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**NCERT TEXTBOOK QUESTIONS SOLVED**

**Page No. 244****EXERCISE 13.1**

[Unless stated otherwise, take Ï€ = 22/7]

**Q 1. Two cubes each of volume 64 cm ^{3} are joined end to end. Find the surface area of the resulting cuboid**

âˆ´ Total volume of the two cubes = 2 Ã— 64 cm

Let the edge of each cube = x

âˆ´ x

â‡’ x =4 cm

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

Breadth of the resulting cuboid b = x cm

Height of the resulting cuboid h = x cm

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

= 2 [(2x Â· x) + (x Â· x) + (x Â· 2x)]

= 2 [(2 Ã— 4 Ã— 4) + (4 Ã— 4) + (4 Ã— 2 Ã— 4)] cm

= 2 [32 + 16 + 32] cm

Radius (r) = 7 cm

Height (h) = 6 cm

âˆ´ Curved surface area

= 2Ï€rh

Radius (r) = 7 cm

âˆ´ Surface area = 2Ï€r

âˆ´ Total surface area

= (264 + 308) cm

âˆ´ h = (15.5 âˆ’ 3.5) cm = 12.0 cm

Surface area of the conical part = Ï€rl

Surface area of the hemispherical part =2Ï€r

**Q 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.****Sol. **Side of the block = 7 cm

â‡’ The greatest diameter of the hemisphere = 7 cm

Surface area of the solid

= [Total S.A. of the cubical block] + [S.A. of the hemisphere] âˆ’ [Base area of the hemisphere]

= (6 Ã— l^{2}) + 2Ï€r^{2} âˆ’ Ï€r^{2}

**Q 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.****Sol.** Let â€˜lâ€™ be the side of the cube.

âˆ´ The greatest diameter of the curved hemisphere = l

â‡’ Radius of the curved hemisphere

âˆ´ Surface area of hemisphere = 2Ï€r^{2}

Base area of the hemisphere

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

âˆ´ Surface area of the remaining solid**Q 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.****Sol.** Radius of the hemispherical part

= 5/2 mm = 2.5 mm

âˆ´ Surface area of one hemispherical part = 2Ï€r^{2}

â‡’ Surface area of both hemispherical parts

Area of cylindrical part =

âˆ´ Total surface area**Page No. 245****Q 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.)**

Radius (r) = 4/2 m = 2 m

Height (h) = 2.1 m

âˆ´ Curved surface area = 2Ï€rh =

For conical part:

Slant height (l) = 2.8 m

Base radius (r)= 2m

âˆ´ Curved surface area

âˆ´

= [Surface area of the cylindrical part] + [Surface area of conical part]

Cost of 1 m

âˆ´ Cost of 44 m

Height = 2.4 cm

Diameter = 1.4 cm

â‡’ Radius (r) = 0.7 cm

Base area (r) = 0.7 cm

Height (h) = 2.4 cm

âˆ´ Slant height

âˆ´ Curved surface area of the conical part

Base area of the conical part

Height of the cylinder (h) = 10 cm

âˆ´ Total surface area = 2Ï€rh + 2Ï€r

Curved surface area of a hemisphere = 2Ï€r

âˆ´ Curved surface area of both hemispheres

Base area of a hemisphere = Ï€r

âˆ´ Base area of both hemispheres = 2Ï€r

âˆ´ Total surface area of the remaining solid

= 297 cm

= (451 âˆ’ 77) cm

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