# NCERT Solutions for Class 10th: Ch 13 Surface Areas and Volumes Math

## NCERT Solutions for Class 10th: Ch 13 Surface Areas and Volumes Math

Page No: 244**Exercise 13.1**

**Unless stated otherwise, take π = 22/7.**

1. 2 cubes each of volume 64 cm

^{3}are joined end to end. Find the surface area of the resulting cuboid.

**Answer**

^{3}) = 64 cm

^{3}

^{}

⇒ a

^{3}= 64 cm

^{3}

^{}

⇒ a = 4 cm

Side of the cube = 4 cm

Length of the resulting cuboid = 4 cm

Breadth of the resulting cuboid = 4 cm

Height of the resulting cuboid = 8 cm

∴ 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**

Radius of the hemisphere(r) = 7 cm

Height of the cylinder(h) = 13 - 7 = 6 cm

Also, radius of the hollow hemisphere = 7 cm

Inner surface area of the vessel = CSA of the cylindrical part + CSA of 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 same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

**Answer**

Total height of the toy = 15.5 cm

Height of the cone(h) = 15.5 - 3.5 = 12 cm

= 275/2 cm

^{2}

Curved Surface Area of hemisphere = 2πr

^{2}

= 2 × 22/7 × (7/2)

^{2}

= 77 cm

^{2}

Total surface area of the toy = CSA of cone + CSA of hemisphere

= (275/2 + 77) cm

^{2}

= (275+154)/2 cm

^{2 }

= 429/2 cm

^{2}= 214.5

^{ }cm

^{2}

The total surface area of the toy is 214.5

^{ }cm

^{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**

∴ Radius of the hemisphere = 7/2 cm

Total surface area of solid = Surface area of cubical block + CSA of hemisphere - Area of base of hemisphere

TSA of solid = 6×(side)

= 6×(side)

= 6×(7)

^{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

= 294 + 38.5 = 332.5 cm

^{2}
The surface area of the solid is 332.5 cm

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.

Diameter of hemisphere = Edge of cube = l

Radius of hemisphere = l/2

Total surface area of solid = Surface area of cube + CSA of hemisphere - Area of base of hemisphere

TSA of remaining solid = 6 (edge)

= 6l

= 6l

= 6l

= l

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.

Surface area of the cylinder = 2πrh

= 2 × 22/7 × 2.5 x 9

= 22/7 × 45

990/7 mm

∴ Required surface area of medicine capsule

= 2 × Surface area of hemisphere + Surface area of cylinder

= (2 × 275/7) × 990/7

= 550/7 + 990/7

= 1540/7 = 220 mm

Page No. 245

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

Tent is combination of cylinder and cone.

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

∴ 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 m

Cost of the canvas of the tent at the rate of ₹500 per m

= Surface area × cost per m

= 44 × 500 = ₹22000

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}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**Radius of hemisphere = l/2

Total surface area of solid = Surface area of cube + CSA of hemisphere - Area of base of hemisphere

TSA of remaining solid = 6 (edge)

^{2 }+ 2πr^{2 }- πr^{2}= 6l

^{2 }+ πr^{2}= 6l

^{2 }+ π(l/2)^{2}= 6l

^{2 }+ πl^{2}/4^{ }= l

^{2}/4^{ }(24 + π) sq units6. 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 hemisphere and one cylinder are given in the figure.

Diameter of the capsule = 5 mm

∴ Radius = 5/2 = 2.5 mm

Length of the capsule = 14 mm

∴ Length of the cylinder = 14 - (2.5 + 2.5) = 9mm

Surface area of a hemisphere = 2πr

^{2}= 2 × 22/7 × 2.5 × 2.5
= 275/7 mm

^{2}Surface area of the cylinder = 2πrh

= 2 × 22/7 × 2.5 x 9

= 22/7 × 45

990/7 mm

^{2}∴ Required surface area of medicine capsule

= 2 × Surface area of hemisphere + Surface area of cylinder

= (2 × 275/7) × 990/7

= 550/7 + 990/7

= 1540/7 = 220 mm

^{2}Page No. 245

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**Tent is combination of cylinder and cone.

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

∴ 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 m

^{2}Cost of the canvas of the tent at the rate of ₹500 per m

^{2}= Surface area × cost per m

^{2}= 44 × 500 = ₹22000

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}.**Answer**
Diameter of cylinder = diameter of conical cavity = 1.4 cm

∴ Radius of cylinder = Radius of conical cavity = 1.4/2 = 0.7

Height of cylinder = Height of conical cavity = 2.4 cm

TSA of remaining solid = Surface area of conical cavity+TSA of 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}**Go Back To NCERT Solutions for Class 10th Math**