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Chapter 1:Real Numbers

Chapter 2:Polynomials

Chapter 3:Pair of Linear Equations in Two Variables

Chapter 4:Quadratic Equations

Chapter 5: Arithmetic Progression

Chapter 6: Triangles

Chapter 7:Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11:Constructions

Chapter 12:Area Related to Circles

Chapter 13:Surface Areas and Volumes

Chapter 14:Statistics

Chapter 15:Probability

NCERT Solutions class 10 maths chapter 13

Surface Areas and Volumes

NCERT Solutions class 10 maths chapter 13 Exercise 13.1

Question 1
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Answer

NCERT Ex-13.1 class 10

Given

The Volume of each cube is = 64 cm3

let's side of the cube =a

a3 = 64 cm3

∴ a = 4 cm

the side of the cube = a = 4 cm

Also, when we joined end to end of each cube then the length and breadth of the resulting cuboid will 8 cm and 4 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

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.

Answer 2
NCERT Ex-13.1 class 10

The diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

the height of the cylinder h = 13 - 7 = 6 cm

the radius of the hollow hemisphere = 7 cm

Now, the inner surface area of the vessel = Coverd Surface area of the cylindrical part + Coverd Surface area of hemispherical part

(2πrh+2πr2) cm2 = 2πr(h+r) cm2

2×(22/7)×7(6+7) cm2 = 572 cm2

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.

Answer 3
NCERT Ex-13.1 class 10

Given that radius of the cone (r) = 3.5 cm
r= 7/2 cm

The total height of the toy = 15.5 cm.

So, the height of the cone = 15.5-3.5 = 12 cm

Total surface area of the toy =curved surface area of cone +curved surface area of the hemisphere = πrl + 2πr2

=(22/7)×(7/2)×(25/2) + 2×(22/7)×(7/2)2

= (275/2)+77 cm2

= (275+154)/2 cm2

= 429/2 cm2 = 214.5cm2

So, the total surface area of the toy is 214.5cm2

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.

Answer 4
NCERT Ex-13.1 class 10

the side of cube = Diameter of hemisphere

So side of cube is 7 cm. and the radius will be 7/2 cm.

We know,

The total surface area of solid = surface area of cubical block +curved surface area 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

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.

Answer 5
NCERT Ex-13.1 class 10

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

Question 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 Ex-13.1 class 10
Answer 6

Two hemisphere and one cylinder are included

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

= 2 × surface area of hemisphere + surface area of the cylinder

= (2×275/7) × 990/7

(550/7) + (990/7) = 1540/7 = 220 mm2

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 Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Answer 7
NCERT Ex-13.1 class 10

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.

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 cm2.

Answer 8
NCERT Ex-13.1 class 10

we know that

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 Total surface area of remaining solid = surface area of conical cavity + Total surface area 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