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 $\angle B$ = ${90}^0$, which group cleaned more area and by how much? Find the total area cleaned by the students.

Answer 1

Area cleaned by first group$=ar\Delta ABC=\frac{1}{2}\times BC\times AB=\frac{1}{2}\times 40\times 9=180m^{2}In,\Delta ABC,AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{9^2+{40}^2}=\sqrt{1681}=41m$
Area cleaned by second group$=ar\Delta ADC=\sqrt{s(s-a)(s-b)(s-c)}s=\frac{a+b+c}{2}=\frac{41+15+28}{2}=\frac{84}{2}=42mar\Delta ADC=\sqrt{42(42-41)(42-15)(42-28)}=\sqrt{42\times 1\times 27\times 14}=\sqrt{14\times 3\times 3\times 9\times 14}=3\times 3\times 14=126m^2$
Area cleaned by first group is more than the area cleaned by second group$=180-126=54m^2$
Total area cleaned $=180+126=306m^2$