TOPICS

Unit-12(Examples)

Areas Related To Circle

**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).

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).

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)

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.

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.

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)

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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