Question

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Answer 1

$•\huge\bigstar{\underline{{\red{A}{\pink{n}{\color{blue}{s}{\color{gold}{w}{\color{aqua}{e}{\color{lime}{r}}}}}}}}}\huge\bigstar•$

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$\huge{\textbf{\textsf{{\purple{have \: a \: great \: day \: }}{\pink{ahead}}}}}$

Answer 2

$AC^2=AB^2+BC^2$       [pythogorus theorem]

$AC^2=9^2+40^2\\\\AC^2=81+1600\\\\AC^2=1681\\\\AC=\sqrt{1681} \\\\AC=41m$

Now,

Area of figure ABC=$\frac{1}{2}b*h$

(b)Base=40 m and h(height)=9m

$\frac{1}{2}*40*9\\\\==>20*9\\\\==>180m^2$

Area of figure ACD= $_\sqrt{s(s-a)(s-b)(s-c)}$

S=Semi perimeter=42 m

Putting value we get=$_\sqrt{42(42-41)(42-15)(42-28)}$

==>$_\sqrt{27*14*27}$

==>$_\sqrt{15876}$

==>$126m^2$

So Figure ABC has large area ,so students who cleaned Figure ABC did more cleanliness .

Total area=$180m^2+126m^2$

==>$_3_0_6_m_^2$

Total area= $_3_0_6_m_^2$

#devanshu1234321