## Mensuration

### Class-10-Mathematics-2-Chapter-7-Maharashtra Board

### Solutions

**Practice set 7.1**

**Question 1.1. Find the volume of a cone if the radius of its base is ****1.5 ****cm and its perpendicular height is ****5 ****cm.**

The radius (r) of the cone = 1.5 cm, Its perpendicular height (h) = 5 cm

Volume of the cone = \(\frac{1}{3}\)πr^{2}h = \(\frac{1}{3}×\frac{22}{7}\) × 1.5 × 1.5 × 5 = \(\frac{82.5}{7}\) = 11.785 ≈ 11.79 cm^{3}

**Answer is : Volume of the cone is 11.79 cm ^{3}.**

**Question 1.2. Find the volume of a sphere of diameter ****6 ****cm.**

Diameter of a sphere = 6 cm, its radius(r) = 3 cm

Volume of the sphere = \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3}\) × 3.14 × 3 × 3 × 3 = 113.04 cm^{3}

**Answer is : Volume of the sphere is 113.04 cm ^{3}.**

**Question 1.3. Find the total surface area of a cylinder if the radius of its base is ****5 ****cm and height is ****40 ****cm.**

The radius (r) of the base of the cylinder = 5 cm, its height (h) = 40 cm

Total surface area of the cylinder = 2πr(r + h)

= 2 × 3.14 × 5(5 + 40)

= 3.14 ×10 × 45

= 1413 cm^{2}

**Answer is : Total surface area of the cylinder is ****1413 cm ^{2}.**

**Question 1.4. Find the surface area of a sphere of radius 7 cm.**

The radius of the sphere (r) = 7 cm.

Surface area of the sphere = 4πr^{2} = 4 × \(\frac{22}{7}\) × 7 × 7 = 88 × 7 = 616 cm^{2}

**Answer is : Surface area of the sphere is 616 cm ^{2}.**

**Question 1.5. 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.**

The length (l) of a cuboid = 44 cm, its breadth (b) = 21 cm, its height (h) = 12 cm

The height of the cone (h_{1}) = 24 cm

Let the radius of the base of the cone be r.

Cuboid is melted and a cone is made.

∴ volume of cuboid = volume of cone

∴ l x b x h = \(\frac{1}{3}\) x πr^{2}h_{1}

∴ 44 x 21 x 12 = \(\frac{1}{3}×\frac{22}{7}\) x r^{2 }x 24

∴ r^{2} = \(\frac{44×21×12×3×7}{22×24}\) = 21 x 21

∴ r = 21 cm ... (Taking square roots of both the sides)

**Answer is : The radius of the base of the cone is 21 cm.**

**Question 1.6. Observe the measures of pots in figure 7.8 and 7.9. How many jugs of water can the cylindrical pot hold ?**

The radius (r) of the conical water jug = 3.5 cm, its height (h) = 10 cm,

The radius (r_{1}) of the cylindrical water pot = 7 cm, its height (h_{1}) = 10 cm.

Number of jugs of water that cylindrical pot can hold = \(\frac{\text{Volume of cylindrical water pot}}{\text{Volume of cylindrical water jug}}\)

\(\frac{πr_1^2h_1}{\frac{1}{3}×πr^2h}\) = \(\frac{7×7×10}{\frac{1}{3}×3.5×3.5×10}\) = 2 × 2 × 3 = 12

**Answer is : ****The cylindrical pot can hold 12 jugs of water.**

**Question 1.7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm ^{2}.The cone is placed upon the cylinder. Volume of the solid figure so formed is 500 cm^{3}. Find the total height of the figure.**

A cylinder and a cone have equal bases. ∴ they have equal radii.

Let the radius be r.

Area of its base = 100 cm^{2}

∴ πr^{2} = 100 cm^{2} ….(1)

Let the height of the cylinder be h_{1} and that of the cone be h_{2}.

h_{1} = 3 cm

Volume of the solid figure = 500 cm^{3}

∴ volume of the cylindrical part + volume of conical part = 500

∴ πr^{2}h_{1} + \(\frac{1}{3}\)πr^{2}h_{2} = 500

∴ 100 × 3 + \(\frac{1}{3}\) × 100 × h_{2} = 500 ….[From (1)]

∴ 300 + \(\frac{100}{3}\)h_{2} = 500

∴ \(\frac{100}{3}\)h_{2} = 500 – 300 = 200

∴ h_{2} = \(\frac{200×3}{100}\) = 6

∴ h_{2} = 6 cm

Height of the figure = h_{1} + h_{2 }= 3 + 6 = 9 cm.

**Answer is : The total height of the figure is 9 cm.**

[Note : The answer in the textbook is given as 5 cm, whereas the correct answer is 9 cm.]

**Question 1.8. In figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.**

A toy is made up of hemisphere, cylinder and a cone.

They have equal bases.

Let their radius be r.

∴ r = 3 cm.

Let the heights of the conical part and cylindrical part be h_{1} and h_{2} respectively.

h_{1} = 4 cm and h_{2} = 40 cm.

Let the slant height of the cone be l.

*l*^{ 2} = r^{2} + h_{1}^{2}

∴ *l*^{ 2} = 3^{2} + 4^{2}

∴ *l*^{ 2} = 9 + 16

∴ *l*^{ 2} = 25

∴ *l* = 5 cm ... (Taking square roots of both the sides)

Total area of the toy = curved surface area of the cone + curved surface area of the cylinder + curved surface area of the hemisphere

= πr*l* + 2πrh_{2} + 2πr^{2}

= πr(*l* + 2h_{2} + 2r)

= π × 3(5 + 2×40 + 2×3)

= π × 3(5 + 80 + 6)

= π × 3 (91)

= 273π cm^{2}

**Answer is : ****Total area of the toy is 273****π ****cm ^{2}**

**Question 1.9. In the 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 ?**

The radius (r) of the tablet = 7 mm, its thickness (h) = 5mm

The diameter of the wrapper = 14 mm, ∴ its radius (r_{1}) = 14/2 = 7 mm

its length (h_{1}) = 10 cm = 100 mm

Number of tablets that can be wrapped in the wrapper = \(\frac{\text{Volume of the wrapper}}{\text{Volume of each tablet}}\)

**=** \(\frac{πr_1^2h_1}{πr^2h}\) = \(\frac{7×7×100}{7×7×5}\) = 20

**Answer is : 20 tablets can be wrapped in the wrapper.**

**Question 1.10. Figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.****(**** π** **= 3.14)**

A toy is made up of a cone and a hemisphere.

They have equal bases.

Let the radius of the base be r.

∴ r = 3 cm

Height of the conical part (h) = 4 cm

Let the slant height of the cone be *l*

*l*^{2} = r^{2} + h^{2}

∴ *l*^{2} = 3^{2} + 4^{2}

∴ *l*^{2} = 9 + 16

∴ *l*^{2} = 25

∴ *l* = 5 ... (Taking square roots of both the sides)

Surface area of the toy = Curved surface area of the conical part + Curved surface area of the hemispherical part

= πr*l* + 2πr^{2}

= πr (*l *+ 2r)

= 3.14 × 3(5 + 2 × 3)

= 3.14 × 3 × 11

= 3.14 × 33 = 103.62 cm^{2}

Volume of the toy = Volume of conical part + Volume of hemispherical part

= \(\frac{1}{3}\)πr^{2}h + \(\frac{2}{3}\)πr^{3} = \(\frac{1}{3}\)πr^{2}(h + 2r) = \(\frac{1}{3}\) × 3.14 × 9(4 + 2 × 3) = 94.2 cm^{3}

**Answer is : ****The surface area of the toy is 103.62 cm ^{2 }and the volume of the toy is 94.2 cm^{3}.**

**Question 1.11. Find the surface area and the volume of a beach ball shown in the figure.**

The diameter of the beach ball = 42 cm, ∴ its radius (r) = 42/2 = 21 cm

Surface area of the beach ball = 4πr^{2} = 4 × 3.14 × 21 × 21 = 5538.96 cm^{2}

Volume of the beach ball = \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3}\) × 3.14 × (21)^{3 }= 38772.72 cm^{3}

**Answer is : ****Surface area of the beach ball is 5538.96 cm ^{2} and volume of the beach ball is 38772.72 cm^{3}.**

**Question 1.12. As showm 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.**

Diameter of the cylindrical glass = 14 cm, ∴ r = 7 cm

Volume of the water with metallic sphere immersed in it = πr^{2}h = π × 49 × 30

= 1470π cm^{3} …..(1)

Diameter of the metallic sphere = 2 cm. ∴ r_{1} = 1 cm

Volume of the metallic sphere = \(\frac{4}{3}\)πr_{1}^{3} = \(\frac{4}{3}\) × π × 1^{3 }= \(\frac{4}{3}\)π = 1.33 π cm^{3} …..(2)

To find the volume of water in the cylinder, subtract the result (2) from the result (1).

∴ Volume of water = (1470 π − 1.33 π) cm^{3 }= 1468.67 π cm^{3}

**Answer is : **** Volume of water is 1468.67**** π**** cm ^{3}.**

**Practice set 7.2**

**Question 2.1. 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 cm**^{3}**)**

The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm.

∴ r_{1} = 14 cm and r_{2} = 7 cm, its height (h) = 30 cm

Volume of the bucket = \(\frac{1}{3}\)π (r_{1}^{2} + r_{2}^{2} + r_{1} x r_{2}) x h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × (142 + 72 + 14 × 7) × 30

= \(\frac{22}{7}\) × 343 × 10

= 10780 cm^{3} = litres ……(∵ 1 litre = 1000 cm^{3})

= 10.78 litres.

**Answer is : ****Bucket can hold 10.78 litres of water.**

**Question 2.2. 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)**

The radii of ends of a frustum are 14 cm and 6 cm.

∴ r_{1} = 14 cm and r_{2} = 6 cm

its height (h) = 6 cm

Let the slant height of the frustum be *l*.

*l* = \(\sqrt{h^2+(r_1-r_2)^2}\)

∴ *l* = \(\sqrt{6^2+(14-6)^2}\)

∴ *l* = \(\sqrt{100}\) = 10 cm

Curved surface area of a frustum = π*l*(r_{1} + r_{2})

= 3.14 × 10 × (14 + 6) = 3.14 × 10 × 20 = 628 cm^{2}

Total surface area of a frustum = π*l*(r_{1} + r_{2}) + πr_{1}^{2} + πr_{2}^{2}

= 628 + π(r_{1}^{2} + r_{2}^{2})

= 628 + 3.14 (14^{2} + 6^{2})

= 628 + 3.14(196 + 36)

= 628 +3.14 × 232

= 628 + 728.48

= 1356.48 cm^{2}

Volume of the frustum = \(\frac{1}{3}\)πh(r_{1}^{2} + r_{2}^{2} + r_{1} x r_{2})

= \(\frac{1}{3}\) x 3.14 × 6 ×(14^{2} + 6^{2} + 14 × 6)

= 3.14 × 2 × (196 + 36 + 84)

= 6.28 × 316

= 1984.48 cm^{3}

**Answer is : ****Curved surface area of a frustum is 628 cm ^{2}, Total surface area of a frustum is 1356.48 cm^{2} and Volume of frustum is 1984.48 cm^{3}.**

**Question 2.3. 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** **= ****22/7 ****).**

circumference_{1} = 2πr_{1} = 132

r_{1 }= \(\frac{132}{2π}\) = […..]

circumference_{2} = 2πr_{2} = 88

r_{2 }= \(\frac{88}{2π}\) = […..]

slant height of frustum, *l* = \(\sqrt{h^2+(r_1-r_2)^2}\) = \(\sqrt{[...]^2+([...])^2}\) = […..] cm

curved surface area of the frustum = π(r_{1} + r_{2})*l*

= π × [….] × [….]

= […..] sq.cm.

circumference_{1} = 2πr_{1} = 132

r_{1 }= \(\frac{132}{2π}\) = [**21 cm**]

circumference_{2} = 2πr_{2} = 88

r_{2 }= \(\frac{88}{2π}\) = [**14 cm**]

slant height of frustum, *l* = \(\sqrt{h^2+(r_1-r_2)^2}\) = \(\sqrt{[24]^2+([7])^2}\) = [**25**] cm

curved surface area of the frustum = π(r_{1} + r_{2})*l*

= π × [**35**] × [**25**]

= [**2750**] sq.cm.

**Practice set 7.3**

**Question 3.1. Radius of a circle is ****10 ****cm. Measure of an arc of the circle is ****54****°****. ****Find the area of the sector associated with the arc. ****(****π** **= 3.14 )**

The radius (r) of a circle =10 cm,

Measure of an arc of the circle (θ) = 54°

Area of the sector = \(\frac{θ}{360}\) × πr^{2} = \(\frac{54}{360}\) × 3.14 × 10^{2} = 3 × 1.57 × 10 = 47.10 cm^{2}

**Answer is : ****Area of the sector associated with the given arc is 47.10 cm ^{2}.**

**Question 3.2. Measure of an arc of a circle is ****80°**** and its radius is ****18 ****cm. Find the length of the arc. ****(****π** **= 3.14 )**

Measure of an arc of a circle (θ) = 80°, its radius (r) = 18 cm.

Length of the arc = \(\frac{θ}{360}\) × 2πr = \(\frac{80}{360}\) × 2 × 3.14 × 18 = 3.14 × 8 = 25.12 cm

**Answer is : Length of the arc is 25.12 cm.**

**Question 3.3. 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.**

The radius (r) of the sector of a circle = 3.5 cm.

Length of the arc = 2.2 cm.

Area of a sector = \(\frac{\text{length of the arc × radius}}{2}\) = \(\frac{2.2×3.5}{2}\) = 3.85 cm^{2}

**Answer is : ****Area of the sector is 3.85 cm ^{2}.**

**Question 3.4. Radius of a circle is ****10 ****cm. Area of a sector of the sector is ****100 ****cm**^{2}**. ****Find the area of its corresponding major sector. ****(****π** **= 3.14)**

The radius (r) of a circle =10 cm.

Area of the circle = πr^{2} = 3.14 × 10 × 10 = 314 cm^{2}

Area of the sector = 100 cm^{2 } ... (Given)

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

= 314 – 100 = 214 cm^{2}

**Answer is : ****Area of the corresponding major sector is 214 cm ^{2}.**

**Question 3.5. Area of a sector of a circle of radius ****15 ****cm is ****30 ****cm**^{2}**. ****Find the length of the arc of the sector.**

The radius (r) of the circle = 15 cm.

Area of the sector = 30 cm^{2}.

Area of the sector = \(\frac{\text{length of the arc × radius}}{2}\)

30 **=** \(\frac{\text{length of the arc × 15}}{2}\)

∴ length of the arc = \(\frac{30×2}{15}\) = 4

∴ length of the arc = 4 cm

**Answer is : Length of the arc is 4 cm.**

**Question 3.6. In the 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).**

The radius (r) of the circle = 7 cm, m(arc MBN) = θ = 60°

(1) Area of the circle = πr^{2 }= \(\frac{22}{7}\) × 7 × 7 = 154 cm^{2} ….(1)

(2) A(O−MBN) = \(\frac{θ}{360}\) × πr^{2} = \(\frac{60}{360}\) × 154 …[from (1)]

** = **\(\frac{154}{6}\) = 25.7 cm^{2}

(3) A(O−MCN) = Area of the circle − A(O−MBN)

= 154 − 25.7 = 128.3 cm^{2}

**Answer is : ****(1) Area of the circle = 154 cm ^{2}, (2) A(O−MBN) = 25.7 cm^{2} and **

**(3) A(O−MCN) = 128.3 cm**

^{2}.

**Question 3.7. In figure****, ****radius of circle is ****3.4 ****cm and perimeter of sector ****P−ABC ****is ****12.8 ****cm****. ****Find ****A(P−ABC).**

The radius (r) of the circle = 3.4 cm.

Perimeter of sector P−ABC = 12.8 cm.

Perimeter of sector P−ABC = r + r + length of arc ABC

∴ 12.8 =3.4 +3.4 + length of arc ABC

∴ 12.8 = 6.8 + length of arc ABC

∴ length of arc ABC = 12.8 − 6.8

∴ length of arc ABC = 6 cm

A(P−ABC) = \(\frac{\text{length of the arc × radius}}{2}\) = \(\frac{6×3.4}{2}\) = 10.2 cm^{2}

**Answer is : ****A(P−ABC) = 10.2 cm ^{2}.**

**Question 3.8. In 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** **= ****22/7****)**

For arc RXQ,

radius(r) = OR = 7 cm

m(arc RXQ) = **∠ **ROQ = θ = 60°

For arc MYN,

radius (r_{1}) = OM = 21 cm

m(arc MYN) = **∠ **MON = θ = 60°

Length of arc RXQ = \(\frac{θ}{360}\) × 2πr = \(\frac{60}{360}\) × 2 × \(\frac{22}{7}\) × 7 = \(\frac{44}{6}\) = 7.3 cm

Length of arc MYN = \(\frac{θ}{360}\) × 2πr_{1 }= \(\frac{60}{360}\) × 2 × \(\frac{22}{7}\) × 21 = 22 cm

**Answer is : ****Length of arc RXQ is 7.3 cm and length of arc MYN is 22 cm.**

**Question 3.9. In figure****, ****if ****A(P−ABC) = 154 ****cm**^{2}**radius of the circle is ****14 ****cm, find ****(1) ****∠** **APC. (2) ***l***(****arc ****ABC)** .

The radius (r) of the circle = 14 cm.

A(P−ABC) = 154 cm^{2}.

Let m(arc ABC) = APC = θ

A(P−ABC) = \(\frac{θ}{360}\) × πr^{2}

∴ 154 = \(\frac{θ}{360}\) × \(\frac{22}{7}\) × (14)^{2}

∴ θ = \(\frac{154×360×7}{22×14×14}\) = 90° ∴ ∠ APC = 90°

Area of the sector = \(\frac{\text{length of the arc × radius}}{2}\)

∴ A(P−ABC) = \(\frac{\text{l(arc ABC) × r}}{2}\)

∴ 154 **= **\(\frac{\text{l(arc ABC) × 14}}{2}\)

∴ *l*(arc ABC) = \(\frac{154}{7}\) = 22 cm

**Answer is : ****∠** **APC = 90° and ***l***(arc ABC) = 22 cm.**

**Question 3.10. 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.**

(i) The radius (r) of a sector of a circle =7 cm.

Measure of an arc of the sector (**θ**) = 30°

Area of the sector = \(\frac{θ}{360}\) × πr^{2} = \(\frac{30}{360}\) × \(\frac{22}{7}\) × (7)^{2} = 12.83 cm^{2}

(ii) The radius (r) of a sector of a circle = 7 cm.

Measure of an arc of the sector (**θ**) = 210°

Area of the sector = \(\frac{θ}{360}\) × πr^{2} = \(\frac{210}{360}\) × \(\frac{22}{7}\) × (7)^{2} = = 89.83 cm^{2}

(iii) The radius (r) of a sector of a circle = 7 cm.

Measure of an arc of the sector (**θ**) = 3 right angles = 3 × 90° = 270°

Area of the sector = \(\frac{θ}{360}\) × πr^{2} = \(\frac{270}{360}\) × \(\frac{22}{7}\) × (7)^{2} = \(\frac{3×22×7}{4}\) = 115.5 cm^{2}

**Answer is : **

**(1) Area of the sector is 12.83 cm ^{2}, when measure of the arc is 30°. **

**(2) Area of the sector is 89.83 cm ^{2}, when measure of the arc is 210°. **

**(3) Area of the sector is 115.5 cm ^{2}, when measure of the arc is equal to three right angles.**

**Question 3.11. ****The ****area of a minor sector of a circle is ****3.85 ****cm**^{2}**and the measure of its central angle is ****36****°. Find the radius of the circle****.**

Area of a minor sector of a circle is 3.85 cm^{2}

Measure of its central angle (θ) = 36°.

Let its radius be r.

Area of the sector = \(\frac{θ}{360}\) × πr^{2}

∴ 3.85 = \(\frac{36}{360}\) × \(\frac{22}{7}\) × r^{2}

∴ r^{2} = \(\frac{38.5×7}{22}\) = 12.25

∴ r = 3.5 cm

**Answer is : The radius of the circle is 3.5 cm.**

**Question 3.12. In figure****, ****□** **PQRS ****is a rectangle. If ****PQ = 14 ****cm, ****QR = 21 ****cm, find the areas of the parts x**

**,**

*y***and**

*z***.**

**□ **PQRS is a rectangle.

PQ = 14 cm and QR = 21 cm ... (Given)

Consider points M and N as shown.

Parts x and y are sectors.

For part x,

radius (r) = PQ = 14 cm

m(arc PT) = **∠ **PQT = θ = 90°

Area of part x = \(\frac{θ}{360}\) × πr^{2} = \(\frac{90}{360}\) × \(\frac{22}{7}\) × 14^{2} = 154 cm^{2}

QT = PQ = 14 cm ... (Radii of the same arc)

QT + TR = QR ... (Q−T−R)

∴ 14 + TR = 21

∴ TR = 21 − 14

∴ TR = 7 cm

**For part y,**

radius (r_{1}) = TR = 7 cm

m(arc TU) = **∠ **TRU = θ_{1} = 90°

Area of part y = \(\frac{θ_1}{360}\) × πr_{1}^{2} = \(\frac{90}{360}\) × \(\frac{22}{7}\) × 7^{2} = 38.5 cm^{2}

Area of rectangle = length x breadth

A(**□ **PQRS) = QR x PQ

∴ A(□ PQRS) = 21 x 14

∴ A(□ PQRS) = 294 cm^{2}

Area of part z,

= A(□ PQRS) − Area of part x − Area of part y

= 294 – 154 − 38.5

= 101.5 cm^{2}

**Answer is : ****Area of part x is 154 cm ^{2}, area of part y is 38.5 cm2 and area of part z is 101.5 cm^{2}.**

**Question 3.13. ****Δ** **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 (****Δ** **LMN)**

**(2) ****Area of any one of the sectors****.**

**(3) ****Total area of all the three sectors.**

**(4) ****Area of the shaded region.**

Δ LMN is equilateral.

Consider points X, Y and Z as shown.

(1) Side of equilateral triangle = LM = 14 cm.

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) x side^{2}

∴ A(Δ LMN)= \(\frac{\sqrt{3}}{4}\) x 14^{2} = \(\frac{\sqrt{3}}{4}\) x 196

∴ A(Δ LMN) = 1.732 × 49 …(as \(\sqrt{3}\) = 1.732)

∴ A(Δ LMN) = 84.868

∴ A(Δ LMN) ~ 84.87 cm^{2}

(2) For a sector L−XZ,

radius (r) = 7 cm

Angular measure of arc (θ) = m(arc XZ) = **∠ **XLZ = 60° ……(Angle of equilateral triangle)

Area of the sector L−XZ = \(\frac{θ}{360}\) × πr^{2} = \(\frac{60}{360}\) × \(\frac{22}{7}\) × 7^{2} = 25.67 cm^{2}

(3) Total area of 3 sectors =3 x Area of the sector L−XZ = 3 × 25.67 = 77.01 cm^{2}

(4) Area of the shaded portion = A(Δ LMN) − Total area of 3 sectors

= 84.87 − 77.01 = 7.86 cm^{2}

**Answer is : **** (1) A(**Δ** LMN) = 84.87 cm ^{2},**

**(2) Area of one of the sectors is 25.67 cm ^{2},**

**(3) Total area of 3 sectors is 77.01 cm ^{2},**

**(4) Area of the shaded portion is 7.86 cm ^{2}.**

**Practice set 7.4**

**Question 4.1. In figure****, A ****is the centre of the circle. ****∠** **ABC = 45****° and ****AC = ****7** ** ****cm. Find the area of segment ****BXC.**

As per data given in question, fig is as below.

The radius (r) of the circle = AC = 7 cm.

In Δ ABC,

seg AB ≅ seg AC …(Radii of the same circle)

∴ **∠ **ABC ≅ **∠ **ACB ... (Isosceles triangle theorem)

**∠ **ABC = **∠ **ACB = 45°

In Δ ABC,

**∠ **ABC + **∠ **ACB + **∠ **BAC =180° ... (Sum of all angles of a triangle is 180°)

∴ 45° + 45° + **∠**BAC = 180°

∴ 90° + **∠ **BAC = 180°

∴ **∠ **BAC =180° − 90°

∴ **∠ **BAC = 90°

m(arc BXC) = **∠ **BAC = θ = 90° ... (Definition of measure of minor arc)

A(segment BXC) = r^{2 }\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

= \((7\sqrt{2})^2[\frac{3.14×90}{360}-\frac{sin\,90}{2}]\)

= 98 × \([\frac{3.14}{4}-\frac{1}{2}]\) …. ( sin 90° = 1)

= 98 × \(\frac{1.14}{4}\) = 27.93 cm^{2}

**Answer is : **** ****A(segment BXC) = ****27.93 cm ^{2}**

[Note : In the question, if **∠ **ABC =45°, then the area of segment BXC is 27.93 cm^{2} and not 3.72 cm^{2} as given in the textbook but if we consider **∠ **BAC = 45° instead of **∠ **ABC = 45° then area of segment BXC will be 3.92 cm^{2} which is near to 3.72 cm^{2} as given in the textbook, if **∠** BAC = 45° then solution would be as follows.]

The radius (r) of the circle = AC = 7 cm.

m(arc BXC) = **∠ **BAC = θ = 45°

A(segment BXC) = r^{2}\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

= \((7\sqrt{2})^2[\frac{3.14×45}{360}-\frac{sin\,45}{2}]\)

= 98 × \([\frac{3.14}{8}-\frac{1}{2\sqrt{2}}]\) …. ( sin 45° = \(\frac{1}{\sqrt{2}}\) )

= 98 × \([\frac{0.32}{8}]\) = 3.92 cm^{2}

**Answer is : **** ****A(segment BXC) = 3****.92 cm ^{2}**

**Question 4.2. In the 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, = 1.73)**

The radius (r) of the circle = OP =10 cm.

m(arc PQR) = θ = 60°.

The shaded region is segment PQR,

A (segment PQR) = r^{2}\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

= 10^{2}\([\frac{3.14×60}{360}-\frac{sin\,60}{2}]\)

= 10^{2}\([\frac{3.14}{6}-\frac{\sqrt{3}}{2×2}]\) …. ( sin 60° = \(\frac{\sqrt{3}}{2}\))

= 100 × \(\frac{1.09}{12}\) … …(as \(\sqrt{3}\) = 1.732)

= \(\frac{109}{12}\) = 9.08 cm^{2}

**Answer is : ****Area of the shaded region is 9.08 cm ^{2}.**

**Question 4.3. In the figure, if ****A ****is the centre of the circle. ****∠** **PAR = 30****°, ****AP = 7.5, ****find the area of the segment ****PQR (****π ****= 3.14)**

The radius (r) of the circle = AP =7.5.

m(arc PQR) = **∠ **PAR = θ =30°.

Area of the segment PQR = r^{2}\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\) = 7.5^{2}\([\frac{3.14×30}{360}-\frac{sin\,30}{2}]\)

**= **7.5^{2}\([\frac{3.14}{12}-\frac{1}{2×2}]\) …. ( sin 30° = \(\frac{1}{2}\))

**= **7.5^{2} \([\frac{0.14}{12}]\) = \(\frac{7.785}{12}\) = 0.65625 sq units

**Answer is : Area of segment PQR is 0.65625 sq units.**

**Question 4.4. In the figure ****7.46, ****if ****O ****is the centre of the circle, ****PQ ****is a chord. ****∠** **POQ = 90****°, area of shaded region is ****114 ****cm**^{2}**, find the radius of the circle. ****(****π** **= 3.14)**

Area of the shaded region is 114 cm2.

The shaded region is segment PRQ.

A (segment PRQ) = 114 cm^{2}

m (arc PRQ) = **∠ **POQ = θ = 90°

A (segment PRQ) = r^{2}\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\)

∴ 114 = r^{2 }\([\frac{3.14×90}{360}-\frac{sin\,90}{2}]\)

∴ 114 = r^{2 }\([\frac{3.14}{4}-\frac{1}{2}]\)

∴ 114 = r^{2 }\([\frac{1.14}{4}]\)

∴ r^{2} = \(\frac{114×4}{1.14}\) = 400

∴ r = 20 cm

**Answer is : The radius of the circle is 20 cm.**

**Question 4.5. 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, ** ** ****= 1.73)**

Let the centre of the circle be O.

Consider points M and N as shown. The radius of

the circle (r) = 15 cm.

m(arc PMQ) = **∠ **POQ = θ = 60°

Area of the minor segment = r^{2}\([\frac{πθ}{360}-\frac{sin\,θ}{2}]\) = 15^{2 }\([\frac{3.14×60}{360}-\frac{sin\,60}{2}]\)

= 225\([\frac{3.14}{6}-\frac{\sqrt{3}}{2×2}]\) = 225\([\frac{3.14}{6}-\frac{1.73}{4}]\) = 225\([\frac{1.09}{12}]\) = 20.43 cm^{2}

A(circle) = πr^{2}

= 3.14 × 15 × 15

= 3.14 × 225

= 706.5 cm^{2}

A(major segment) = A(circle) − A(minor segment)

= 706.5 − 20.43

= 686.07 cm^{2}

**Answer is : A(minor segment) is 20.43 cm ^{2} and A (major segment) is 686.07 cm^{2}.**

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