﻿ Class 10 NCERT Math Solution
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TOPICS
Unit-12(Examples)

Example-1 :-  The cost of fencing a circular field at the rate of ₹ 24 per metre is ₹ 5280. The field is to be ploughed at the rate of ₹ 0.50 per m². Find the cost of ploughing the field (Take π = 22/7).

Solution :-
```  Length of the fence (in metres) = total cost/rate = 5280/24 = 220
So, circumference of the field = 220m
Therefore, if r meteres is the radius of the field, then
2πr = 220
2 x 22/7 x r = 220
44/7 x r = 220
r = 1540/44
r = 35m
Radius of the field = 35m
Therefore, Area of the field = πr² = 22/7 x 35 x 35 = 22 x 5 x 35 = 3850 m²
Now, cost of ploughing 1 m² of the field = ₹ 0.50
So, total cost of ploughing the field = 3850 × 0.50 = ₹ 1925
```

Example-2 :-  Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14). Solution :-
```  Given sector is OAPB
Area of the sector = θ/360 x πr²
= 30/360 x 3.14 x 4 x 4
= 12.56/3 cm²
= 4.19 cm²(approx.)

Area of the corresponding major sector
= πr² - Area of the sector OAPB
= 3.14 x 4 x 4 - 4.19
= 3.14 x 16 - 4.19
= 50.24 - 4.19
= 46.05 cm²
= 46.1 cm²(approx.)
```

Example-3 :-  Find the area of the segment AYB shown in Figure, if radius of the circle is 21 cm and ∠ AOB = 120°. (Use π = 22/7) Solution :-
```  Area of the segment AYB = Area of sector OAYB – Area of Δ OAB ......(i) Now, area of the sector OAYB = θ/360 x πr²
= 120/360 x 22/7 x 21 x 21
= 462 cm² .....(ii)

For finding the area of Δ OAB, draw OM ⊥ AB
Note that OA = OB.
Therefore, by RHS congruence, Δ AMO ≅ Δ BMO.
So, M is the mid-point of AB and ∠ AOM = ∠ BOM = 1/2 x 120 = 60°
Let OM = x cm
So, from Δ OMA, OM/OA = cos 60°
x/21 = 1/2
x = 21/2
So, OM = 21/2 cm
Also, AM/OA = sin 60°
AM/OA = √3/2
So, AM = 21√3/2 cm
Therefore, AB = 2AM = (2 x 21√3)/2 cm = 21√3 cm
So, area of Δ OAB = 1/2 x AB x OM = 1/2 x 21√3 x 21/2 = 441√3/4 cm² .....(iii)
Therefore, area of the segment AYB = (464 - 441/4) [from (i), (ii) & (iii)]
= 21(88 - 21√3)/4 cm²
```

Example-4 :-  In Figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds. Solution :-
```  Total area = Area of sector OAB + Area of sector ODC + Area of Δ OAD + Area of Δ OBC
= (90/360 x 22/7 x 28 x 56 + 90/360 x 22/7 x 28 x 56 + 1/4 x 56 x 56 + 1/4 x 56 x 56)
= 1/4 x 28 x 56 (22/7 + 22/7 + 2 + 2)m²
= (7 x 56)/7 [22 + 22 + 14 + 14]m²
= 56 x 72
= 4032 m²
```

Example-5 :-  Find the area of the shaded region in Figure, where ABCD is a square of side 14 cm. Solution :-
```  Area of square ABCD = 14 × 14 cm² = 196 cm²
Diameter of each circle = 14/2cm = 7cm
So, radius of each circle = 7/2 cm
So, area of one circle = πr²
= 22/7 x 7/2 x 7/2 cm²
= 154/4 cm²
= 77/2 cm²
Therefore, area of the four circles = 4 x 77/2 cm² = 154 cm²
Hence, area of the shaded region = (196 – 154) cm² = 42 cm².
```

Example-6 :-  Find the area of the shaded design in Figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (Use π = 3.14) Solution :-
```  Let us mark the four unshaded regions as I, II, III and IV.
Area of I + Area of III = Area of ABCD – Areas of two semicircles of each of radius 5 cm
= (10 x 10 - 2 x 1/2 x π x 5²)
= (100 - 3.14 x 25)
= 100 - 78.5 = 21.5 cm²

Similarly, Area of II + Area of IV = 21.5 cm²
So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) cm²
= (100 – 43) cm² = 57 cm²
```
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