FRAC{15}{\pi }\] \[\therefore \] Area \[=\pi {{r}^{2}}=\pi \times \frac{{{\left( 15 \RIGHT)}^{2}}}{{{\pi }^{2}}}\] \[=\frac{225}{\pi }\] No. of plant \[=\frac{Area}{Area\,\,per\,\,plant}=\frac{225/\pi }{4}=\frac{225}{4\pi }\APPROX 18\]

"> FRAC{15}{\pi }\] \[\therefore \] Area \[=\pi {{r}^{2}}=\pi \times \frac{{{\left( 15 \RIGHT)}^{2}}}{{{\pi }^{2}}}\] \[=\frac{225}{\pi }\] No. of plant \[=\frac{Area}{Area\,\,per\,\,plant}=\frac{225/\pi }{4}=\frac{225}{4\pi }\APPROX 18\]

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A How many plants (approximately) will be there in a circular bed whose outer edge measures 30 cm allowing \[4\text{ }c{{m}^{2}}\]for each plant?

7th Class Mathematics in 7th Class . 10 months ago

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(a): From data it is clear that 30 cm = perimeter \[=2\pi r\] \[\Rightarrow \,\,r=\FRAC{15}{\pi }\] \[\therefore \] Area \[=\pi {{r}^{2}}=\pi \times \frac{{{\left( 15 \RIGHT)}^{2}}}{{{\pi }^{2}}}\] \[=\frac{225}{\pi }\] No. of plant \[=\frac{Area}{Area\,\,per\,\,plant}=\frac{225/\pi }{4}=\frac{225}{4\pi }\APPROX 18\]

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