The sides of a triangular field are 41, m,40, | vedclass

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(A) Let the sides of the triangular field be $a = 41\, m$,$b = 40\, m$,and $c = 9\, m$.The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \

Vedclass · Accepted Answer

$2000$

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(A) Let the sides of the triangular field be $a = 41\, m$,$b = 40\, m$,and $c = 9\, m$.The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{41 + 40 + 9}{2} = \frac{90}{2} = 45\, m$.Using Heron's formula,the area of the triangular field is $\sqrt{s(s - a)(s - b)(s - c)}$.Area $= \sqrt{45(45 - 41)(45 - 40)(45 - 9)} = \sqrt{45 	imes 4 	imes 5 	imes 36}$.Area $= \sqrt{9 	imes 5 	imes 4 	imes 5 	imes 36} = \sqrt{32400} = 180\, m^2$.Each rose bed needs $900\, cm^2$ space. Converting this to square meters: $900\, cm^2 = \frac{900}{10000}\, m^2 = 0.09\, m^2$.The number of rose beds $= \frac{	ext{Total Area}}{	ext{Area per bed}} = \frac{180}{0.09} = 2000$.

The sides of a triangular field are $41\, m$,$40\, m$,and $9\, m$. Find the number of rose beds that can be prepared in the field,if each rose bed,on an average,needs $900\, cm^2$ space.

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