Mathematics Part ii Solutions Solutions for Class 10 Math Chapter 7 - Mensuration

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Mathematics Part ii Solutions Solutions for Class 10 Math Chapter 7 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among class 10 students for Math Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 10 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 145:

Question 1:

Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.

Answer:

Radius of the cone, r = 1.5 cm

Height of the cone, h = 5 cm

∴ Volume of the cone, V = $\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times {(1.5)}^{2} \times 5 =$ 11.79 cm³

Thus, the volume of the cone is 11.79 cm³.

Page No 145:

Question 2:

Find the volume of a sphere of diameter 6 cm.

Answer:

Radius of the sphere, r = $\frac{6}{2}$ = 3 cm

∴ Volume of the sphere, V = $\frac{4}{3} π r^{3} = \frac{4}{3} \times 3.14 \times {(3)}^{3}$ = 113.04 cm³

Thus, the volume of sphere is 113.04 cm³.

Page No 145:

Question 3:

Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm.

Answer:

Radius of the cylinder, r = 5 cm

Height of the cylinder, h = 40 cm

∴ Total surface area of cylinder, S = $2 π r (r + h) = 2 \times 3.14 \times 5 \times (5 + 40) = 2 \times 3.14 \times 5 \times 45$ = 1413 cm²

Thus, the total surface area of cylinder is 1413 cm².

Page No 145:

Question 4:

Find the surface area of a sphere of radius 7 cm.

Answer:

Radius of the sphere, r = 7 cm

∴ Surface area of the sphere, S = $4 π r^{2} = 4 \times \frac{22}{7} \times {(7)}^{2}$ = 616 cm²

Thus, the surface area of sphere is 616 cm².

Page No 145:

Question 5:

The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.

Answer:

The dimensions of the cuboid are 44 cm, 21 cm and 12 cm.

Let the radius of the cone be r cm.

Height of the cone, h = 24 cm

It is given that cuboid is melted to form a cone.

∴ Volume of metal in cone = Volume of metal in cuboid

$\Rightarrow \frac{1}{3} π r^{2} h = 44 \times 21 \times 12 (Volume of cuboid = Length \times Breadth \times Height) \Rightarrow \frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24 = 44 \times 21 \times 12 \Rightarrow r = \sqrt{\frac{44 \times 21 \times 12 \times 21}{22 \times 24}} = \sqrt{21 \times 21} = 21 cm$
Thus, the radius of the base of cone is 21 cm.

Page No 145:

Question 6:

Observe the measures of pots In the given figure. How many jugs of water can the cylindrical pot hold?

Answer:

Radius of conical water jug, r = 3.5 cm

Height of conical water jug, h = 10 cm

Radius of cylindrical pot, R = 7 cm

Height of cylindrical pot, H = 10 cm

Let n be the number of jugs of water that cylindrical pot can hold.

$∴ n = \frac{Volume of cylindrical water pot}{Volume of conical water jug} \Rightarrow n = \frac{π R^{2} H}{\frac{1}{3} π r^{2} h} \Rightarrow n = \frac{3 \times {(7)}^{2} \times 10}{{(3.5)}^{2} \times 10} = 12$
Thus, the cylindrical water pot can hold water of 12 conical water jugs.

Page No 145:

Question 7:

A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm² .The cone is placed
upon the cylinder. Volume of the solid figure so formed is 500 cm³. Find the total height of the figure.

Answer:

Height of the cylinder, h = 3 cm

Let the radius of the cylinder be r cm and the height of the cone be H cm.

Area of the base of cylinder = 100 cm²

$∴ π r^{2} = 100$ .....(1)

Volume of the solid figure = 500 cm³

∴ Volume of the cylinder + Volume of the cone = 500 cm³

$\Rightarrow π r^{2} h + \frac{1}{3} π r^{2} H = 500 \Rightarrow π r^{2} (h + \frac{H}{3}) = 500 \Rightarrow 100 (3 + \frac{H}{3}) = 500 [Using (1)] \Rightarrow 3 + \frac{H}{3} = \frac{500}{100} = 5 \Rightarrow \frac{H}{3} = 5 - 3 = 2 \Rightarrow H = 6 cm$
∴ Total height of the figure = h + H = 3 + 6 = 9 cm

Thus, the total height of the figure is 9 cm.

Page No 145:

Question 8:

In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

Answer:

Radius of the sphere = Radius of the cylinder = Radius of cone = r = 3 cm

Height of the cone, h = 4 cm

Height of the cylinder, H = 40 cm

Let the slant height of the cone be l cm.

$∴ l = \sqrt{r^{2} + h^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 cm$

Total area of the toy

= Curved surface area of hemisphere + Curved surface area of cylinder + Curved surface area of cone

$= 2 π r^{2} + 2 π r H + π r l = 2 π \times {(3)}^{2} + 2 π \times 3 \times 40 + π \times 3 \times 5 = 18 π + 240 π + 15 π = 273 π {cm}^{2}$
Thus, the total area of the toy is 273 $π$ cm².

Page No 145:

Question 9:

In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

Answer:

Radius of a tablet, r = 7 mm

Thickness of a tablet, h = 5 mm

Radius of cylindrical wrapper, R = $\frac{14}{2}$ = 7 mm

Height of cylindrical wrapper, H = 10 cm = 100 mm (1 cm = 10 mm)

Let n be the number of tablets wrapped in the wrapper.

$n = \frac{Volume of the cylindrical wrapper}{Volume of each tablet} \Rightarrow n = \frac{π R^{2} H}{π r^{2} h} \Rightarrow n = \frac{{(7)}^{2} \times 100}{{(7)}^{2} \times 5} = 20$
Thus, 20 tablets are wrapped in the wrapper.

Page No 145:

Question 10:

In the given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and surface area of the toy from the measures shown in the figure ( $π = 3.14$ )

Answer:

Radius of the hemisphere = Radius of the cone = r = 3 cm

Height of the cone, h = 4 cm

Let l be the slant height of the cone.

$l = \sqrt{r^{2} + h^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 cm$

Volume of the toy = Volume of the hemisphere + Volume of the cone
$= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{2}{3} \times 3.14 \times {(3)}^{3} + \frac{1}{3} \times 3.14 \times {(3)}^{2} \times 4 = 56.52 + 37.68 = 94.20 {cm}^{3}$
Surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere
$= π r l + 2 π r^{2} = 3.14 \times 3 \times 5 + 2 \times 3.14 \times {(3)}^{2} = 47.10 + 56.52 = 103.62 {cm}^{2}$
Thus, the volume and surface area of the toy are 94.20 cm³and 103.62 cm², respectively.

Page No 146:

Question 11:

Find the surface area and the volume of a beach ball shown in the figure

Answer:

Radius of the beach ball, r = $\frac{42}{2}$ = 21 cm

Surface area of the beach ball, S = $4 π r^{2} = 4 \times 3.14 \times {(21)}^{2}$ = 5538.96 cm²

Volume of the beach ball, V = $\frac{4}{3} π r^{3} = \frac{4}{3} \times 3.14 \times {(21)}^{3}$ = 38772.72 cm³

Thus, the surface area and volume of the beach ball are 5538.96 cm²and 38772.72 cm³, respectively.

Page No 146:

Question 12:

As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water .

Answer:

Radius of the metallic sphere, r = $\frac{2}{2}$ = 1 cm

Radius of the cylindrical glass, R = $\frac{14}{2}$ = 7 cm

Height of water in cylindrical glass, H = 30 cm

∴ Volume of the water = Volume of water in the cylindrical glass − Volume of the metallic sphere
$= π R^{2} H - \frac{4}{3} π r^{3} = π {(7)}^{2} \times 30 - \frac{4}{3} π \times {(1)}^{3} = 1470 π - 1.33 π = 1468.67 π {cm}^{3}$
Thus, the volume of the water is 1468.67 $π$ cm³.

Page No 148:

Question 1:

The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many liters of water it can hold ?

(1 litre = 1000 cm³)

Answer:

Radius of one circular end, r₁ = 14 cm

Radius of other circular end, r₂= 7 cm

Height of the bucket, h = 30 cm

∴ Volume of water in the bucket = Volume of frustum of cone
$= \frac{1}{3} π h (r_{1}^{2} + r_{1} r_{2} + r_{2}^{2}) = \frac{1}{3} \times \frac{22}{7} \times 30 \times (14^{2} + 14 \times 7 + 7^{2}) = \frac{1}{3} \times \frac{22}{7} \times 30 \times 343 = 10780 {cm}^{3}$
$= \frac{10780}{1000} = 10.780 L$
Thus, the bucket can hold 10.780 litres of water.

Page No 148:

Question 2:

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its i) curved surface area ii) total surface area
iii ) volume

(

π

= 3.14)

Answer:

Here, r₁ = 14 cm, r₂ = 6 cm and h = 6 cm.

Slant height of the frustum, l = $\sqrt{h^{2} + {(r_{2} - r_{1})}^{2}} = \sqrt{6^{2} + {(14 - 6)}^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100}$ = 10 cm

i)
Curved surface area of frustrum
$= π (r_{1} + r_{2}) l = 3.14 \times (14 + 6) \times 10 = 3.14 \times 20 \times 10 = 628 {cm}^{2}$
ii)
Total surface area of frustrum
$= π (r_{1} + r_{2}) l + π r_{1}^{2} + π r_{2}^{2} = 3.14 \times (14 + 6) \times 10 + 3.14 \times 14^{2} + 3.14 \times 6^{2} = 3.14 \times 20 \times 10 + 3.14 \times 196 + 3.14 \times 36 = 628 + 615.44 + 113.04 = 1356.48 {cm}^{2}$
iii)
Volume of the frustum
$= \frac{1}{3} π h (r_{1}^{2} + r_{1} r_{2} + r_{2}^{2}) = \frac{1}{3} \times 3.14 \times 6 \times (14^{2} + 14 \times 6 + 6^{2}) = 3.14 \times 2 \times (196 + 84 + 36) = 6.28 \times 316 = 1984.48 {cm}^{3}$

Page No 148:

Question 3:

The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.( $π = \frac{22}{7}$ )

circumference₁ = 2 $π$ r₁= 132

r₁= $\frac{132}{π} = 134$

  circumference₂ =2 $π$ r₂= 88

   r₂= $\frac{88}{2 π} = 134$

slant height of frustum, l = $\sqrt{h^{2} + {(r_{1} - r_{2})}^{2}} \sqrt{1223^{2} + 1234^{2}} = 1234 cm$

curved surface area of the frustum = $π$ (r₁+ r₂ ) l

= $π \times 123 \times 132 = 123 sq . cm .$

Answer:

Circumference₁ = 2 $π$ r₁= 132

r₁= $\frac{132}{2 π} = \frac{132 \times 7}{2 \times 22} = 21 cm$

Circumference₂ = 2 $π$ r₂= 88

r₂= $\frac{88}{2 π} = \frac{88 \times 7}{2 \times 22} = 14 cm$

Slant height of frustum,
$l = \sqrt{h^{2} + {(r_{1} - r_{2})}^{2}} = \sqrt{24^{2} + {(21 - 14)}^{2}} = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 cm$
Curved surface area of the frustum = $π$ (r₁+ r₂ ) l
$= π \times (21 + 14) \times 25 = π \times 35 \times 25 = \frac{22}{7} \times 35 \times 25 = 2750 sq . cm .$

Page No 154:

Question 1:

Radius of a circle is 10 cm. Measure of an arc of the crcleis 54^°. Find the area of the sector associated with the arc. (

π

= 3.14 )

Answer:

Radius of the circle, r = 10 cm

Measure of the arc, θ = 54º

∴ Area of the sector = $\frac{θ}{360 °} \times π r^{2} = \frac{54 °}{360 °} \times 3.14 \times {(10)}^{2}$ = 47.1 cm²

Thus, the area of the sector is 47.1 cm².

Page No 154:

Question 2:

Measure of an arc of a circle is 80 cm and its radius is 18 cm. Find the length of the arc. (

π

= 3.14 )

Answer:

Radius of the arc, r = 18 cm

Measure of the arc, θ = 80º

∴ Length of the arc = $\frac{θ}{360 °} \times 2 π r = \frac{80 °}{360 °} \times 2 \times 3.14 \times 18$ = 25.12 cm

Thus, the length of the arc is 25.12 cm.

Page No 154:

Question 3:

Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.

Answer:

Radius of the sector, r = 3.5 cm

Length of the arc, l = 2.2 cm

∴ Area of the sector = $\frac{1}{2} l r = \frac{1}{2} \times 2.2 \times 3.5$ = 3.85 cm²

Thus, the area of the sector is 3.85 cm².

Page No 154:

Question 4:

Radius of a circle is 10 cm. Area of a sector of the sector is 100 cm² . Find the area of its corresponding major sector. (

π

= 3.14 )

Answer:

Radius of the circle, r = 10 cm

Area of the sector = 100 cm²

∴ Area of the corresponding major sector = Area of the circle − Area of the sector

$= π r^{2} - 100 = 3.14 \times {(10)}^{2} - 100 = 314 - 100 = 214 {cm}^{2}$
Thus, the area of the corresponding major sector is 214 cm².

Page No 154:

Question 5:

Area of a sector of a circle of radius 15 cm is 30 cm² . Find the length of the arc of the sector.

Answer:

Radius of the sector, r = 15 cm

Area of the sector = 30 cm²

∴ Length of the arc = $\frac{2 \times Area of the sector}{Radius of the sector}$ $= \frac{2 \times 30}{15}$ = 4 cm (Area of the sector = $\frac{1}{2}$ × Length of the arc × Radius)

Thus, the length of the arc of the sector is 4 cm.

Page No 154:

Question 6:

In the given figure, radius of the circle is 7 cm and m ( arc MBN) = 60^°,

find (1) Area of the circle .

(2) A(O - MBN) .

(3) A(O - MCN) .

Answer:

Radius of the circle, r = 7 cm

m(arc MBN) = ∠MON = θ = 60º

(1)
Area of the circle = $π r^{2} = \frac{22}{7} \times {(7)}^{2}$ = 154 cm²

(2)
A(O-MBN) = Area of the sector OMBN = $\frac{θ}{360 °} \times π r^{2} = \frac{60 °}{360 °} \times \frac{22}{7} \times {(7)}^{2}$ = 25.7 cm²

(3)
A(O-MCN) = Area of the sector OMCN

= Area of the circle − Area of the sector OMBN

= 154 − 25.7

= 128.3 cm²

Page No 155:

Question 7:

In the given figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm . Find A(P-ABC).

Answer:

Radius of the circle, r = 3.4 cm

Perimeter of sector P-ABC = 12.8 cm

Let l be the length of the arc ABC.

∴ l + 2r = 12.8 cm

⇒ l + 2 × 3.4 = 12.8

⇒ l = 12.8 − 6.8 = 6 cm

∴ A(P-ABC) = Area of the sector PABC = $\frac{1}{2} l r = \frac{1}{2} \times 6 \times 3.4$ = 10.2 cm²

Thus, A(P-ABC) is 10.2 cm².

Page No 155:

Question 8:

In the given figure, O is the centre of the sector. $∠$ ROQ = $∠$ MON = 60^° . OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and arc MYN. ( $π = \frac{22}{7}$ )

Answer:

In the given figure, ∠ROQ = ∠MON = θ = 60º

Radius of the sector ORXQ = OR = 7 cm

∴ Length of the arc RXQ = $\frac{θ}{360 °} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 7$ = 7.3 cm

Radius of the sector OMYN = OM = 21 cm

∴ Length of the arc MYN = $\frac{θ}{360 °} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21$ = 22 cm

Thus, the lengths of the arc RXQ and arc MYN are 7.3 cm and 22 cm, respectively.

Page No 155:

Question 9:

In the given figure, if A(P $-$ ABC) = 154 cm²radius of the circle is 14 cm,

find
(1)

∠

APC.

(2) l ( arc ABC) .

Answer:

Radius of the circle, r = 14 cm

(1)
A(P $-$ ABC) = Area of the sector PABC = 154 cm²

$∴ \frac{θ}{360 °} \times π r^{2} = 154 \Rightarrow \frac{∠ APC}{360 °} \times \frac{22}{7} \times {(14)}^{2} = 154 \Rightarrow ∠ APC = \frac{154 \times 7 \times 360 °}{22 \times 196} = 90 °$
Thus, the measure of $∠$ APC is 90º.

(2)
l(arc ABC) = Length of the arc ABC $= \frac{∠ APC}{360 °} \times 2 π r = \frac{90 °}{360 °} \times 2 \times \frac{22}{7} \times 14$ = 22 cm

Thus, the length of arc ABC is 22 cm.

Page No 155:

Question 10:

Radius of a sector of a circle is 7 cm. If measure of arc of the sector is -

(1) 30^° (2) 210^° (3 ) three right angles;

find the area of the sector in each case.

Answer:

Radius of the sector of the circle, r = 7 cm

(1)
Measure of arc of the sector = θ = 30º

∴ Area of the sector = $\frac{θ}{360 °} \times π r^{2} = \frac{30 °}{360 °} \times \frac{22}{7} \times {(7)}^{2}$ = 12.83 cm²

(2)
Measure of arc of the sector = θ = 210º

∴ Area of the sector = $\frac{θ}{360 °} \times π r^{2} = \frac{210 °}{360 °} \times \frac{22}{7} \times {(7)}^{2}$ = 89.83 cm²

(3)
Measure of arc of the sector = θ = 3 × 90º = 270º

∴ Area of the sector = $\frac{θ}{360 °} \times π r^{2} = \frac{270 °}{360 °} \times \frac{22}{7} \times {(7)}^{2}$ = 115.5 cm²

Page No 155:

Question 11:

The area of a minor sector of a circle is 3.85 cm²and the measure of its central angle is 36^°. Find the radius of the circle .

Answer:

Area of minor sector of the circle = 3.85 cm²

Measure of central angle, θ = 36º

Let the radius of the circle be r cm.

Now,

Area of minor sector = 3.85 cm²

$\Rightarrow \frac{θ}{360 °} \times π r^{2} = 3.85 \Rightarrow \frac{36 °}{360 °} \times \frac{22}{7} \times r^{2} = 3.85 \Rightarrow r = \sqrt{\frac{3.85 \times 360 ° \times 7}{36 ° \times 22}} \Rightarrow r = \sqrt{12.25} = 3.5 cm$
Thus, the radius of the circle is 3.5 cm.

Page No 155:

Question 12:

In the given figure, $□$ PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z .

Answer:

PQRS is a rectangle.

∴ ∠Q = ∠R = 90º

Radius of sector PTQ = PQ = 14 cm

∴ Area of part x = Area of the sector PTQ = $\frac{θ}{360 °} \times π r^{2} = \frac{∠ Q}{360 °} \times π \times {(PQ)}^{2} = \frac{90 °}{360 °} \times \frac{22}{7} \times {(14)}^{2}$ = 154 cm²

Radius of sector TUR = TR = QR − QT = QR − PQ = 21 − 14 = 7 cm (QT = PQ)

∴ Area of part y = Area of the sector TUR = $\frac{θ}{360 °} \times π r^{2} = \frac{∠ R}{360 °} \times π \times {(TR)}^{2} = \frac{90 °}{360 °} \times \frac{22}{7} \times {(7)}^{2}$ = 38.5 cm²

Now,

Area of rectangle PQRS = QR × PQ = 21 × 14 = 294 cm²

∴ Area of part z = Area of rectangle PQRS − Area of part x − Area of part y = 294 − 154 − 38.5 = 101.5 cm²

Page No 155:

Question 13:

∆

LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm.

Find,

(1) A (

∆

LMN)

(2) Area of any one of the sectors .

(3) Total area of all the three sectors.

(4) Area of the shaded region.

Answer:

âˆ†LMN is an equilateral triangle.

∴ LM = MN = LN = 14 cm

∠L = ∠M = ∠N = 90º

(1)
Area of âˆ†LMN = $\frac{\sqrt{3}}{4} {(Side)}^{2} = \frac{\sqrt{3}}{4} \times {(14)}^{2} = \frac{1.732}{4} \times 196$ = 84.87 cm²

(2)
Radius of the each sector, r = 7 cm

Area of any one of the sectors = $\frac{θ}{360 °} \times π r^{2} = \frac{60 °}{360 °} \times \frac{22}{7} \times {(7)}^{2}$ = 25.67 cm²

(3)
Total area of all the three sectors = 3 × Area of any one of the sectors = 3 × 25.67 = 77.01 cm²

(4)
Area of the shaded region = Area of âˆ†LMN − Total area of all the three sectors = 84.87 − 77.01 = 7.86 cm²

Page No 159:

Question 1:

In the given figure, A is the centre of the circle. $∠$ ABC = 45^°and AC = 7 $\sqrt{2}$ cm. Find the area of segment BXC.

Answer:

In âˆ†ABC,

AB = AC (Radii of the circle)

∴ ∠ACB = ∠ABC = 45º (Equal sides have equal angles opposite to them)

Using angle sum property, we have

∠ACB + ∠ABC + ∠BAC = 180º

∴ 45º + 45º + ∠BAC = 180º

⇒ ∠BAC = 180º − 90º = 90º

Here,

Radius of the circle, r = $7 \sqrt{2}$ cm

Measure of arc BXC, θ = 90º

∴ Area of segment BXC

$= r^{2} (\frac{π θ}{360 °} - \frac{\sin θ}{2}) = {(7 \sqrt{2})}^{2} (\frac{22}{7} \times \frac{90 °}{360 °} - \frac{\sin 90 °}{2}) = 98 \times \frac{22}{7} \times \frac{1}{4} - 98 \times \frac{1}{2} = 77 - 49 = 28 {cm}^{2}$
Thus, the area of the segment BXC is 28 cm².

Page No 159:

Question 2:

In the given figure, O is the centre of the circle. m ( arc PQR) = 60^°OP = 10 cm. Find the area of the shaded region.

(

π

= 3.14,

\sqrt{3}

= 1.73)

Answer:

Radius of the circle, r = 10 cm

m(arc PQR) = ∠POR = θ = 60º

∴ Area of the shaded region = Area of segment PQR

$= r^{2} (\frac{π θ}{360 °} - \frac{\sin θ}{2}) = {(10)}^{2} (\frac{3.14 \times 60 °}{360 °} - \frac{\sin 60 °}{2}) = 100 \times (\frac{3.14}{6} - \frac{\sqrt{3}}{4}) = \frac{314}{6} - \frac{173}{4} (\sqrt{3} = 1.73) = 52.33 - 43.25 = 9.08 {cm}^{2}$
Thus, the area of the shaded region is 9.08 cm².

Page No 160:

Question 3:

In the given figure, if A is the centre of the circle. $∠$ PAR = 30^°, AP = 7.5, find the area of the segment PQR ( $π$ = 3.14)

Answer:

Radius of the circle, r = 7.5 cm

∠â€‹PAR = θ = 30º

∴ Area of segment PQR

$= r^{2} (\frac{π θ}{360 °} - \frac{\sin θ}{2}) = {(7.5)}^{2} (\frac{3.14 \times 30 °}{360 °} - \frac{\sin 30 °}{2}) = 56.25 \times (\frac{3.14}{12} - \frac{1}{4}) = 56.25 \times 0.01167 = 0.65625 {cm}^{2}$
Thus, the area of the segment PQR is 0.65625 cm².

Page No 160:

Question 4:

In the given figure, if O is the centre of the circle, PQ is a chord. $∠$ POQ = 90^°, area of shaded region is 114 cm² , find the radius of the circle.
( $π$ = 3.14)

Answer:

∠POQ = θ = 90º

Let the radius of the circle be r cm.

Area of the shaded region = Area of the segment PRQ = 114 cm²

$∴ r^{2} (\frac{π θ}{360 °} - \frac{\sin θ}{2}) = 114 \Rightarrow r^{2} (\frac{3.14 \times 90 °}{360 °} - \frac{\sin 90 °}{2}) = 114 \Rightarrow r^{2} (\frac{3.14}{4} - \frac{1}{2}) = 114 \Rightarrow r^{2} \times (0.785 - 0.5) = 114 \Rightarrow r = \sqrt{\frac{114}{0.285}} \Rightarrow r = \sqrt{400} = 20 cm$
Thus, the radius of the circle is 20 cm.

Page No 160:

Question 5:

A chord PQ of a circle with radius 15 cm subtends an angle of 60^° with the centre of the circle. Find the area of the minor as well as the major segment. (

π

= 3.14,

\sqrt{3}

= 1.73)

Answer:

Radius of the circle, r = 15 cm

Let O be the centre and PQ be the chord of the circle.

∠POQ = θ = 60º

Area of the minor segment = Area of the shaded region

$= r^{2} (\frac{π θ}{360 °} - \frac{\sin θ}{2}) = {(15)}^{2} \times (\frac{3.14 \times 60 °}{360 °} - \frac{\sin 60 °}{2}) = 225 \times \frac{3.14}{6} - 225 \times \frac{\sqrt{3}}{4} = 117.75 - 97.31 = 20.44 {cm}^{2}$
Now,

Area of the circle = $π r^{2} = 3.14 \times {(15)}^{2} = 3.14 \times 225$ = 706.5 cm²

∴ Area of the major segment = Area of the circle − Area of the minor segment = 706.5 − 20.44 = 686.06 cm²

Thus, the areas of the minor segment and major segment are 20.44 cm² and 686.06 cm², respectively.

Page No 160:

Question 1:

Choose the correct alternative answer for each of the following questions.

(1) The ratio of circumference and area of a circle is 2:7. Find its circumference.
(A)

14 π

(B)

\frac{7}{π}

π

(D)

\frac{14}{π}

(2) If measure of an arc of a circle is 160^° and its length is 44 cm, find the circumference of the circle.

(A) 66 cm (B) 44 cm (C) 160 cm (D) 99 cm

(3) Find the perimeter of a sector of a circle if its measure is 90^° and radius is 7 cm .

(A) 44 cm (B) 25 cm (C) 36 cm (D) 56 cm

(4) Find the curved surface area of a cone of radius 7 cm and height 24 cm.

(A) 440 cm² (B) 550 cm² (C) 330 cm²(D) 110 cm²

(5) The curved surface area of a cylinder is 440 cm 2 and its radius is 5 cm. Find its height.
(A)

\frac{44}{π}

cm (B) 22

π

cm (C) 44

π

cm (D)

\frac{22}{π}

(6) A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.

(A) 15 cm (B) 10 cm (C) 18 cm (D) 5 cm

(7) Find the volume of a cube of side 0.01 cm.

(A) 1 cm³ (B) 0.001 cm³ (C) 0.0001 cm³ (D) 0.000001 cm³

(8) Find the side of a cube of volume 1 m³.

(A) 1 cm (B) 10 cm (C) 100 cm (D)1000 cm

Answer:

(1)
Let r be the radius of the circle.

Circumference of the circle : Area of the circle = 2 : 7

$\Rightarrow \frac{2 π r}{π r^{2}} = \frac{2}{7} \Rightarrow \frac{1}{r} = \frac{1}{7} \Rightarrow r = 7$
∴ Circumference of the circle = $2 π r = 2 π \times 7 = 14 π$

Hence, the correct answer is option (A).

(2)
Measure of arc of circle, θ = 160º

Let r be the radius of the circle.

Length of the arc = 44 cm

$∴ \frac{θ}{360 °} \times 2 π r = 44 \Rightarrow \frac{160 °}{360 °} \times 2 π r = 44 \Rightarrow 2 π r = \frac{44 \times 360 °}{160 °} = 99 cm$
Thus, the circumference of the circle is 99 cm.

Hence, the correct answer is option (D).

(3)
Here, r = 7 cm and θ = 90º.

∴ Perimeter of the sector = $\frac{θ}{360 °} \times 2 π r + 2 r = \frac{90 °}{360 °} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7$ = 11 + 14 = 25 cm

Hence, the correct answer is option (B).

(4)
Radius of the cone, r = 7 cm

Height of the cone, h = 24 cm

Let l be the slant height of the cone.

$l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625}$ = 25 cm

∴ Curved surface area of the cone = $π r l = \frac{22}{7} \times 7 \times 25$ = 550 cm²

Hence, the correct answer is option (B).

(5)
Radius of the cylinder, r = 5 cm

Let the height of the cylinder be h cm.

Curved surface area of the cylinder = 440 cm²

$∴ 2 π r h = 440 \Rightarrow h = \frac{440}{2 π r} \Rightarrow h = \frac{440}{2 π \times 5} = \frac{44}{π} cm$
Thus, the height of the cylinder is $\frac{44}{π}$ cm.

Hence, the correct answer is option (A).

(6)
Radius of the cone = Radius of the cylinder = r cm (Say)

Height of the cylinder, h = 5 cm

Let the height of the cone be H cm.

It is given that the cone melted and recasted into a cylinder.

∴ Volume of the cone = Volume of the cylinder

$\Rightarrow \frac{1}{3} π r^{2} H = π r^{2} h \Rightarrow H = 3 h \Rightarrow H = 3 \times 5 = 15 cm$
Thus, the height of the cone is 15 cm.

Hence, the correct answer is option (A).

(7)
Side of the cube = 0.01 cm

∴ Volume of the cube = (Side)³ = (0.01 cm)³ = 0.000001 cm³

Hence, the correct answer is option (D).

(8)
Volume of the cube = 1 m³

∴ (Side)³= 1 m³ = (1 m)³

⇒ Side = 1 m = 100 cm

Thus, the side of the cube is 100 cm.

Hence, the correct answer is option (C).

Page No 161:

Question 2:

A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub ? (

π = \frac{22}{7}

)

Answer:

Radius of the circular top of washing tub, r₁ = 20 cm

Radius of the circular bottom of washing tub, r₂ = 15 cm

Height of the washing tub, h = 21 cm

∴ Capacity of the washing tub = Volume of frustum of cone

$= \frac{1}{3} π h (r_{1}^{2} + r_{1} r_{2} + r_{2}^{2}) = \frac{1}{3} \times \frac{22}{7} \times 21 \times (20^{2} + 20 \times 15 + 15^{2}) = 22 \times (400 + 300 + 225) = 22 \times 925 = 20350 {cm}^{3}$
$= \frac{20350}{1000} L (1 L = 1000 {cm}^{3}) = 20.35 L$
Thus, the capacity of the tub is 20.35 litres.

Page No 161:

Question 3:

Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm , 90 cm and 30 cm respectively. How many balls were melted to make the tube?

Answer:

Radius of each plastic ball, R = 1 cm

Outer radius of the tube, r₂ = 30 cm

Thickness of the tube = 2 cm

∴ Inner radius of the tube, r₁ = Outer radius of the tube − Thickness of the tube = 30 − 2 = 28 cm

Length of the tube, h = 90 cm

Let the number of plastic balls melted to make the tube be n.

It given that plastic balls are melted to form the tube.

∴ n × Volume of each plastic ball = Volume of the tube

$\Rightarrow n = \frac{Volume of the tube}{Volume of each plastic ball} \Rightarrow n = \frac{π (r_{2}^{2} - r_{1}^{2}) h}{\frac{4}{3} π R^{3}} \Rightarrow n = \frac{(30^{2} - 28^{2}) \times 90}{\frac{4}{3} \times 1^{3}} \Rightarrow n = \frac{116 \times 3 \times 90}{4} = 7830$
Thus, the number of plastic balls melted to make the tube are 7830.

Page No 161:

Question 4:

A metal parallelopiped of measures 16 cm

\times

11 cm

\times

10 cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively ?

Answer:

Radius of each coin, r = $\frac{2}{2}$ = 1 cm

Thickness of each coin, h = 2 mm = $\frac{2}{10}$ = 0.2 cm (1 cm = 10 mm)

Let the number of coins made be n.

It is given that a metal parallelopiped is melted to make the coins.

∴ n × Volume of metal in each coin = Volume of the metal parallelopiped

$\Rightarrow n = \frac{Volume of the metal parallelopiped}{Volume of metal in each coin} \Rightarrow n = \frac{16 \times 11 \times 10}{π r^{2} h} \Rightarrow n = \frac{16 \times 11 \times 10}{\frac{22}{7} \times {(1)}^{2} \times 0.2} = 2800$
Thus, the number of coins made are 2800.

Page No 161:

Question 5:

The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of Rs. 10 per sq.m.

Answer:

Radius of the roller, r = $\frac{120}{2}$ = 60 cm

Length of the roller, h = 84 cm

Area of the ground levelled in one rotation of the roller = Curved surface area of the roller = $2 π r h = 2 \times \frac{22}{7} \times 60 \times 84$ = 31680 cm²

∴ Area of the ground levelled in 200 rotations of the roller

= 200 × Area of the ground levelled in one rotation of the roller

= 200 × 31680

= 6336000 cm²

= $\frac{6336000}{10000}$ (1 m² = 10000 cm²)

= 633.6 m²

Rate to level the ground = Rs 10/m²

∴ Expenditure (or total cost) to level the ground

= Total area of the ground × Rate to level the ground

= Area of the ground levelled in 200 rotations of the roller × Rate to level the ground

= 633.6 × 10

= Rs 6,336

Thus, the expenditure to level the ground is Rs 6,336.

Page No 161:

Question 6:

The diameter and thickness of a hollow metals sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³. Find the outer surface area and mass of the sphere.

Answer:

Outer radius of the sphere, R = $\frac{12}{2}$ = 6 cm

Thickness of the sphere = 0.01 m = 0.01 × 100 cm = 1 cm (1 m = 100 cm)

∴ Inner radius of the sphere, r = R − 1 = 6 − 1 = 5 cm

Outer surface area of the sphere = $4 π R^{2} = 4 \times 3.14 \times {(6)}^{2} = 4 \times 3.14 \times 36$ = 452.16 cm²

Volume of metal in the sphere = $\frac{4}{3} π (R^{3} - r^{3}) = \frac{4}{3} \times 3.14 \times (6^{3} - 5^{3}) = \frac{4}{3} \times 3.14 \times (216 - 125) = \frac{4}{3} \times 3.14 \times 91$ = 380.97 cm³

Density of the metal = 8.88 g/cm³

∴ Mass of the sphere = Volume of metal in the sphere × Density of the metal = 380.97 × 8.88 = 3383.01 g

Thus, the outer surface area and mass of the sphere are 452.16 cm² and 3383.01 g, respectively.

Page No 161:

Question 7:

A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone ?

Answer:

Radius of the bucket, r = $\frac{28}{2}$ = 14 cm

Height of the bucket, h = 20 cm

Height of the cone, H = 14 cm

Let the radius of the base of the cone be R cm.

∴ Area of the base of the cone = $π$ R²

Now,

Volume of sand in the cone = Volume of sand in the cylindrical bucket

$∴ \frac{1}{3} π R^{2} H = π r^{2} h \Rightarrow π R^{2} = \frac{3 π r^{2} h}{H} \Rightarrow π R^{2} = \frac{3 \times \frac{22}{7} \times {(14)}^{2} \times 20}{14} \Rightarrow π R^{2} = 2640 {cm}^{2}$
Thus, the base area of the cone is 2640 cm².

Page No 161:

Question 8:

The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.

Answer:

Radius of the sphere, R = 9 cm

Radius of the wire, r = $\frac{4}{2}$ = 2 mm = $\frac{2}{10}$ = 0.2 cm (1 cm = 10 mm)

Let the length of the wire be l cm.

It is given that the metallic sphere is melted to make the wire.

∴ Volume of metal in the wire = Volume of the metallic sphere

$\Rightarrow π r^{2} l = \frac{4}{3} π R^{3} \Rightarrow l = \frac{\frac{4}{3} R^{3}}{r^{2}} \Rightarrow l = \frac{\frac{4}{3} \times {(9)}^{3}}{{(0.2)}^{2}} \Rightarrow l = 24300 cm$
$\Rightarrow l = \frac{24300}{100} \Rightarrow l = 243 m (1 m = 100 cm)$
Thus, the length of the wire is 243 m.

Page No 161:

Question 9:

The area of a sector of a circle of 6 cm radius is 15

π

sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.

Answer:

Radius of the sector, r = 6 cm

Let the measure of the arc be θ and the length of the arc corresponding to the sector be l cm.

Area of the sector = 15 $π$ cm² (Given)

$∴ \frac{1}{2} l r = 15 π \Rightarrow \frac{1}{2} \times l \times 6 = 15 π \Rightarrow l = \frac{15 π}{3} = 5 π cm$
Length of the arc = $5 π cm$

$∴ \frac{θ}{360 °} \times 2 π r = 5 π \Rightarrow θ = \frac{5 \times 360 °}{2 r} \Rightarrow θ = \frac{5 \times 360 °}{2 \times 6} \Rightarrow θ = 150 °$
Thus, the measure of the arc and length of the arc corresponding to the sector are 150º and 5 $π$ cm, respectively.

Page No 161:

Question 10:

In the given figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion. (

π

= 3.14,

\sqrt{3}

= 1.73 )

Answer:

Draw PQ ⊥ AB.

∴ AQ = QB (Perpendicular from the centre of the circle to the chord bisects the chord)

In right âˆ†APQ,

$AQ = \sqrt{{AP}^{2} - {PQ}^{2}} \Rightarrow AQ = \sqrt{8^{2} - 4^{2}} \Rightarrow AQ = \sqrt{64 - 16} \Rightarrow AQ = \sqrt{48} \Rightarrow AQ = 4 \sqrt{3} cm$
∴ AB = 2AQ = $2 \times 4 \sqrt{3} = 8 \sqrt{3} cm$

Also,
$\sin ∠ APQ = \frac{AQ}{AP} \Rightarrow \sin ∠ APQ = \frac{4 \sqrt{3}}{8} \Rightarrow \sin ∠ APQ = \frac{\sqrt{3}}{2} = \sin 60 ° \Rightarrow ∠ APQ = 60 °$
Similarly,

$∠ BPQ = 60 °$

∴ ∠APB = ∠APQ + ∠BPQ = 60º + 60º = 120º

Radius of the circle, r = 8 cm

Measure of arc AB, θ = 120º

∴ Area of the shaded portion = Area of the sector ABP − Area of âˆ†APB
$= \frac{θ}{360 °} \times π r^{2} - \frac{1}{2} \times AB \times PQ = \frac{120 °}{360 °} \times 3.14 \times 8^{2} - \frac{1}{2} \times 8 \sqrt{3} \times 4 = 66.97 - 27.68 = 39.29 {cm}^{2}$
Thus, the area of the shaded portion is 39.29 cm².

Page No 162:

Question 11:

In the given figure, square ABCD is inscribed in the sector A - PCQ. The radius of sector C - BXD is 20 cm. Complete the following activity to find the area of shaded region

Answer:

Side of square ABCD = radius of sector C-BXD = $20$ cm

Area of square = (side)² = $20^{2}$ = $400$ cm²

Area of shaded region inside the square

= Area of square ABCD − Area of sector C-BXD

= $400$ $- \frac{θ}{360 °} \times π r^{2}$

= $400$ $- \frac{90 °}{360 °} \times \frac{3.14}{1} \times \frac{400}{1}$

= $400$ − 314

= $86$ cm²

Radius of bigger sector = Length of diagonal of square ABCD = $20 \sqrt{2}$ cm

Area of the shaded regions outside the square

= Area of sector A-PCQ − Area of square ABCD

= A(A-PCQ) − A( ABCD)

= $\frac{θ}{360 °} \times π \times r^{2}$ − $20^{2}$

= $\frac{90 °}{360 °} \times 3.14 \times {(20 \sqrt{2})}^{2}$ − ${(20)}^{2}$

= $628$ − $400$

= $228$ cm²

∴ Total surface area of the shaded region = 86 + 228 = 314 cm²

Page No 163:

Question 12:

In the given figure , two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.

Answer:

Let the radius of the bigger circle be R and that of smaller circle be r.

OA, OB, OC and OD are the radii of the bigger circle.

∴ OA = OB = OC = OD = R

PQ = PA = r

OQ = OB − BQ = $R - 9$

OE = OD − DE = $R - 5$

As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,

OQ × OA = OE × OF

$R - 9$ × R = $R - 5$ × $R - 5$ .....(âˆµ OE = OF)

R² − 9R = R² − 10R + 25

R = $25$

AQ = 2r = AB − BQ (âˆµAB = 2R)

2r = 50 − 9 = 41

r = $\frac{41}{2}$ = $20.5$

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