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