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Case Study - 1
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 figure and its base ABCD is shown from the front side. The rate of silver plating is ₹20 per cm$^2$.
Based on the above, answer the following questions:
(i) What is the area of the quadrant ODCO?
(ii) Find the area of $\triangle AOB$.
(iii) (a) What is the total cost of silver plating the shaded part ABCD?
OR
(iii) (b) What is the length of arc CD?
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 figure and its base ABCD is shown from the front side. The rate of silver plating is ₹20 per cm$^2$.
Based on the above, answer the following questions:
(i) What is the area of the quadrant ODCO?
(ii) Find the area of $\triangle AOB$.
(iii) (a) What is the total cost of silver plating the shaded part ABCD?
OR
(iii) (b) What is the length of arc CD?
Show SolutionHide Solution↓
(i)Area of sector ODCO = $\frac{22}{7} \times 7 \times 7 \times \frac{90}{360} = \frac{77}{2}$ or $38.5$
$\therefore$ Area of sector ODCO is $\frac{77}{2}$ or $38.5$ cm$^2$
(ii) ar ($\triangle AOB$) = $\frac{1}{2} \times 10 \times 10 = 50$
$\therefore$ ar ($\triangle AOB$) is $50$ cm$^2$
(iii) (a) Required cost = $(50 – 38.5) \times 20$
$= 230$
$\therefore$ required cost is ₹230.
OR
(iii) (b) Length of arc CD = $\frac{90}{360} \times 2 \times \frac{22}{7} \times 7$
$= 11$
$\therefore$ Length of arc CD is $11$ cm.
$\therefore$ Area of sector ODCO is $\frac{77}{2}$ or $38.5$ cm$^2$
(ii) ar ($\triangle AOB$) = $\frac{1}{2} \times 10 \times 10 = 50$
$\therefore$ ar ($\triangle AOB$) is $50$ cm$^2$
(iii) (a) Required cost = $(50 – 38.5) \times 20$
$= 230$
$\therefore$ required cost is ₹230.
OR
(iii) (b) Length of arc CD = $\frac{90}{360} \times 2 \times \frac{22}{7} \times 7$
$= 11$
$\therefore$ Length of arc CD is $11$ cm.