Question

In an annual day function of a school, the organizers wanted to give a cash prize along with a memento to their best students. Each memento is made as shown in the given figure and its base ABCD is shown from the front side. The rate of silver plating is ₹ 20 per \(\mathrm{cm}^{2}\).

Based on the above information, answer the following questions.

1. What is the area of the quadrant ODCO?
2. Find the area of the \(\triangle AOB\).
3. What is the total cost of silver plating the shaded part ABCD?
4. What is the length of \(\operatorname{arc} CD\)?

Answer 1

i. Area of quadrant ODCO

Radius r = 7 cm.

\[\text{Area} = \frac{1}{4}\pi r^2\]

\[= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7\]

\[= 38.5\ \text{cm}^2\]

ii. Area of \(\triangle AOB\)

Given AO = OB = 10 cm, right-angled at O.

\[ \text{Area} = \frac{1}{2} \times AO \times OB\]

\[= \frac{1}{2} \times 10 \times 10\]

\[= 50\ \text{cm}^2\]

iii. Cost of silver plating the shaded part ABCD

Shaded area \(ABCD = \text{Area of } \triangle AOB - \text{Area of quadrant } ODCO\)

\[= 50 - 38.5\]

\[= 11.5\ \text{cm}^2\] 

\[\text{Cost} = 11.5 \times 20\]

\[= ₹,230\]

iv. Length of arc CD

\[\text{Arc length} = \frac{1}{4} \times 2\pi r\]

\[= \frac{1}{4} \times 2 \times \frac{22}{7} \times 7\]

\[= 11\ \text{cm}\]