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 cm2.
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?
(iii) (b) What is the length of arc CD?
(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 cm2
(ii) Area of Δ AOB:
Line OA = OD + AD = 7 + 3 = 10 cm
Line OB = OC + CB = 7 + 3 = 10 cm
Since ∠ OAB = 900
Area of Δ AOB = x OA x OB = x 10 x 10
- Area of Δ AOB = 50 cm2
(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 cm2
Since, Cost of silver plating is Rs. 20/cm2
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