In this post, we will share NCERT Class 10th Maths Book Solutions ex 13.1 class 10 Chapter 13 Surface Areas and Volumes. These solutions of maths chapter 13 exercise 13.1 are based on the new NCERT Syllabus.

You can also Download our Android App for Class 10 Solutions in English Medium from Google Play here

## Ex 13.1 Class 10 Chapter 13 Surface Areas and Volumes NCERT Maths Solutions

Unless stated otherwise, take π = \(\frac{22}{7}\)

Ex 13.1 Class 10 Question 1.

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

Solution:

Volume of each cube = 64 cm^{3}

Let the edge of each cube = x

∴ x^{3} = 64 cm^{3}

⇒ x = 4 cm

Now, Length of the resulting cuboid (l) = 2x cm = 8 cm

Breadth of the resulting cuboid (b) = x cm = 4 cm

Height of the resulting cuboid (h) = x cm = 4 cm

∴ Surface area of the cuboid = 2 (lb + bh + hl)

= 2 [(8 × 4) + (4 × 4) + (4 × 8)] cm^{2}

= 2 [32 + 16 + 32] cm^{2} = 2 [80] cm^{2}

= 160 cm^{2}.

Ex 13.1 Class 10 Question 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.

Solution:

For hemispherical part,

radius (r)= \(\frac{14}{2}\) = 7cm

∴ Curved surface area = 2πr^{2}

= 2 × \(\frac{22}{7}\) × 7 × 7cm^{2}

= 308cm^{2} 7

Total height of vessel = 13 cm

∴ Height of cylinder = (13 – 7)cm = 6 cm and radius(r) = 7 cm

∴ Curved surface area of cylinder = 2πrh

= 2 × \(\frac{22}{7}\) × 7 × 6cm^{2} = 264cm^{2} 7

∴ Inner surface area of vessel = (308 + 264)cm^{2} = 572 cm^{2}

Ex 13.1 Class 10 Question 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.

Solution:

Let h be the height of cone and r be the radius of cone and hemisphere.

∴ h = [height of toy – radius of hemi sphere]

= (15.5 – 3.5) cm = 12 cm

Also l^{2} = h^{2} + r^{2} = 12^{2} + (3.5)^{2} = 156.25 cm^{2}

∴ l = 12.5 cm

Curved surface area of the conical part = πrl

Curved surface area of the hemispherical part = 2πr^{2}

∴ Total surface area of the toy = πrl + 2πr^{2}

= πr(l + 2 r)

= \(\frac{22}{7} \times \frac{35}{10}\) (12.5 + 2 × 3.5) cm^{2}

= 11 × (12.5 + 7) cm^{2} = 11 × 19.5 cm^{2}

= 214.5 cm^{2}

Ex 13.1 Class 10 Question 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.

Solution:

Let side of the block, l = 7 cm

∴ The greatest diameter of the hemisphere = 7 cm

Surface area of the solid = [Total surface area of the cubical block] + [C.S.A. of the hemisphere] – [Base area of the hemisphere]

= 6 × l^{2} + 2πr^{2} – πr^{2}

[where l = 7 cm and r = \(\frac{7}{2}\) cm]

Ex 13.1 Class 10 Question 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.

Solution:

Let l be the side of the cube.

∴ Diameter of the hemisphere = l

⇒ Radius of the hemisphere (r) = \(\frac{l}{2}\)

Curved surface area of hemisphere = 2πr^{2}

= 2 × π × \(\frac{l}{2} \times \frac{l}{2}=\frac{\pi l^{2}}{2}\)

Base area of the hemisphere = πr^{2}

= \(\pi\left(\frac{l}{2}\right)^{2}=\frac{\pi l^{2}}{4}\)

Surface area of the cube = 6 × l^{2} = 6l^{2}

∴ Surface area of the remaining solid = [Total surface area of cube + C.S.A. of hemisphere – base area of hemisphere]

Ex 13.1 Class 10 Question 6.

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

Solution:

Radius of the hemispherical part

Curved surface area of one hemispherical part = 2πr^{2}

∴ Surface area of both hemispherical parts

= 2(2πr^{2}) = 4πr^{2} = [4 × \(\frac{22}{7} \times\left(\frac{25}{10}\right)^{2}\)] mm^{2}

= \(\left(4 \times \frac{22}{7} \times \frac{25}{10} \times \frac{25}{10}\right)\) mm^{2}

Entire length of capsule = 14 mm

∴ Length of cylindrical part = [Length of capsule – Radius of two hemispherical part]

= (14 – 2 × 2.5)mm = 9mm Area of cylindrical part = 2πrh

= (2 × \(\frac{22}{7}\) × 2.5 × 9 ]mm^{2} = (2 × \(\frac{22}{7} \times \frac{25}{10}\) × 9) mm^{2}

Total surface area

= [Surface area of cylindrical part + Surface area of both hemispherical parts]

Ex 13.1 Class 10 Question 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 ₹ 500 per m^{2}. (Note that the base of the tent will not be covered with canvas.)

Solution:

For cylindrical part:

Radius (r) = \(\frac{4}{2}\) m = 2m and height (h) = 2.1 m

∴ Curved surface area = 2πrh = (2 × \(\frac{22}{7}\) × 2 × \(\frac{21}{10}\))m^{2}

For conical part:

Slant height (l) = 2.8 m

and base radius (r) = 2 m

∴ Curved surface area

= πrl = (\(\frac{22}{7}\) × 2 × \(\frac{28}{10}\)) m^{2}

∴ Total surface area = [Curved surface area of the cylindrical part] + [Curved surface area of conical part]

Cost of the canvas used :

Cost of 1 m^{2} of canvas = ₹ 500

∴ Cost of 44 m^{2} of canvas = ₹ (500 × 44)

= ₹ 22000.

Question 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 cm^{2}.

Solution:

For cylindrical part :

Height (h) = 2.4 cm and diameter = 1.4 cm

⇒ Radius (r) = 0.7 cm

∴ Total surface area of the cylindrical part

= 2πrh + 2πr^{2} = 2πr [h + r]

= 2 × \(\frac{22}{7} \times \frac{7}{10}\) [2.4 + 0.7]

= \(\frac{44}{10}\) × 3.1 = \(\frac{44 \times 31}{100}\) = \(\frac{1364}{100}\) cm^{2}

For conical part:

Base radius (r) = 0.7 cm and height (h) = 2.4 cm

Base area of the conical part

= \(\pi r^{2}=\frac{22}{7} \times\left(\frac{7}{10}\right)^{2}=\frac{22 \times 7}{100} \mathrm{cm}^{2}=\frac{154}{100} \mathrm{cm}^{2}\)

Total surface area of the remaining solid = [(Total surface area of cylindrical part) + (Curved surface area of conical part) – (Base area of the conical part)]

= \(\left[\frac{1364}{100}+\frac{550}{100}-\frac{154}{100}\right] \mathrm{cm}^{2}=\frac{1760}{100} \mathrm{cm}^{2}\)

Hence, total surface area to the nearest cm^{2} is 18cm^{2}.

maths chapter 13 surface Question 9.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

Radius of the cylinder (r) = 3.5 cm

Height of the cylinder (h) = 10 cm

∴ Curved surface area = 2πrh

= 2 × \(\frac{22}{7} \times \frac{35}{10}\) × 10cm^{2} = 220cm^{2}

Curved surface area of a hemisphere = 2πr^{2}

∴ Curved surface area of both hemispheres

= 2 × 2πr^{2} = 4πr^{2} = 4 × \(\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}\) cm^{2}

= 154 cm^{2}

Total surface area of the remaining solid = (220 + 154) cm^{2} = 374 cm^{2}.