Practice Set 7.3 Mensuration Class 10th Mathematics Part 2 MHB Solution

Practice Set 7.3

1. Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of…
2. Measure of an arc of a circle is 80 cm and its radius is 18 cm. Find the length of the…
3. Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area…
4. Radius of a circle is 10 cm. Area of a sector of the sector is 100 cm^2 . Find the area…
5. Area of a sector of a circle of radius 15 cm is 30 cm^2 . Find the length of the arc of…
6. In the figure 7.31, radius of the circle is 7 cm and m(arc MBN) = 60°,find (1) Area of…
7. In figure 7.32, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm.…
8. In figure 7.33 O is the centre of the sector.∠ ROQ = ∠ MON = 60°. OR = 7 cm, and OM =…
9. In figure 7.34, if A(P-ABC) = 154 cm^2 radius of the circle is 14cm, find (1) ∠ APC.…
10. Radius of a sector of a circle is 7 cm. If measure of arc of the sector is - (l) 30°…
11. The area of a minor sector of a circle is 3.85 cm^2 and the measure of its central…
12. In figure 7.35, □PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the…
13. ∆ LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are…

Practice 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)

Answer:

Radius of circle, r = 10 cm

Angle made between the arc ,θ = 54°

⇒ Area of sector = 47.1 sq. cm

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:

Measure of an arc of circle, θ = 80°

Radius of circle, r = 18 cm

⇒ Area of sector = 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.

Answer:

Radius of circle, r = 3.5 cm

Length of arc, l = 2.2 cm

As we know,

Also,

⇒ Area of sector = 3.85 sq. cm

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 circle, r = 10 cm

Area of sector (minor sector) = 100 sq. cm

Area of circle, AC = πr2

⇒ AC = 3.14× 102

⇒ AC = 314 sq. cm

Area of major sector, AM = area of circle – area of minor sector

⇒ AM= AC – 100

⇒ AM = 314 -100 = 214 sq. cm

∴ area of major sector is 214 sq. cm

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 circle, r = 15 cm

Also,

On substituting the values, we get,

⇒ Length of arc = 4 cm

Question 6.

In the figure 7.31, radius of the circle is 7 cm and m(arc MBN) = 60°,find



(1) Area of circle

(2) A(O – MBN)

(3) A(O - MCN)

Answer:

(1) Radius of circle, r = 7cm

Area of circle,AC = πr2

⇒ AC = 154 sq. cm

∴ area of circle is 154 sq.cm

(2) Angle subtended by the arc = 60°

As we know,

⇒ A(O- MBN) = 25.7 sq. cm

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

⇒ Area(O – MCN) = AC – 25.7

⇒ Area(O – MCN) = 154 – 25.7

∴ Area of major sector is 128.3 sq. cm

Question 7.

In figure 7.32, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A (P-ABC).

Answer:

Radius of circle, r = 3.4 cm

Perimeter of sector, P = 12.8

⇒ P = length of arc + 2× radius

⇒ Length of arc, l = P – 2×r

⇒ l = 12.8 – 2(3.4)

⇒ l = 6 cm

Let the ∠ APC be θ

As we know that,

Also ,

On Substituting the value of theta from above equation,

⇒ A = 10.2 sq. cm

∴ Area of Sector is 10.2 sq. cm

Question 8.

In figure 7.33 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. 

Answer:

Let the ∠ ROQ= ∠ MON = θ = 60°

As we know that,

⇒ Length(RXQ) = 7.6 cm

Similarly, 

⇒ Length (MYN) = 22 cm

Question 9.

In figure 7.34, if A(P-ABC) = 154 cm2radius of the circle is 14cm, find

(1) ∠ APC.

(2) l (arc ABC).

Answer:

As we know that,

(1) Let the ∠APC be θ

Radius of circle, r =14 cm

Area of sector, A = 154 cm2

⇒ θ = 90°

(2) Since the angle formed is 90° , which is one–fourth of the perimeter of circle

⇒ l = 22 cm

Question 10.

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

(l) 30°

(ll) 210°

(lll) three right angles

Find the area of sector in each case.

Answer:

Radius of circle, r = 7cm

(l) Angle subtended by arc, θ = 30°

As we know,

⇒ Al = 12. 83 sq. cm

(ll) Angle subtended by arc, θ = 210°

Similarly,

⇒ All = 89.83 sq. cm

(lll) Angle subtended by arc, θ = (3× 90)° = 270°

Similarly,

⇒ Alll = 115.5 sq. cm

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.

Answer:

As we know that,

Given area of sector, A = 3.85 sq.cm

Radius of circle= r

Central angle, θ = 36°

On substituting the values, we get,

⇒ r = 3.5 cm

∴ Radius of circle is 3.5 cm

Question 12.

In figure 7.35, □PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Answer:

Since part x is a sector of a circle with radius, r = 14 cm and the central angle is 90°, so the area of x will be equal to one-fourth of the area of circle with PQ as radius.

Area of circle with PQ as radius = π (PQ)2

⇒ x = 154 sq. cm

Similarly, area y is also equal to one-fourth od area of circle with radius, r = QR – PQ

⇒ r = 21 – 14 = 7 cm

Area of circle with r as radius = π (r)2

⇒ y = 38.5 sq. cm

Also,

z = Area of rectangle(PQRS) – x – y

Area of rectangle = PQ× QR

⇒ Area of rectangle = 14× 21 = 294 sq. cm

⇒ z = 294 – 154 – 38.5

⇒ z = 101.5 sq.cm

∴ the area of x, y and z are 154 sq.cm, 38.5 sq.cm and 101.5 sq.cm respectively

Question 13.

∆ LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centre and radius7 cm. Find,



(1) A (Δ LMN)

(2) Area of any one of the sectors

(3) Total area of all three sectors

(4) Area of shaded region

Answer:

(1) Side of triangle = LM = a = 14 cm

Since Δ LMN is an equilateral triangle, so the area of the triangle is given by:

⇒ AT = 84.87 sq.cm

(2) Angle subtended by the corner = θ = 60°

As we know,

Here ,

⇒ AS = 25.67 sq. cm

(3) Total area of all sector, ATS = 3× AS

⇒ ATS = 3× 25.67

⇒ ATS = 77.01 sq.cm

(4) Area of shaded region, AR = Area of triangle – Area of all three sectors

⇒ AS = AT - ATS

⇒ AS = 84.87 – 77.01

⇒ AS = 7.86 sq. cm

∴ area of shaded region is 7.86 sq. cm