Mathematics Part I Solutions Solutions for Class 10 Math Chapter 5 Solids are provided here with simple step-by-step explanations. These solutions for Solids are extremely popular among Class 10 students for Math Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 100:

Question 1:

A square of side 5 centimetres and four isosceles triangles, each of one side 5 centimetres and the height to the opposite vertex 8 centimetres; these are to be joined to make a square pyramind. How much paper is needed for the job?

Answer:

Side of the square = 5 cm

Base of the isosceles triangle = 5 cm

Height of the isosceles triangle = 8 cm

Area of the square = (side)2 = (5 cm)2 = 25 cm2

Area of one isosceles triangle =

Area of four isosceles triangles = 20 × 4 cm2 = 80 cm2

Area of the square pyramid = Area of the square + Area of four isosceles triangles

= (25 + 80) cm2 

= 105 cm2

Thus, we need 105 cm2 paper for the job.

Page No 100:

Question 2:

A toy in the shape of a square pyramid has base edge 16 centimetres and slant height 10 centimetres. 500 of these are to be painted and the cost is 80 rupees per square metre. What would be the total cost?

Answer:

Base edge of the square pyramid = 16 cm

Slant height of the square pyramid = Height of the triangle = 10 cm

Area of the square = (side)2 = (16 cm)2 = 256 cm2

Area of triangle =

Area of one square pyramid = Area of square + Area of four triangles

= (256 + 4 × 80) cm2 

= 576 cm2

Area of 500 square pyramids = 576 × 500 cm2 

= 288000 cm2

= 28.8 m2 (1 m2 = 10000 cm2)

Cost of painting for 1 square metre = Rs.80

Total cost of painting 28.8 m2 area = Rs.80 × 28.8 = Rs.2304

Page No 100:

Question 3:

A lateral faces of a square pyramid are equilateral triangles of side 30 centimetres. What is its surface area?

Answer:

Side of the equilateral triangle = 30 cm

Side of the square = Side of the equilateral triangle = 30 cm

Area of the square = (side)2 = (30 cm)2 = 900 cm2

Area of one equilateral triangle =

Area of four equilateral triangles =

Surface area of the square pyramid = Area of square + Area of four equilateral triangles

= (900 + ) cm2 



Page No 102:

Question 1:

A square pyramid is made using a square and four triangles with dimensions as shown below.

What would be the height of the pyramid?

What if the square and triangles are like these?

Answer:

(1)

Edge of the square base, b = 24 cm

Slant height of the square pyramid, l = Height of the triangle = 18 cm

We know that , where h is the height of the square pyramid.

Height of the square pyramid is


 

(2)

Side of the square = 24 cm

Length of diagonal of the square, d =

Length of lateral edge, e = 30 cm

We know that, where h is the height of the square pyramid.

h =

Thus, the height of the square pyramid is

Page No 102:

Question 2:

We want to make a paper pyramid with base a square of side 10 centimetres and height 12 centimetres. What should be the lengths of the sides of the triangles?

Answer:

Side of the square base, b = 10 cm

Height of the square pyramid, h = 12 cm

Base of the triangle = Side of the square = 10 cm

We know that, where l is the slant height of the square pyramid.

l

Now, , where e is the lateral edge of the square pyramid.

Thus, the lengths of sides of the triangles are 10 cm,

Page No 102:

Question 3:

Prove that in any square pyramid, the squares of the height, slant height and lateral edge are in arithmetic sequence.

Answer:

Let the height, slant height, lateral edge and base edge of the square pyramid be h units, l units, e units and b units respectively.

We know that

Thus, we have the sequence as h2, l2, e2 =

Now,

Since the common difference is same, the terms h2, l2, e2 form an arithmetic sequence.



Page No 104:

Question 1:

What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?

Answer:

Base edge of the square pyramid, b = 10 cm

Slant height of the square pyramid, l = 15 cm

Area of the base = (side)2 = (10 cm)2 = 100 cm2

We know that , where h is the height of the square pyramid.

Volume of the square pyramid =

Page No 104:

Question 2:

In two square pyramids of the same volume, the base edge of one is half the base edge of the other. How many times the height of the pyramid with larger base is the height of the other?

Answer:

Let the base edge of the second pyramid be b units.

Base edge of the first pyramid = units

Let the height of the first and the second pyramid be h1 units and h2 units respectively.

Base area of the first pyramid = ( units)2 = sq. units

Base area of the second pyramid = (b units)2 = b2 sq. units

Volume of square pyramid = Base area × Height 

According to given condition:

Volume of the first pyramid = Volume of the second pyramid

Therefore, the height of the pyramid with larger base is one-fourth times the height of the other.

Page No 104:

Question 3:

The base edges of two square pyramid are in the ratio 1 : 2 and their heights are in the ratio 1 : 3. The volume of the first pyramid is 180 cubic centimetres. What is the volume of the second?

Answer:

Let the base edge of the first and the second pyramid be x units and 2x units respectively.

Let the height of the first and the second pyramid be y units and 3y units respectively.

Base area of the first pyramid = (x units)2 = x2 sq. units

Base area of the second pyramid = (2x units)2 = 4x2 sq. units

Volume of square pyramid = Base area × Height 

Volume of the first pyramid =

Volume of the second pyramid =



Page No 106:

Question 1:

What is the base-radius and slant height of the cone made by rolling up a sector of radius 10 centimetres and central angle 60°?

Answer:

Given: Central angle of the sector = 60°

Radius of the sector = 10 cm

Slant height of the cone = Radius of the sector = 10 cm

Central angle of the sector is 60°, which is

Base radius of the small circle = of the radius of the larger circle

Page No 106:

Question 2:

What is the central angle of the sector needed to make a cone of base-radius 10 centimetres and slant height 25 centimetres?

Answer:

Base radius of the cone, i.e., radius of the smaller circle = 10 cm

Slant height of the cone = Radius of the larger circle = 25 cm

Let the base radius of the small circle be of the radius of the larger circle.

Central angle =

Thus, the central angle of the sector needed to make the cone is 144°.

Page No 106:

Question 3:

What is the ratio of the base-radius and slant height of a cone made by rolling up a semicircle?

Answer:

Let the radius of the bigger circle be R units.

Slant height of the cone = Radius of the bigger circle = R units

Since the cone is made from a semi-circular sector, the central angle = = 180°.

Radius of the small circle = of the radius of the larger circle

Base-radius of the cone

Therefore, ratio of the base radius to the slant height of the cone is 1:2.



Page No 108:

Question 1:

What is the curved surface area of a cone of base radius 12 centimetres and slant height 25 centimetres?

Answer:

Base-radius of the cone = 12 cm

Slant height of the cone = 25 cm 

Curved surface area of the cone = πrl

Page No 108:

Question 2:

What is the curved surface area of a cone of base diameter 30 centimetres and height 40 centimetres?

Answer:

Base-diametre of the cone = 30 cm

Base-radius of the cone = = 15 cm

Height of the cone = 40 cm

We know that the slant height of the cone = .

Curved surface area of the cone = πrl

Page No 108:

Question 3:

A cone shaped firework is of base-diametre 10 centimetres and height 12 centimetres. 10000 such firework are to be wrapped in colour paper. The price of paper is 2 rupees per square metre. What is the total cost of wrapping?

Answer:

Base-diametre of the cone = 10 cm

Base-radius of the cone = = 5 cm

Height of the cone = 12 cm

We know that the slant height of the cone = .

Curved surface area of one conical firework = πrl 

Curved surface area of 10000 fireworks =

Price of paper required for 1 square metre = Rs.2

Total cost of wrapping = Rs.2 × 204.1 = Rs.408.20

Page No 108:

Question 4:

Prove that for a cone made by bending a semicircle, the curved surface area is double the base area.

Answer:

Let the radius of the bigger circle be R units.

Slant height of the cone = Radius of the bigger circle = R units

Since the cone is made from a semi-circular sector, the central angle = = 180°.

Radius of the small circle = of the radius of the larger circle

Base-radius of the cone

Curved surface area of the cone = πrl 

Therefore, curved surface area of the cone is double the base area.



Page No 109:

Question 1:

The base-radius of a cylindrical block of wood is 15 centimetres and its height 40 centimetres. What is the volume of the largest cone that can be carved out from this?

Answer:

Volume of the largest cone that can be carved out from the cylindrical block 

= of the volume of the cylindrical block

Base-radius of the cylindrical block = 15 cm 

Height of the cylindrical block = 40 cm

Volume of the cylindrical block = Base area × Height

Volume of the largest cone =

Page No 109:

Question 2:

A solid metal cylinder is of base-radius 12 centimetres and height 20 centimetres. By melting and recasting, how many cones of base-radius 4 centimetres and height 5 centimetres can be made?

Answer:

Base-radius of the cylinder = 12 cm

Height of the cylinder = 20 cm

Base-radius of the cone = 4 cm

Height of the cone = 5 cm

Volume of the cylinder = Base area × height 

= π × (12 cm)2 × 20 cm

= 2880 π cm3

Volume of the cone

Number of cones =

Therefore, 108 cones can be made from the given cylinder.

Page No 109:

Question 3:

A sector of central and 216° is cut out from a circle of radius 25 centimetres and it is rolled up into a cone. What is the base-radius and height of this cone? What is its volume?

Answer:

Central angle of the sector = 216°, which is .

Radius of the large circle = 25 cm

Radius of the small circle = of the radius of the larger circle

Slant height of the cone = Radius of the larger circle = 25 cm

We know that (Slant height)2 = (Height)2 + (Base-radius)

Volume of the cone =

Page No 109:

Question 4:

The ratio of the base-radii of two cones is 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?

Answer:

Let the base-radius of the first and the second cone be 3x units and 5x units respectively.

Let the height of the first and the second cone be 2y units and 3y units respectively.

Volume of the first cone =

Volume of the second cone

Ratio of the volumes =

Therefore, the ratio of the volume of the first cone to the second cone is 6:25.

Page No 109:

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 the base-radius of the first and the second cone be 4x and 5x respectively.

Let the height of the first and the second cone be h1and h2 respectively.

Volume of the first cone =

Volume of the second cone

According to the given condition:

Volume of the first cone = Volume of the second cone 

Therefore, the ratios of their heights is 25:16.



Page No 112:

Question 1:

The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii?

Answer:

Let the radius of the first and the second sphere be r1 units and r2 units respectively.

Volume of the sphere =

Ratio of the volume =

Therefore, the ratio of their radii is 3:4.



Page No 113:

Question 1:

A metal cylinder of base-radius 4 centimetres and height 10 centimetres is melted and recast into spheres of radius 2 centimetres. How many such spheres are got?

Answer:

Base-radius of the cylinder = 4 cm

Height of the cylinder = 10 cm

Radius of the sphere = 2 cm

Volume of the cylinder = Base area × height 

= π × (4 cm)2 × 10 cm

= 160π cm3

Volume of the sphere =

Number of spheres =

Therefore, 15 spheres can be made from the given cylinder.

Page No 113:

Question 2:

The picture below shows a petrol tank.

How many litres does it hold?

Answer:

Radius of the hemisphere = 1 m

Radius of the cylinder = 1 m

Height of the cylinder = 6 m

Volume of one hemisphere =

Volume of the two hemispheres = 2 × 2.1 m3 = 4.2 m3

Volume of the cylinder = Base area × height 

= π × (1 m)2 × 6 m

= 6 × 3.14 m3

= 18.84 m3

Volume of the petrol tank = Volume of the two hemispheres + Volume of the cylinder

= (4.2 + 18.84) m3 

= 23.04 m3

We know that 1 m3 = 1000 L

23.04 m3 = 1000 × 23.04 L = 23040 L

Therefore, the petrol tank can hold 23040 litres of petrol.



View NCERT Solutions for all chapters of Class 10