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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 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)What is the total area of parking and the two quadrants ?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.
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)
What is the total area of parking and the two quadrants ?
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 \(\displaystyle =\frac{1}{2} \pi \mathrm{r}^{2}=\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{4}\)
\[\text { Area of quadrants }=2 \cdot \frac{22}{7} \times 2 \times 2 \times \frac{1}{4}=\frac{44}{7}
\]
\[\text { Total Area }=\frac{77}{4}+\frac{44}{7}=\frac{715}{28} \text { or } 25 \cdot 54 \text { sq. units }
\]
(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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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.