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(C) Since $AB = 9 \, m$ and $BC = 40 \, m$ with $\angle B = 90^{\circ}$,we use the Pythagorean theorem to find $AC$:$AC = \sqrt{AB^2 + BC^2} = \sqrt{9

Vedclass · Accepted Answer

$306$

Vedclass · Answer

(C) Since $AB = 9 \, m$ and $BC = 40 \, m$ with $\angle B = 90^{\circ}$,we use the Pythagorean theorem to find $AC$:$AC = \sqrt{AB^2 + BC^2} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \, m$.The first group cleans the area of the right-angled triangle $ABC$:Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 40 	imes 9 = 180 \, m^2$.The second group cleans the area of triangle $ACD$ with sides $41 \, m, 15 \, m$ and $28 \, m$. Using Heron's formula,the semi-perimeter $s = \frac{41 + 15 + 28}{2} = \frac{84}{2} = 42 \, m$.Area of $\Delta ACD = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{42(42 - 41)(42 - 15)(42 - 28)} = \sqrt{42 	imes 1 	imes 27 	imes 14} = \sqrt{15876} = 126 \, m^2$.The first group cleaned $180 \, m^2$ and the second group cleaned $126 \, m^2$. The first group cleaned more area by $(180 - 126) = 54 \, m^2$.Total area cleaned by the students $= 180 + 126 = 306 \, m^2$.

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