**Chapter 7 Mensuration Set 7.3**

## Chapter 7 Mensuration Set 7.3

**Question 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)Given : Radius (r) = 10 cm,Measure of the arc (θ) = 54°To find : Area of the sector.**

= 3 × 3.14 × 5

= 15 × 3.14

= 47.1 cm

^{2}

∴ The area of the sector is 47.1 cm

^{2}.

**Question 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)Given: Radius (r) = 18 cm,Measure of the arc (θ) = 80°To find: Length of the arc.**

= 2 × 2 × 3.14 × 2 = 25.12 cm

∴ The length of the arc is 25.12 cm.

**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.Solution:**

Given: Radius (r) = 3.5 cm,

length of arc (l) = 2.2 cm

To find: Area of the sector.

Solution:

= 1.1 × 3.5 = 3.85 cm

^{2}

∴ The area of the sector is 3.85 cm

^{2}.

**Question 4.Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm ^{2}. Find the area of its corresponding major sector, (π = 3.14)Given: Radius (r) = 10 cm,area of minor sector =100 cm^{2}To find: Area of maj or sector.Solution:**

Area of circle = πr

^{2}

= 3.14 × (10)

^{2}

= 3.14 × 100 = 314 cm

^{2}

Now, area of major sector

= area of circle – area of minor sector

= 314 – 100

= 214 cm

^{2}

∴ The area of the corresponding major sector is 214 cm

^{2}.

**Question 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.Given: Radius (r) =15 cm,area of sector = 30 cm^{2}To find: Length of the arc (l).Solution:**

∴ The length of the arc is 4 cm.

**Question 6.In the adjoining figure, radius of the circle is 7 cm and m (arc MBN) = 60°, findi. Area of the circle.ii. A(O-MBN).iii. A(O-MCN).Given: radius (r) = 7 cm,m(arc MBN) = θ = 60°Solution:**

i. Area of circle = πr

^{2}

= 22/7 × (7)

^{2}

= 22 × 7

= 154 cm

^{2}

∴ The area of the circle is 154 cm

^{2}

= 25.67 cm^{2}

= 25.7 cm^{2}

∴ A(O-MBN) = 25.7 cm^{2}

iii. Area of major sector = area of circle – area of minor sector

∴ A(O-MCN) = Area of circle – A(O-MBN)

= 154 – 25.7

∴ A(O-MCN) = 128.3 cm^{2}

**Question 7.In the adjoining figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC).Given: Radius (r) = 3.4 cm,perimeter of sector 12.8 cmTo find: A(P-ABC)Solution:**
Perimeter of sector

= Iength of arc ABC + AP + CP

∴ 12.8 = l + 3.4 + 3.4

∴ 12.8 = l + 6.8

∴ l = 12.8 – 6.8 = 6cm

∴ A(P-ABC) = 10.2 cm

^{2}

Question 8.

In the adjoining 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 (π = 22/7)

Given: ∠ROQ = ∠MON = 60°,

radius (r) = OR = 7 cm, radius (R) = OM = 21 cm

To find: Lengths of arc RXQ and arc MYN.

∴ The lengths of arc RXQ and arc MYN are 7.33 cm and 22 cm respectively.

**Question 9.In the adjoining figure, if A(P-ABC) = 154 cm ^{2}, radius of the circle is 14 cm, findi. ∠APC,ii. l(arc ABC).Given: A(P – ABC) = 154 cm^{2},radius (r) = 14 cmSolution:**

i. Let ∠APC = θ

**Question 10.Radius of a sector of a circle is 7 cm. If measure of arc of the sector isi. 30°ii. 210°iii. three right angles, find the area of the sector in each case.Given: Radius (r) = 7 cmTo find: Area of the sector.Solution:**

i. Measure of the arc (θ) = 30°

∴ Area of the sector is 12.83 cm

^{2}.

ii. Measure of the arc (θ) = 210°

∴ Area of the sector is 89.83 cm

^{2}.

iii. Measure of the arc (θ) = 3 right angle

= 3 × 90° = 270°

∴ Area of the sector is 115.50 cm

^{2}.

**Question 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.Given: Area of minor sector = 3.85 cm^{2},central angle (θ) = 36°To find: Radius of the circle (r).Solution:**

Area of minor sector = θ360×πr2

∴ The radius of the circle ¡s 3.5 cm.

**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.Given: In rectangle PQRS,PQ = 14 cm, QR = 21 cmTo find: Areas of the parts x, y and z.Solution:**

∠Q = ∠R = θ = 90° …[Angles of a rectangle]

For the sector (Q – PA),

PQ = QA …[Radii of the same circle]

∴ QA = 14 cm

Now, QR = QA + AR … [Q – A – R]

∴ 21 = 14 + AR

∴ AR = 7 cm

Area of rectangle = length × breadth

area of ꠸PQRS = PQ × QR

= 14 × 21

= 294 cm

^{2}

Area of part z = area of ꠸PQRS

– area of part x – area of part y

= 294 – 154 – 38.5

= 101.5 cm

^{2}

∴ The area of part x is 154 cm

^{2}, the area of part y is 38,5 cm

^{2}and the area of part z is 101.5 cm

^{2}.

**Question 13.∆ALMN is an equilat triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm. Find,i. A (∆ LMN).ii. Area of any one of the sectors.iii. Total area of all the three sectors.iv. Area of the shaded region. (√3 = 1.732 )Given: In equilateral triangle LMN, LM =14 cm,radius of sectors (r) = 7 cmSolution:**

i. ∆LMN is an equilateral triangle.

ii. Central angle (θ) = 60° …[Angle of an equilateral triangle]

∴ Area of one sector = 25.67 cm

^{2}

iii. Total area of all three sectors

= 3 × Area of one sector

= 3 × 25.67

= 77.01 cm

^{2}

∴ Total area of all three sectors = 77.01 cm

^{2}

iv. Area of shaded region

= A(∆LMN) – total area of all three sectors

= 84.87 – 77.01

= 7.86 cm

^{2}

∴ Area of shaded region = 7.86 cm

**Intext Questions and Activities**

**Question 1.Complete the following table with the help of given figure. (Textbook pg. no. 149)Solution:**

**Question 2.Observe the figures below. Radii of all circles are equal. Observe the areas of the shaded regions and complete the following table. (Textbook pg. no. 150)Solution:**

Thus, if measure of an arc of a circle is θ, then

**Question 3.In the following figures, radii of all circles are equal. Observe the length of arc in each figure and complete the table. (Textbook pg. no. 151)Solution:**