Question

Students of a school staged a rally for cleanliness campaign in two groups. Group A walked through the lanes AB, BC and CA, while Group B walked through AC, CD and DA. They cleaned the area enclosed within their lanes. If $\text{AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and}\angle \text{B = 90°}$, which group cleaned more area. Also, find the total area cleaned by the students?

• A. Group A and 306 m²
• B. Group A and 180 m²
• C. Group B and 126 m²
• D. Group B and 54 m²

Answer 1

The correct option is A

Group A and 306 m²

Since $\text{AB = 9 m, BC = 40 m and}\angle \text{B = 90°}$
Using Pythagoras theorem, we have $A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(9{)}^2+(40{)}^2$
$AC=\sqrt{(}81+1600)$
$AC=\sqrt{(}1681)m=41m$
Group A has to clean the triangle ABC which is a right angled triangle.
Area covered by $△$ABC = $\frac{1}{2}$$\times$Base$\times$Height
= $\frac{1}{2}$$\times$(BC)$\times$(AB)
= $\frac{1}{2}\times 40\times 9$ = $180m^2$

Area covered by Group B is the area covered by $△ACD$, which is a scalene triangle having sides $\text{41 m, 28 m and 15 m}$.
For finding the area of the $△ACD$, we use Heron’s formula.
Semi perimeter of the $△ACD$, $s=\frac{(a+b+c)}{2}=42m$
Area of the $△ACD$ using Heron’s formula = $\sqrt{(}s(s-a)(s-b)(s-c))$
= $\sqrt{(}42\left)\right(42-41\left)\right(42-28\left)\right(42-15)$
= $\sqrt{(}42\left)\right(1\left)\right(14\left)\right(27)$ = $126m^2$
So, Group A cleaned $180m^2$ whereas Group B cleaned $126m^2$.
Clearly, Group A cleaned more area than Group B.
Total area cleaned by all the students = Area cleaned by Group A + Area cleaned by Group B = $180m^2$ + $126m^2$ = $306m^2$