surface area of a solid sphere is 120 square centimetres. If it is cut into two halves, what would be the surface area of each hemisphere?
Answer:
4πr² = 120
⇒ πr² = 30
Surface area of hemisphere = 3 × πr²
= 3 × 30
= 90 cm²

Question 2.
The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii? And the ratio of their surface areas?
Answer:
Volume of a sphere = \(\frac{4}{3} \pi r^3\)
In volume radius is in the third degree.
27 = 3³, 64 = 4³
So, the ratio of their radii = 3 : 4
Surface area of a sphere is 4πr².
In surface area, radius is in the second degree.
So, ratio of their surface area = 3² : 4² = 9 : 16

Question 3.
The base radius and length of a metal cylinder are 4 centimetres and 10 centimetres. If it is melted and recast into spheres of radius 2 centimetres each, how many spheres can be made?
Answer:
Number of spheres = \(\frac{\text { Volume of cylinder melted }}{\text { Volume of sphere formed by recasting }}\)
= \(\frac{\pi \times 4^2 \times 10}{\frac{4}{3} \pi \times 2^3}\)
= 15

Question 4.
A metal sphere of radius 12 centimetres is melted and recast into 27 small spheres of equal size. What is the radius of each small sphere?
Answer:
Volume of the small sphere = \(\frac{\text { Volume of melted sphere }}{27}\)
= \(\frac{\frac{4}{3} \pi \times 12^3}{27}\)
= \(\frac {4}{3}\) × π × 64
= \(\frac {4}{3}\) × π × 4³
So, the radius of the small sphere is 4 cm.

Question 5.
From a solid sphere of radius 10 centimetres, the largest cone of height 16 centimetres is carved out. What fraction of the volume of the sphere is the volume of the cone?
Answer:

From the figure,
10² = 6² + r²
⇒ r = 8 cm
Volume of the cone = \(\frac {1}{3}\) × π × 8² × 16
Volume of the sphere = \(\frac {4}{3}\) × π × 10³
∴ Ratio of Volumes = \(\frac{\frac{1}{3} \pi \times 8^2 \times 16}{\frac{4}{3} \pi \times 10^3}=\frac{32}{125}\)

Question 6.
A solid sphere is cut into two hemispheres. From one, a square pyramid, and from the other a cone, each of maximum possible size, are carved out. What is the ratio of their volumes?
Answer:
Consider the figure,

The figure shows the base of a square pyramid and a cone.
From the figure,
Base edge of the square pyramid = √2r
Base radius of the cone is r.
The height of both the square pyramid and the cone is equal to the radius r.
Volume of the square pyramid = \(\frac{1}{3} \times(\sqrt{2} r)^2 \times r\)
Volume of the cone = \(\frac {1}{3}\) × π × r² × r
Ratio of volumes = \(\frac{1}{3} \times(\sqrt{2} r)^2 \times r: \frac{1}{3} \pi \times r^2 \times r\) = 2 : π

Solids Class 10 Notes Pdf

Class 10 Maths Chapter 12 Solids Notes Kerala Syllabus

→ Surface area of a square pyramid = base area + area of lateral faces

→ Volume of a square pyramid = \(\frac {1}{3}\) × base area × height

→ We can make a cone by rolling up a sector of a circle.

→ Radius of the sector = Slant height of the cone.

→ Curved surface area of a cone is the area of the sector used to make it.

→ Curved surface area (CSA) = πrl, where, r = base radius of the cone, l = slant height of the cone.

→ Volume of a cone = \(\frac {1}{3}\) × base area × height

→ The surface of a sphere is curved. The distance from the centre of the sphere to its surface is the radius of the sphere.

→ Surface area of a sphere = 4πr²

→ Surface area of a hemisphere = 3πr²

→ Volume of a sphere is \(\frac{4}{3} \pi r^3\)

→ Volume of a hemisphere is \(\frac{2}{3} \pi r^3\)

Introduction
Have you ever wondered how much ice cream your favourite cone can hold or how the ancient Egyptians built such massive pyramids? The answer lies in understanding solids, the three-dimensional shapes that surround us in the real world. In this chapter, we’ll embark on a journey into the fascinating world of 3D, focusing on three specific shapes: cones, square pyramids, and spheres.

Square Pyramids
Just think about the Egyptian pyramids. Each one has four triangle-shaped walls that meet at a point. We will learn how to spot square pyramids and what makes them unique, such as their height and lateral lines. We will also learn how much space they can hold (volume) and how much “skin” they have (surface area).

Cones
Think of your favourite ice cream cone. The top is pointy, and the bottom is round. We’ll learn how to recognise cones, understand their parts, like the slant and bottom, and figure out how much ice cream or juice they can hold.

Spheres
Imagine a ball – it’s perfectly round, no matter where you look at it. Spheres are cool shapes found in many things, like planets and marbles. We’ll learn all about spheres and discover how much space they hold (volume) and how much “skin” they have (surface area). We will examine the distinct characteristics of every form, mastering the skills to recognise them, compute their volume-the quantity of space they occupy and even investigate their surface area, which is the sum of their faces. Get ready to discover the mysteries behind these fascinating 3D wonders as you buckle in!

Solids with a base and a sharp apex opposite to it are called pyramids. Its base must be a polygon. The pyramid is named on the basis of the number of sides of its base. If base is a square, then it is a square pyramid. The sides of the polygon forming the base of a pyramid are called base edges, and the other sides of the triangles are called lateral edges. The topmost point of a pyramid is called its apex.

Pyramids and Area
Solids with a base and a sharp apex opposite to it are called pyramids. Its base must be a polygon. The pyramid is named on the basis of the number of sides of its base. If base is a square, then it is a square pyramid. Lateral faces of the pyramid are triangles. It can be an isosceles or an equilateral triangle. One edge of the triangle is the base edge of the pyramid. The other two edges of the triangle are the lateral edges. A solid with a square base and lateral faces is a pyramid. So, we have to calculate the perimeter and area of a square, and the perimeter and area of a triangle.

We can make square pyramids by cutting paper and pasting the edges.

The first picture shows the picture drawn on a paper to make the square pyramid. Cut and paste through the outer edges to make the square pyramid. The second picture shows the completed pyramid. In both pictures, A is the apex of the pyramid. B is the centre of the base edge, and C is the vertex of the base. The length of AB is the slant height of the pyramid. BC is the half of the base edge, and AC is the length of the lateral edge. In both pictures, ABC is a right triangle.

Question 1.
The figure shows the sketch to make a square pyramid from a square paper. The base edge of the pyramid is 10 cm, and the lateral edge is 12 cm.

(a) What is the slant height?
(b) What is the length of the side of the square paper?
Answer:
13² = 5² + (Slant height)²
Slant height = \(\sqrt{13^2-5^2}\)
= √144
= 12 cm
(b) Base edge = \(\frac {40}{4}\) = 10 cm

Question 2.
Divide a 96 cm long rod into 8 equal parts, join the edges to make a square pyramid.
(a) Calculate the slant height.
(b) Find the area of the paper required to cover its lateral face.
Answer:
(a) Length of an edge = \(\frac {96}{8}\) = 12 cm
Lateral faces are equilateral triangles.
Length of one side is 12 cm.
Slant height = 6√3 cm
(b) Area of one lateral face = \(\frac {1}{2}\) × 12 × 6√3 = 36√3 sq.cm
Total lateral surface area is the sum of the area of four triangles = 4 × 36√3 = 144√3 sq.cm

Question 3.
Four equal triangles are cut off from a segment of a circle with a central angle of 240° and radius 12 cm, as shown in the figure. Make a square pyramid with these triangles as lateral faces.
(a) What is the base edge of the pyramid?
(b) What is the slant height?
(c) Find the lateral surface area.
Answer:
Central angle of a segment = \(\frac {240}{4}\) = 60°
All triangles cut off are equilateral triangles.
(a) Length of base edge = 12 cm
(b) Slant height = 6√3 cm
(c) Lateral surface area = 4 × \(\frac {1}{2}\) × 12 × 6√3 = 144√3 sq.cm

Height and Slant Height
The height of a pyramid is the perpendicular distance from the apex to the base. In the case of a square pyramid, the foot of the perpendicular is the point where the diagonals of the base meet.

Both of the shaded triangles in the picture are right triangles.
Height of the pyramid, half of the base edge, and slant height are joined to form a right triangle.
The height of the pyramid, half of the diagonal, and the lateral edge are joined to form a right triangle.
A right triangle is also formed by joining the slant height, half of the base edge, and the lateral edge.

Question 1.
Base perimeter of a square pyramid is 8 metres, length of the lateral edge is 8 metres.
(a) What is the length of the base edge?
(b) Find the base diagonal.
(c) What is the height of the pyramid?
Answer:
(a) Length of base edge = \(\frac {8}{4}\) = 2 m
(b) Base diagonal = 2√2 cm
(c) Half of base diagonal = √2 m
Height = \(\sqrt{8^2-\sqrt{2}^2}\) = √62 m

Question 2.
The base edge of a square pyramid is double the height. If the base perimeter is 40 cm, then
(a) Find slant height.
(b) Find lateral surface area.
Answer:
(a) Let the height be x, then base edge is 2x.
4 × 2x = 40
⇒ x = 5
Height = 5
⇒ Half of the base edge = 5 cm
Slant height = 5√2 cm
(b) Lateral surface area = 4 × \(\frac {1}{2}\) × 10 × 5√2 = 100√2 sq.cm

Question 3.
The base area of a square pyramid is 100 sq cm, slant height is 13 cm.
(a) What is the length of the base edge?
(b) What is the surface area?
(c) What is the height of the pyramid?
Answer:
(a) Length of base edge = √100 = 10 cm
Slant height = 13 cm
(b) Surface area = 100 + 4 × \(\frac {1}{2}\) × 10 × 13 = 360 sq. cm
(c) Height of the pyramid = \(\sqrt{13^2-5^2}\) = √144 = 12 cm

Question 4.
The total surface area of a square pyramid is 360 sq.cm, slant height is 13 cm.
(a) If base edge is x, form the equation.
(b) What is the base edge?
(c) Find the height of the pyramid.
Answer:
(a) Lateral surface area = 4 × \(\frac {1}{2}\) × x × 13 = 26x
⇒ x² + 26x = 360
⇒ x² + 26x – 360 = 0
(b) x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-26 \pm \sqrt{676+1440}}{2}\)
= 10
(c) Height = 12 cm

Volume of a Pyramid
The volume of a solid is the measure of the space it occupies. The volume of any prism is equal to the product of the base area and the height. But, the volume of a square pyramid is equal to a third of the base area and the height. That is, the volume of a pyramid is a third of the volume of the prism of the same base area and height.
For example, let the base area of a prism be 100 sq cm and height 30 cm.
Then its volume is 100 × 30 = 3000 cubic centimetres.
The volume of a pyramid with the same base area and height is 1000 cubic centimetres.

Question 1.
The base area of a square pyramid is 100 sq cm, and the slant height is 13 cm.
(a) Find the height of the pyramid.
(b) Find the volume of the pyramid.
Answer:
(a) Length of base edge = √100 = 10 cm
Half of the base edge, height, and slant height together form a right triangle.
Height = \(\sqrt{13^2-5^2}\) = 12 cm
(b) Volume = \(\frac {1}{3}\) × base area × height
= \(\frac {1}{3}\) × 100 × 12
= 400 cm³

Question 2.
From a wooden square prism, a square pyramid of the same base area and height is carved out. Base area 400 sq cm and height 24 cm.
(a) What is the volume of the pyramid?
(b) What is the slant height?
(c) Find the total surface area.
Answer:
(a) Volume = \(\frac {1}{3}\) × base area × height
= \(\frac {1}{3}\) × 400 × 24
= 3200 cm³
(b) Base edge = √400 = 20 cm, Height = 24 cm
Slant height = \(\sqrt{10^2+24^2}\)
= √676
= 26 cm
(c) Total surface area = Base area + 4 × area of triangle in one lateral face
= 400 + 4 × \(\frac {1}{2}\) × 20 × 26
= 400 + 1040
= 1440 cm²

Question 3.
The base perimeter and height of a square prism are 32 cm and 3 cm, respectively. From this, a square pyramid of maximum size is carved out.
(a) What is the length of the base edge of the pyramid?
(b) What is the slant height?
(c) Find the total surface area.
(d) Find the volume.
Answer:
(a) Base perimeter = 32 cm
So, base edge = \(\frac {32}{4}\) = 8 cm
(b) Slant height = \(\sqrt{4^2+3^2}\) = 5 cm
(c) Base area = 8² = 64 cm
Total surface area = Base area + 4 × Area of one isosceles triangle
= 64 + 4 × \(\frac {1}{2}\) × 8 × 5
= 64 + 80
= 144 sq. cm
(d) Volume = \(\frac {1}{3}\) × 64 × 3 = 64 cm³

Cone
Like this, we can make a cone by rolling up a sector of a circle.

Basic measures of a sector are the central angle and the radius of the sector. Using these measures, we can calculate the length of the arc and the area. What part of 360° is the central angle of the sector? That much of the perimeter is the length of the arc. That much of the area is the area of the sector. From a circular plate, cut out a semicircle, rolling up to form a cone. Half of 360°, that is 180°, is the central angle of the sector. The base perimeter of the cone formed by rolling up a semicircle is half of the circumference of the circle.

Question 1.
A circular paper of radius 12 cm is cut into 4 equal sectors. If one of these sectors is rolling up to form a cone.
(a) What is the central angle of one sector?
(b) What is the radius and slant height of the cone?
Answer:
(a) 360 × \(\frac {1}{4}\) = 90°
(b) Slant height of the cone = 12 cm
Radius = 12 × \(\frac {1}{4}\) = 3 cm

Question 2.
From a circular plate, a sector of central angle 120° is cut off. The radius of the sector formed by rolling up the sector is 10 cm.
(a) What is the radius of the circular plate?
(b) What is the slant height of the cone?
Answer:
(a) 30 cm
(b) 30 cm

Question 3.
A circular plate of circumference 36π cm is divided into six equal sectors. One of them rolled up to form a cone.
(a) What is the radius of the plate?
(b) What is the radius of the cone?
Answer:
(a) 2πr = 36π
⇒ r = 18 cm
(b) Radius of the cone = \(\frac {18}{6}\) = 3 cm

Curved Surface Area
A sector rolling up to form a cone, the area of the sector is the lateral surface area of the cone. The arc length of the sector is the base area of the cone. The length of the perpendicular from the apex of the cone to the centre of the base is the height. We can see a right triangle with the slant height of the cone as the hypotenuse, base radius, and height as perpendicular sides.

Question 1.
A circular plate of radius 20 cm is divided into 4 sectors. One of them is rolled up to form a cone.
(a) What is the slant height?
(b) What is the radius of the cone?
(c) What is the height of the cone?
Answer:
(a) 20 cm
(b) Central angle of the sector is \(\frac {1}{4}\) part of 360°.
Radius of the cone is 5 cm.
Height = \(\sqrt{20^2-5^2}\)
= √375
= 5√15 cm

Question 2.
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 base radius of the cone?
(b) What is the slant height of the cone?
Answer:
(a) Half of the base edge is the radius of the base circle.
Radius = 15 cm
(b) Slant height = \(\sqrt{20^2+15^2}\)
= √625
= 25 cm

Question 3.
A circular plate of radius 12 cm is divided into two sectors. Its central angles are 120° and 240°. Both sectors rolled up to form cones.
(a) What is the common measure for both cones?
(b) What is the radius of the cone formed using a sector of a central angle?
(c) What is the radius of the cone formed using a sector of central angle?
(d) Which cone is highest?
Answer:
(a) Common measure is the slant height.
(b) \(\frac{120}{360}=\frac{1}{3}\)
\(\frac {1}{3}\) × 12 = 4 cm
(c) \(\frac{240}{360}=\frac{2}{3}\)
\(\frac {2}{3}\) × 12 = 8 cm
(d) Slant height of both cones is are same.
Slant height² = radius² + height²
∴ If the radius increases, the height decreases. The second cone is the shortest.

Question 4.
Find the lateral surface area of a cone of base perimeter 24π and height 5 cm.
Answer:
2π × radius = 24π
⇒ radius = 12 cm
Height = 5 cm
Slant height = \(\sqrt{12^2+5^2}\) = 13 cm
Radius of the sector used to form the cone = 13 cm
The area of the sector is \(\frac {12}{13}\) part of the area of the circle. So is the lateral surface area.
∴ Lateral surface area = π × 13² × \(\frac {12}{13}\) = 156π cm²

Question 5.
If the radius of a cone is r and the slant height is l, then prove that the lateral surface area is πrl.
Answer:
A sector rolled up to form a cone.
If the radius of the sector is l (which can be the slant height of the cone), the area of the circle is πl2.
Circumference is 2πl.
The central angle of the sector is x.
\(\frac{x}{360}=\frac{p}{2 \pi l}\)
p is the length of the arc.
Length of the arc, p = 2πl × \(\frac {x}{360}\), which is the base perimeter of the cone.
2πl × \(\frac {x}{360}\) = 2πr, r is the radius of the cone.
From this, lx = 360r
Lateral surface area is the area of the sector.
∴ Lateral surface area = πl² × \(\frac {x}{360}\)
= \(\frac{\pi \times l \times l x}{360}\)
= \(\frac{\pi \times l \times 360 r}{360}\)
= πrl sq. unit

Volume of a Cone
The volume of a cone is the product of base area and height.
The volume of a cone is \(\frac {1}{3}\) part of the volume of a prism with the same base area and height.
That is, the volume of a cone is one-third of the product of base area and height.

Question 1.
The volume of a cylinder is 144π cubic centimetres. What is the volume of a cone with the same base area and height?
Answer:
Volume must be one-third.
Volume = \(\frac {1}{3}\) × 144π = 48π cubic centimetres.

Question 2.
There is a conical vessel of radius 6 cm and slant height 10 cm.
(a) What is the height of this vessel?
(b) Find the volume of the vessel.
(c) How many times is water to be poured to fill a cylindrical vessel of the same radius and twice the height?
Answer:
(a) Height of the vessel = \(\sqrt{10^2-6^2}\) = 8 cm
(b) Volume = \(\frac {1}{3}\) × π × 6² × 8
= 12 × 8 × π
= 96π cubic centimetres.
(c) Volume of the cylindrical vessel with the same radius and twice the height = π × 6² × 16 = 576π cubic centimetres = 96π × 6
Hence, the water must be poured 6 times to fill the cylindrical vessel.

Question 3.
The volume of a cone is 60 cm³. Find the volume of a cone with the same height and half the radius.
Answer:
If the radius becomes half, the volume becomes \(\frac {1}{4}\)th.
Volume = \(\frac {1}{4}\) × 60 = 15 cm³

Question 4.
Radii of two cones are in the ratio 1 : 2 and the ratio of their heights is 2 : 1. Then, find the ratio of their volumes.
Answer:
Radii are r, 2r, and heights are 2h, h.
Then, volume = \(\frac{1}{3} \pi r^2 \times 2 h: \frac{1}{3} \pi \times(2 r)^2 \times h\)
= 2r²h : 4r²h
= 1 : 2

Sphere
The surface of a sphere is curved. The distance from the centre of the sphere to its surface is the radius of the sphere.
Consider a sphere of radius 4 cm. Four circular papers of radius 4 cm can cover the whole surface of the sphere.
If the radius is 4 cm, the surface area of a sphere is 4πr².
The surface area of a sphere is equal to the square of its radius multiplied by 4π.
A solid sphere cut into two equal halves exactly through the middle, we get two hemispheres.
It has one curved surface and one flat circular face.
Curved surface area of a solid sphere is 2πr², and its total surface area is 3πr².
Volume of a sphere is \(\frac{4}{3} \pi r^3\) and volume of a hemisphere is \(\frac{2}{3} \pi r^3\).

Question 1.
Find the radius of a sphere with volume is equal to its surface area. What is the surface area of this sphere?
Answer:
4πr² = \(\frac{4}{3} \pi r^3\)
⇒ r = 3
Surface area = 4π × 3² = 36π sq.unit

Question 2.
If the radius of a sphere becomes twice, what is the change in its volume?
Answer:
To calculate surface area, we have to multiply the square of its radius by 4π.
If the radius becomes twice, the surface area becomes 4 times.
Volume of a sphere is \(\frac{4}{3} \pi r^3\).
So, if the radius becomes twice, then the volume becomes 2³ = 8 times.

Question 3.
How many solid spheres of radius 1 cm can be formed by melting and recasting a solid metal sphere of radius 4 cm?
Answer:
Number of Spheres = \(\frac{\text { Volume of larger sphere }}{\text { Volume of smaller sphere }}\)
= \(\frac{\frac{4}{3} \pi \times 4^3}{\frac{4}{3} \pi \times 1^3}\)
= 64

Question 4.
Find the surface area of the sphere formed by joining two solid hemispheres of surface area 75π cm². What is the radius of the sphere? Find the volume.
Answer:
3πr² = 75π
⇒ r² = 25
⇒ r = 5 cm
Surface area of the sphere = 4π × 5² = 100π cm²
Volume of the sphere = \(\frac {4}{3}\) × π × 5³ = \(\frac{500 \pi}{3}\) cm³

Question 5.
What is the slant height of a cone of maximum size that can be carved out from a solid hemisphere of radius 12 cm? Find surface area and volume.
Answer:

Slant height = 12√2 cm
Surface area = π × 12² + π × 12 × 12√2 = 144π + 144√2π
Volume = \(\frac {1}{3}\) × π × 12² × 12
= 4 × 144π
= 576π cm³