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 cm2
∴ The area of the sector is 47.1 cm2.
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 cm2
∴ The area of the sector is 3.85 cm2.
Question 4.
Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm2. Find the area of its corresponding major sector, (π = 3.14)
Given: Radius (r) = 10 cm,
area of minor sector =100 cm2
To find: Area of maj or sector.
Solution:
Area of circle = πr2
= 3.14 × (10)2
= 3.14 × 100 = 314 cm2
Now, area of major sector
= area of circle – area of minor sector
= 314 – 100
= 214 cm2
∴ The area of the corresponding major sector is 214 cm2.
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.
Given: Radius (r) =15 cm,
area of sector = 30 cm2
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°, find
i. 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 = πr2
= 22/7 × (7)2
= 22 × 7
= 154 cm2
∴ The area of the circle is 154 cm2
= 25.67 cm2
= 25.7 cm2
∴ A(O-MBN) = 25.7 cm2
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 cm2
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 cm
To 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 cm2
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 cm2, radius of the circle is 14 cm, find
i. ∠APC,
ii. l(arc ABC).
Given: A(P – ABC) = 154 cm2,
radius (r) = 14 cm
Solution:
i. Let ∠APC = θ
Question 10.
Radius of a sector of a circle is 7 cm. If measure of arc of the sector is
i. 30°
ii. 210°
iii. three right angles, find the area of the sector in each case.
Given: Radius (r) = 7 cm
To find: Area of the sector.
Solution:
i. Measure of the arc (θ) = 30°
∴ Area of the sector is 12.83 cm2.
ii. Measure of the arc (θ) = 210°
∴ Area of the sector is 89.83 cm2.
iii. Measure of the arc (θ) = 3 right angle
= 3 × 90° = 270°
∴ Area of the sector is 115.50 cm2.
Question 11.
The area of a minor sector of a circle is 3.85 cm2 and the measure of its central angle is 36°. Find the radius of the circle.
Given: Area of minor sector = 3.85 cm2,
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 cm
To 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 cm2
Area of part z = area of ꠸PQRS
– area of part x – area of part y
= 294 – 154 – 38.5
= 101.5 cm2
∴ The area of part x is 154 cm2, the area of part y is 38,5 cm2 and the area of part z is 101.5 cm2.
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 cm
Solution:
i. ∆LMN is an equilateral triangle.
ii. Central angle (θ) = 60° …[Angle of an equilateral triangle]
∴ Area of one sector = 25.67 cm2
iii. Total area of all three sectors
= 3 × Area of one sector
= 3 × 25.67
= 77.01 cm2
∴ Total area of all three sectors = 77.01 cm2
iv. Area of shaded region
= A(∆LMN) – total area of all three sectors
= 84.87 – 77.01
= 7.86 cm2
∴ 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
![](https://mhboardsolutions.xyz/wp-content/uploads/2021/12/word-image-8305-31.png)
![Maharashtra Board Class 10 Maths Solutions Chapter 7 Mensuration Practice Set 7.3 24](https://mhboardsolutions.xyz/wp-content/uploads/2021/12/maharashtra-board-class-10-maths-solutions-chapter-1845.png)
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: