**Q) **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 Rs. 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 **Δ** 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?

**Ans: **

**VIDEO SOLUTION**

**STEP BY STEP SOLUTION**

**(i) Area of the quadrant ODCO:**

We know that, Area of a sector of a circle =

Area of the quadrant ODCO = x 7 x7 x

= 22 x 7 x

**= 38.5 cm ^{2}**

**(ii) ****Area of ****Δ**** AOB:**

Line OA = OD + AD = 7 + 3 = 10 cm

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

Since ∠ OAB = 90^{0}

Area of Δ AOB = x OA x OB = x 10 x 10

**Area of****Δ****AOB = 50 cm**^{2}

**(iii) (a) ****Total cost of silver plating the shaded part ABCD**:

Area of shaded area ABCD = Area of Triangle ABC – Area of sector ODCO

Therefore, Area of shaded area ABCD = 50 – 38.5 = 11.5 cm^{2}

Since, Cost of silver plating is Rs. 20/cm^{2}

Hence, cost of silver plating area ABCD = 11.5 x 20 = Rs. 230

**Therefore, Hence, cost of silver plating area ABCD = Rs. 230/-**

**(iii) (b)** **Length of Arc CD:**

Length of arc is given by = 2

Length of arc CD = 2 x x 7 x

= 2 x 22 x

= 11 cm

**Therefore, ****Length of arc CD is 11 cm**