Question

Students of a school staged a rally for a cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes $AB$, $BC$, and $CA$; where the other through $AC$, $CD$, and $DA$. Then they cleaned the area enclosed within their lanes. If $AB=9m$,$BC=40m$,$CD=15m$,$DA=28m$ and $\angle B=90°$, which group cleaned more area and by how much? Find the total area cleaned by the students.

Answer 1

Step 1: Solve for the area covered by group walked through lanes $AB$, $BC$, and $CA$

Given that $AB=9m$,$BC=40m$ and $\angle B=90°$

We know that area of right angled triangle is $\frac{1}{2}\times base\times height$

$\therefore$ Area of $△ABC=\frac{1}{2}\times BC\times AB$

$\begin{array}{cl}=& \frac{1}{2}\times 40\times 9\\ =& 20\times 9\end{array}$

$\Rightarrow$Area of $△ABC=180m^2$

Step 2: Solve for the area covered by group walked through lanes $AC$, $CD$, and $DA$

According to Pythagoras theorem,

$\begin{array}{crcl}& {\left(AC\right)}^2& =& {\left(AB\right)}^2+{\left(BC\right)}^2\\ & & =& 9^2+{\left(40\right)}^2\\ & & =& 81+1600\\ \Rightarrow & {\left(AC\right)}^2& =& 1681\\ \Rightarrow & AC& =& \sqrt{1681}\\ \Rightarrow & AC& =& 41m\end{array}$

We know that area of triangle with sides $a,b,c$ is $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where $s=\frac{a+b+c}{2}$

Here say $a=28,b=15,c=41\Rightarrow s=\frac{28+15+41}{2}=42$

$\therefore$ Area of $△ADC=\sqrt{42\left(42-28\right)\left(42-15\right)\left(42-41\right)}$

$\begin{array}{cc}=& \sqrt{42\times 14\times 27\times 1}\\ =& \sqrt{15876}\end{array}$

$\Rightarrow$Area of $△ADC=126m^2$

Step 3: Determine which group cleaned more area

Area of $△ABC=180m^2$ and area of $△ADC=126m^2$

Area of $△ABC>$ Area of $△ADC$

Area of $△ABC-$ Area of $△ADC=180-126=54m^2$

Thus the area cleaned by the group walked through the lanes $AB$, $BC$, and $CA$ is $54m^2$ more than the area cleaned by the group walked through the lanes $AC$, $CD$, and $DA$.

Step 4: Solve for the total area cleaned by the students

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

$\begin{array}{cl}=& 180+126\\ =& 306m^2\end{array}$

Hence the group walked through the lanes $AB$, $BC$, and $CA$ cleaned $54m^2$more area and the total area cleaned by the students is $306m^2$