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

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.