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Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.
After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are $14$ units and $7$ units, respectively. There are two quadrants of radius $2$ units on one side for special seats.
Based on the above information, answer the following questions :
(i) What is the total perimeter of the parking area?
(ii) (a) What is the total area of parking and the two quadrants ?
OR
(b) What is the ratio of area of playground to the area of parking area ?
(iii) Find the cost of fencing the playground and parking area at the rate of ₹$2$ per unit.
After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are $14$ units and $7$ units, respectively. There are two quadrants of radius $2$ units on one side for special seats.
Based on the above information, answer the following questions :
(i) What is the total perimeter of the parking area?
(ii) (a) What is the total area of parking and the two quadrants ?
OR
(b) What is the ratio of area of playground to the area of parking area ?
(iii) Find the cost of fencing the playground and parking area at the rate of ₹$2$ per unit.
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Sol. (i) Total perimeter = $\pi r + 2r$
$= \frac{22}{7} \times \frac{7}{2} + 7 = 18$ units
(ii) (a) Area of parking $= \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{4}$
Area of quadrants = $2 . \frac{1}{4} \pi r^2 = 2 \times \frac{1}{4} \times \frac{22}{7} \times 2 \times 2 = \frac{44}{7}$
Total Area = $\frac{77}{4} + \frac{44}{7} = \frac{715}{28}$ or $25.54$ sq. units
OR
(ii) (b) $\frac{\text{Area of playground}}{\text{Area of parking}} = \frac{98}{77/4} = \frac{56}{11} = 56:11$
(iii) Required Perimeter = $2(l + b) + \frac{2\pi r}{2}$
$= 2(14 + 7) + \frac{22}{7} \times \frac{7}{2} = 53$ units
Cost of fencing = $53 \times 2 = \text{Rs} 106$
$= \frac{22}{7} \times \frac{7}{2} + 7 = 18$ units
(ii) (a) Area of parking $= \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{4}$
Area of quadrants = $2 . \frac{1}{4} \pi r^2 = 2 \times \frac{1}{4} \times \frac{22}{7} \times 2 \times 2 = \frac{44}{7}$
Total Area = $\frac{77}{4} + \frac{44}{7} = \frac{715}{28}$ or $25.54$ sq. units
OR
(ii) (b) $\frac{\text{Area of playground}}{\text{Area of parking}} = \frac{98}{77/4} = \frac{56}{11} = 56:11$
(iii) Required Perimeter = $2(l + b) + \frac{2\pi r}{2}$
$= 2(14 + 7) + \frac{22}{7} \times \frac{7}{2} = 53$ units
Cost of fencing = $53 \times 2 = \text{Rs} 106$