## Maharashtra State Board Class 10 Maths Solutions Chapter 7 Mensuration Practice Set 7.3

Practice Set 7.3 Geometry Class 10 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.

Solution:

Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\)

= \(\frac { 54 }{ 360 } \) × 3.14 × (10)^{2}

= \(\frac { 3 }{ 20 } \) × 3.14 × 100

= 3 × 3.14 × 5

= 15 × 3.14

= 47.1 cm^{2}

∴ The area of the sector is 47.1 cm^{2}.

Mensuration Practice Set 7.3 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.

Solution:

Length of arc = \(\frac{\theta}{360} \times 2 \pi r\) × 2πr

= \(\frac { 8 }{ 360 } \) × 2 × 3.14 × 18

= \(\frac { 2 }{ 9 } \) × 2 × 3.14 × 18

= 2 × 2 × 3.14 × 2 = 25.12 cm

∴ The length of the arc is 25.12 cm.

Practice Set 7.3 Geometry 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:

Area of sector = \(\frac{l \times \mathrm{r}}{2}\)

= \(\frac{2.2 \times 3.5}{2}\)

= 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.

Practice Set 7.3 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 = πr^{2}

= \(\frac { 22 }{ 7 } \) × (7)^{2}

= 22 × 7

= 154 cm^{2}

∴ The area of the circle is 154 cm^{2}

ii. Central angle (θ) = ∠MON = 60°

Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\)

∴ A(O – MBN) = \(\frac { 60 }{ 360 } \) × \(\frac { 22 }{ 7 } \) × (7)^{2}

\(\frac { 1 }{ 6 } \) × 22 × 7

= 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 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 cm^{2}

7.3 Class 10 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 = \(\frac { 22 }{ 7 } \))

Given: ∠ROQ = ∠MON = 60°,

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

To find: Lengths of arc RXQ and arc MYN.

Solution:

i. Length of arc RXQ = \(\frac{\theta}{360} \times 2 \pi r\)

= \(\frac { 60 }{ 2 } \) × 2 × \(\frac { 22 }{ 7 } \) × 7

= \(\frac { 1 }{ 6 } \) × 2 × 22

= 7.33 cm

ii. Length of arc MYN = \(\frac{\theta}{360} \times 2 \pi R\)

= \(\frac { 60 }{ 2 } \) × 2 × \(\frac { 22 }{ 7 } \) × 21

= \(\frac { 1 }{ 6 } \) × 2 × 22 × 3

= 22 cm

∴ 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, find

i. ∠APC,

ii. l(arc ABC).

Given: A(P – ABC) = 154 cm^{2},

radius (r) = 14 cm

Solution:

i. Let ∠APC = θ

Std 10 Geometry Mensuration 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 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}.

Mensuration Practice 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 = \(\frac{\theta}{360} \times \pi r^{2}\)

∴ The radius of the circle ¡s 3.5 cm.

10th Geometry Practice Set 7.3 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 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. (\(\sqrt { 3 }\) = 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 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

**Maharashtra Board Class 10 Maths Chapter 7 Mensuration Intext Questions and Activities**

Mensuration Practice Set 7.3 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

Area of sector (A) = \(\frac{\theta}{360}\) × Area of circle

∴ Area of sector (A) = \(\frac{\theta}{360}\) × πr^{2
}

Mensuration In Maths 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:

Thus, if the measure of an arc of a circle is 0, then

Length of arc (l) = \(\frac{\theta}{360}\) × circumference of circle

∴ Length of arc (l) = \(\frac{\theta}{360}\) × 2πr