Question

Students of school stages a rally for cleanliness campaign. They walk through the lane in two groups. Once group walk through the lanes $AB,BC$and $CA$, while the other through $AC,CD$ and $DA$(see fig.) Then the cleaned an $\angle B={90}^{\circ }$, which group clean the more area and by how much? Find the total area cleaned by the students (neglecting the width of the lanes).

Answer 1

Area of first group$=$Area of $△ABC$

Area of second group$=$Area of $△ADC$

Total Area$=$Area of $△ABC+$Area of $△ADC$

Area of $△ABC$

Since $AB=9m$ and $BC=40m\angle B={90}^{\circ }$

Then $△ABC$ is a right angled triangle.

Area of $△ABC=\frac{1}{2}\times base\times height=\frac{1}{2}\times 40\times 9=180m^2$

Area of first group$=$Area of $△ABC=180m^2$

Area of $△ADC$

Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

Here $s$ is the semiperimeter$=\frac{a+b+c}{2}=\frac{28+15+c}{2}$

Since $\angle B={90}^{\circ }$

Applying Pythagoras theorem in $△ABC$

${AC}^2={AB}^2+{BC}^2$

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

Thus,$AC=41m$

$\therefore c=41m$

$s=\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}=42m$

Area of $△ADC=\sqrt{41(42-28)(42-15)(42-41)}{m}^2$

$=\sqrt{14\times 3\times 14\times 9\times 3}{m}^2$

$=14\times 9=126m^2$

Thus,Area second group$=$Area of $△ADC=126m^2$

Area of first group$=$Area of $△ABC=180m^2$

So,first group cleaned$=180-126=54m^2$

Total area cleaned by all the students$=$Area of $△ABC+$Area of $△ADC$

$=180+126=306m^2$