NCERT Solutions for Class 10 Maths Chapter 13 - Surface Areas and Volumes Exercise 13.1

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 12.

NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes Exercise 13.1, are available here in PDF format, which students can easily download. The questions on this exercise in NCERT Solutions have been solved by our Maths experts to help students understand the concepts in a better way.

Questions present in these materials are in accordance with the NCERT syllabus and guidelines. Therefore, these NCERT Solutions of Class 10 Maths are very helpful for students who are doing the preparation for the board exams to score well.

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.1

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Access Other Exercise Solutions of Class 10 Maths Chapter 13 – Surface Areas and Volumes

Exercise 13.2 Solutions 8 Questions (7 long, 1 short)

Exercise 13.3 Solutions 9 Questions (9 long)

Exercise 13.4 Solutions 5 Questions (5 long)

Exercise 13.5 Solutions 7 Questions (7 long)

Access Answers to NCERT Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.1

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

Answer:

The diagram is given as

Given,

The Volume (V) of each cube is = 64 cm³

This implies that a³ = 64 cm³

∴ a = 4 cm

Now, the side of the cube = a = 4 cm

Also, the length and breadth of the resulting cuboid will be 4 cm each, while its height will be 8 cm.

So, the surface area of the cuboid = 2(lb+bh+lh)

= 2(8×4+4×4+4×8) cm²

= 2(32+16+32) cm²

= (2×80) cm² = 160 cm²

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer:

The diagram is as follows:

Now, the given parameters are

The diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

Also, the height of the cylinder = h = (13-7) = 6 cm

And the radius of the hollow hemisphere = 7 cm

Now, the inner surface area of the vessel = CSA of the cylindrical part + CSA of the hemispherical part

(2πrh+2πr²) cm² = 2πr(h+r) cm²

2×(22/7)×7(6+7) cm² = 572 cm²

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer:

The diagram is as follows:

Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7/2 cm

The total height of the toy is given as 15.5 cm.

So, the height of the cone (h) = 15.5-3.5 = 12 cm

∴ The curved surface area of the cone = πrl

(22/7)×(7/2)×(25/2) = 275/2 cm²

Also, the curved surface area of the hemisphere = 2πr²

2×(22/7)×(7/2)²

= 77 cm²

Now, the total surface area of the toy = CSA of the cone + CSA of the hemisphere

= (275/2)+77 cm²

= (275+154)/2 cm²

= 429/2 cm² = 214.5cm²

So, the total surface area (TSA) of the toy is 214.5cm²

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer:

It is given that each side of the cube is 7 cm. So, the radius will be 7/2 cm.

We know,

The total surface area of solid (TSA) = surface area of the cubical block + CSA of the hemisphere – Area of the base of the hemisphere

∴ TSA of solid = 6×(side)²+2πr²-πr²

= 6×(side)²+πr²

= 6×(7)²+(22/7)×(7/2)×(7/2)

= (6×49)+(77/2)

= 294+38.5 = 332.5 cm²

So, the surface area of the solid is 332.5 cm²

5. A hemispherical depression is cut out from one face of a cubical wooden block, such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer:

The diagram is as follows:

Now, the diameter of the hemisphere = Edge of the cube = l

So, the radius of the hemisphere = l/2

∴ The total surface area of solid = surface area of the cube + CSA of the hemisphere – Area of the base of the hemisphere

TSA of the remaining solid = 6 (edge)²+2πr²-πr²

= 6l² + πr²

= 6l²+π(l/2)²

= 6l²+πl²/4

= l²/4(24+π) sq. units

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find its surface area.

Answer:

Two hemispheres and one cylinder are shown in the figure given below.

Here, the diameter of the capsule = 5 mm

∴ Radius = 5/2 = 2.5 mm

Now, the length of the capsule = 14 mm

So, the length of the cylinder = 14-(2.5+2.5) = 9 mm

∴ The surface area of a hemisphere = 2πr² = 2×(22/7)×2.5×2.5

= 275/7 mm²

Now, the surface area of the cylinder = 2πrh

= 2×(22/7)×2.5×9

(22/7)×45 = 990/7 mm²

Thus, the required surface area of the medicine capsule will be

= 2×surface area of hemisphere + surface area of the cylinder

= (2×275/7) × 990/7

= (550/7) + (990/7) = 1540/7 = 220 mm²

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m, respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)

Answer:

It is known that a tent is a combination of a cylinder and a cone.

From the question, we know that

Diameter = 4 m

Slant height of the cone (l) = 2.8 m

Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m

Height of the cylinder (h) = 2.1 m

So, the required surface area of the tent = surface area of the cone + surface area of the cylinder

= πrl+2πrh

= πr(l+2h)

= (22/7)×2(2.8+2×2.1)

= (44/7)(2.8+4.2)

= (44/7)×7 = 44 m²

∴ The cost of the canvas of the tent at the rate of ₹500 per m² will be

= Surface area × cost per m²

44×500 = ₹22000

So, Rs. 22,000 will be the total cost of the canvas.

8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the

same height and same diameter is hollowed out. Find the total surface area of the

remaining solid to the nearest cm².

Answer:

The diagram for the question is as follows:

From the question, we know the following:

The diameter of the cylinder = diameter of the conical cavity = 1.4 cm

So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

Also, the height of the cylinder = height of the conical cavity = 2.4 cm

Now, the TSA of the remaining solid = surface area of the conical cavity + TSA of the cylinder

= πrl+(2πrh+πr²)

= πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm²

So, the total surface area of the remaining solid is 17.6 cm²

Exercise 13.1 of Class 10 Maths has problems based on finding

• The surface area of a given cuboid
• The surface area of a cylindrical vessel
• The total surface area of a cone-shaped object mounted on a hemisphere
• The surface area of the solid surmounted by a hemisphere
• The surface area of the remained solid when a cone-shaped is from a solid cylinder

To get complete solutions for Chapter 13 of Class 10 Maths, students can visit BYJU’S website. Moreover, they can get other study materials as well, such as notes, books, question papers and some tips and tricks to effectively prepare for the Maths exam.

The questions in Exercise 13.1 are prepared as per the topics covered in the chapter before this exercise and also some solved examples. The methods present in the NCERT Class 10 solutions help students to grasp the given concepts of finding surface areas for different shapes easily.