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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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