Class 10th, Area Related Circle. Exe-12.2 (Solution3)

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Maths NCERT Class 10 Chapter 12 – Areas Related to Circles

 Class 10 Maths Chapter 12 Exercise: 12.2

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11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

Given,

Radius (r) = 25 cm

Sector angle (θ) = 115°

Since there are 2 blades,

The total area of the sector made by wiper = 2 × (θ/360°) × π r2

=2 × 115/360 × 22/7 × 252

= 2 × 158125/252 cm2

= 158125/126 = 1254.96 cm2

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use π = 3.14)

Solution:

Let O bet the position of Lighthouse.

Here the radius will be the distance over which light spreads.

Given, radius (r) = 16.5 km

Ssector angle (θ) = 80°

Now, the total area of the sea over which the ships are warned = Area made by the sector

Or, Area of sector = (θ/360°) × π r2

= (80°/360°) × π r2 km2

= 189.97 km2

13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2 . (Use √3 = 1.7)

Solution:

Total number of equal designs = 6

∠AOB = 360°/6 = 60°

Radius of the cover = 28 cm

Cost of making design = ₹ 0.35 per cm2

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be √3/4 × a2

Here, a = OA

∴ Area of equilateral ΔAOB = √3/4 × 282 = 333.2 cm2

Area of sector ACB = (60°/360°) × π r2 cm2

= 410.66 cm2

So, area of a single design = area of sector ACB – area of ΔAOB

= 410.66 cm2 – 333.2 cm2 = 77.46 cm2

∴ Area of 6 designs = 6 × 77.46 cm2 = 464.76 cm2

So, total cost of making design = 464.76 cm2 × Rs. 0.35 per cm2

= Rs. 162.66

14. Tick the correct solution in the following :

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) p/180 × 2πR

(B) p/180 × π R2

(C) p/360 × 2πR

(D) p/720 × 2πR2

Solution:

The area of a sector = (θ/360°) × π r2

Given, θ = p

So, area of sector = p/360 × π R2

Multiplying and dividing by 2 simultaneously,

= p/360 × 2/2 × π R2

= 2p/720 × 2πR2

So, option (D) is correct.