Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1 NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 12, Exercise 12.1, focus on Surface Areas and Volumes of solid shapes like cones, cylinders, spheres, hemispheres, and combinations of these, as covered in the CBSE Class 10th syllabus. This exercise will help you understand the application of formulas to calculate curved, lateral, and total surface areas and volumes of these shapes.

Through regular practise and the support of detailed NCERT Solutions for class 10 maths , you can develop a clearer approach to solving area-based questions and improve both speed and accuracy in exams.

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

1. 2 cubes each of volume 64 cm 3 are joined end to end. Find the surface area of the resulting cuboid.

Answer :

Given, The Volume (V) of each cube is = 64 cm 3 

This implies that a 3 = 64 cm 3 

∴ 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) cm 2 

= 2(32+16+32) cm 2 

= (2×80) cm 2 

= 160 cm 2

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:

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+2πr 2 ) cm 2 

= 2πr(h+r) cm 2 2×(22/7)×7(6+7) cm 2 

= 572 cm 2

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:

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 cm 2 

Also, the curved surface area of the hemisphere = 2πr 2 2×(22/7)×(7/2) 2 = 77 cm 2 

Now, the Total surface area of the toy = CSA of the cone + CSA of the hemisphere = (275/2)+77 cm 2 

= (275+154)/2 cm 2 

= 429/2 cm 2 

= 214.5cm 2 

So, the total surface area (TSA) of the toy is 214.5cm 2

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:

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 +2πr 2 -πr 2 

= 6×(side) 2 +πr 2 

= 6×(7) 2 +(22/7)×(7/2)×(7/2) 

= (6×49)+(77/2) 

= 294+38.5 

= 332.5 cm 2 

So, the surface area of the solid is 332.5 cm 2

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:

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 cube + CSA of the hemisphere – Area of the base of the hemisphere 

The surface area of the remaining solid = 6 (edge) 2 +2πr 2 -πr 2 = 6l 2 + πr 2 

= 6l 2 +π(l/2) 2 

= 6l 2 +πl 2 /4 

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:

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 = 2πr 2 

= 2×(22/7)×2.5×2.5 = 275/7 mm 2 

Now, the surface area of the cylinder = 2πrh 

= 2×(22/7)×2.5×9 (22/7)×45 

= 990/7 mm 2 

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 mm 2

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 m ² . (Note that the base of the tent will not be covered with canvas.)

Answer:

From the question, we know that Diameter = 4 m 

The 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 m 2 

∴ The cost of the canvas of the tent at the rate of ₹500 per m 2 will be = Surface area × cost per m 2 

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 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 cm ² .

Answer:

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

Now, the TSA of the remaining solid = surface area of conical cavity + TSA of the cylinder = πrl+(2πrh+πr 2 ) 

= πr(l+2h+r) = (22/7)× 0.7(2.5+4.8+0.7) 

= 2.2×8 

= 17.6 cm 2 

So, the total surface area of the remaining solid is 17.6 cm 2 

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising CBSE Class 10 Previous Year Question Papers (PYQs) helps you understand question patterns and important topics.