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

1800

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2000

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2200

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2400

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Solution

The correct option is A $2000$
Let $a = 41 m, b = 40 m, c = 9 m$
$s = \frac{a + b + c}{2} = \frac{41 + 40 + 9}{2} = 45 m$
Area of the triangular field $= \sqrt{s (s - a) (s - b) (s - c)}$
$= \sqrt{45 (45 - 41) (45 - 40) (45 - 9)}$
$= \sqrt{45 \times 4 \times 5 \times 36} = 180 m^{2}$ .... $(∵ 900 c m^{2} = 0.09 m^{2})$

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The sides of a triangle field are 41m,40m and 9m. Find the number of rose beds can be prepared in the field, if each rose bed, on an average needs 900cm2 space.

The correct option is A 2000Let a=41m,b=40m,c=9ms=a+b+c2=41+40+92=45mArea of the triangular field =√s(s−a)(s−b)(s−c)=√45(45−41)(45−40)(45−9)=√45×4×5×36=180m2 ....(∵900cm2=0.09m2)

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