Example 5
Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA. Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠ B = 90°, which group cleaned more area and by how much? Find the total area cleaned by the students (neglecting the width of the lanes).
Area first group = Area of ∆ABC
Area second group = Area of ∆ADC
Total Area = Area ∆ABC + Area ∆ADC
Area ∆ABC
Since AB = 9 m and BC = 40 m, ∠ B = 90°,
Then ΔABC is a right angled triangle
Area of ∆ABC = 1/2 × base × height
= 1/2 × 40 × 9 m2
= 20 × 9 m2
= 180 m2
Area first group = Area of ∆ABC = 180 m2
Area ΔADC
Area of triangle = √(s(s−a)(s−b)(s −c))
Here, s is the semi-perimeter,
and a, b, c are the sides of the triangle
Here, a = 28 m , b = 15 m, c = AC
Since ∠ B = 90° ,
Applying Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
AC = √(AB2+BC2)
AC = √(92+402) m
= √(81+1600) m
= √1681 m
= √412 m
= 41m
Thus, AC = 41
∴ c = 41 m
s = (𝑎 + 𝑏 + 𝑐)/2
Area of Δ ADC = √(42(42 −28)(42 −15) (42 −41) ) m2
Putting a = 28 , b = 15, c = 41 & s = 42
= √(42×14×27×1) m2
= √((14×3)×(14)×(9×3) ) m2
= √((14×14)×(9)×(3×3)) m2
= √((14×14)×(9)×(9)) m2
= √((14×14)×(9)×(9)) m2
= √((142)×(92) ) m2
= √((14)2) × √((9)2)
= 14× 9
= 126 m2
Thus, Area second group = Area of ∆ADC = 126 m2
Area first group = Area of ∆ABC = 180 m2
So first group cleaned (180 – 126) m2,
i.e., 54 m2 more than the area cleaned by the second group.
Total area cleaned by all the students
= Area ∆ABC + Area ∆ADC
= (180 + 126) m2
= 306 m2.

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.