Circumference & Area of a Circle — Case-Study Based Questions

Case Study
Mr Ranveer lives in Agra. He purchased a rectangular plot ABCD to build a house. He leaves a rectangular area ADEF for parking and two congruent semicircular areas to make lower beds, as shown in the figure.

Based on the above information, answer the following questions:

1. Area of the plot left for parking is:
   (a) 24 m²
   (b) 48 m²
   (c) 60 m²
   (d) 63 m²

2. Diameter of each semi-circle is:
   (a) 21 m
   (b) 10.5 m
   (c) 7 m
   (d) 3.5 m

3. Total area of the two semi-circular flower beds is :
   (a) 80.6 m²
   (b) 82.625 m²
   (c) 86.625 m²
   (d) 90.625 m²

4. Area of the plot BCEF is :
   (a) 441 m²
   (b) 400 m²
   (c) 380 m²
   (d) 350 m²

5. Total length of the two semi-circular arcs is :
   (a) 33 m
   (b) 35 m
   (c) 16.5 m
   (d) 54 m

Answer

1. Area of plot left for parking is : ADEF

Area of rectangle ADEF = length × breadth

= 3 × 21 = 63 m².

Hence, option (d) is the correct option.

2. The vertical line FE = 21 m.

Two semicircles are placed one above the other and are congruent.

Diameter of each = $\dfrac{21}{2}$ = 10.5 m.

Hence, option (b) is the correct option.

3. Two semi-circles = one full circle

Diameter = 10.5

Radius = $\dfrac{10.5}{2}$ = 5.25 m

Area of circle = πr²

= $\dfrac{22}{7}$ × (5.25)²

= $\dfrac{22}{7}$ × 27.5625

= 86.625 m².

Hence, option (c) is the correct option.

4. Since,

BF = FE = EC = BC = 21 m.

Thus, BCEF is a square.

Area of square BCEF = (side)²

= (21)²

= 441 m².

Hence, option (a) is the correct option.

5. Two semicircles = one circle

∴ Total length of two semicircular arcs = Circumference of one circle = 2πr = πd.

= $\dfrac{22}{7}$ × 10.5

= 22 × 1.5

= 33 m.

Hence, option (a) is the correct option.

Case Study
Some mementos are ordered by a school for awarding their students on the occasion of Annual Day. Each memento is designed as shown in the figure, where its base ABCD is silver plated from the front side at the rate of ₹50 per cm².

Based on the above information, answer the following questions:

1. Area of △AOB is :
   (a) 50 cm²
   (b) 52 cm²
   (c) 56 cm²
   (d) 60 cm²

2. Length of the arc CD is :
   (a) 44 cm
   (b) 22 cm
   (c) 15 cm
   (d) 11 cm

3. Area of quadrant OCDO is :
   (a) 154 cm²
   (b) 77 cm²
   (c) 38.5 cm²
   (d) 30.5 cm²

4. Area of major sector formed in the figure is :
   (a) 154 cm²
   (b) 77 cm²
   (c) 100.5 cm²
   (d) 115.5 cm²

5. Total cost of silver plating is :
   (a) ₹575
   (b) ₹500
   (c) ₹450
   (d) ₹400

Answer

1. OA = OD + AD = 7 + 3 = 10 cm

OB = OC + CB = 7 + 3 = 10 cm

Area of right triangle △AOB = $\dfrac{1}{2} × OA × OB$

= $\dfrac{1}{2}$ × 10 × 10

= 50 cm².

Hence, option (a) is the correct option.

2. Arc CD subtends 90° at the centre.

∴ Arc length CD = $\dfrac{90°}{360°}$ × 2πr

= $\dfrac{1}{4} × 2 × \dfrac{22}{7}$ × 7

= 11 cm.

Hence, option (d) is the correct option.

3. Calculating,

$\Rightarrow \text{Area of quadrant OCD} = \dfrac{90°}{360°} \times πr^2 \\[1em] \Rightarrow \dfrac{1}{4} \times \dfrac{22}{7} \times 7^2 \\[1em] \Rightarrow \dfrac{11 × 7}{2} \\[1em] \Rightarrow 38.5 \text{ cm}^2.$

Hence, option (c) is the correct option.

4. Major sector angle:

360° - 90° = 270°

$\Rightarrow \text{Area of major sector} = \dfrac{270}{360} \times πr^2 \\[1em] = \dfrac{3}{4} \times \dfrac{22}{7} \times 7^2 \\[1em] = \dfrac{3}{4} \times 22 \times 7 \\[1em] = \dfrac{3 × 154}{4} \\[1em] = 115.5 \text{ cm}^2.$

Hence, option (d) is the correct option.

5. Area of plated region = Area of triangle AOB - Area of quadrant OCD

= 50 - 38.5

= 11.5 cm²

Total cost of silver plating = ₹50 × 11.5 = ₹ 575.

Hence, option (a) is the correct option.