NCERT Solutions for class 10 Maths Chapter 13- Exercise 13.1

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 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 NCERT syllabus and guidelines. Therefore, these solutions of class 10 Maths, are very helpful for students who are doing the preparation for the exams and to score well.

Download PDF of 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 Question (7 long, 1 short)

Exercise 13.3 Solutions 9 Question (9 long)

Exercise 13.4 Solutions 5 Question (5 long)

Exercise 13.5 Solutions 7 Question (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:

Ncert solutions class 10 chapter 13-1

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:

Ncert solutions class 10 chapter 13-2

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 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 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:

Ncert solutions class 10 chapter 13-3

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

Ncert solutions class 10 chapter 13-4

∴ The curved surface area of 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 cone + CSA of 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 cube is 7 cm. So, the radius will be 7/2 cm.

Ncert solutions class 10 chapter 13-5

We know,

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

Ncert solutions class 10 chapter 13-6

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

So, the radius of hemisphere = l/2

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

TSA of 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.

Ncert solutions class 10 chapter 13-7

Answer:

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

Ncert solutions class 10 chapter 13-8

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 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 cylinder and a cone.

Ncert solutions class 10 chapter 13-9

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 tent = surface area of cone + surface area of 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. 22000 will be the total cost of the canvas.

8. From a solid cylinder whose height is 2.4 cm and diameter 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:

Ncert solutions class 10 chapter 13-10

From the question we know the following:

The diameter of the cylinder = diameter of 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

Ncert solutions class 10 chapter 13-11

Now, the TSA of remaining solid = surface area of 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 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 us. Moreover, they can get other study materials as well such as notes, books, questions 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 are the generic methods present on the solutions, which students can relate to easily with the given concepts of finding surface areas for different shapes.

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