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Mathematics · 2023 · 4 marks
CBSE 2023 · Region 4 · Set 1 · Q38
Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will…
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 $\displaystyle 14$ units and $\displaystyle 7$ units, respectively. There are two quadrants of radius $\displaystyle 2$ 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 ₹ $\displaystyle 2$ per unit.
Marking-scheme solution
(i) Total perimeter \(\displaystyle =\pi \mathrm{r}+2 \mathrm{r}\)
\[=\frac{22}{7} \times \frac{7}{2}+7=18 \text { units }
\]
(ii) (a) Area of parking \(\displaystyle =\frac{1}{2} \pi \mathrm{r}^{2}=\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{4}\)
\[\text { Area of quadrants }=2 \cdot \frac{22}{7} \times 2 \times 2 \times \frac{1}{4}=\frac{44}{7}
\]
\[\text { Total Area }=\frac{77}{4}+\frac{44}{7}=\frac{715}{28} \text { or } 25 \cdot 54 \text { sq. units }
\]
(ii)(b)
\(\displaystyle \frac{\text { Area of playground }}{\text { Area of parking }}=\frac{98}{77 / 4}=\frac{56}{11}=56: 11\)
(iii)
Required Perimeter \(\displaystyle =2(l+b)+\frac{2 \pi r}{2}\)
\[=2(14+7)+\frac{22}{7} \times \frac{7}{2}=53 \text { units }
\]
Cost of fencing \(\displaystyle =53 \times 2=₹ 106\)
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