Question

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 (see figure). Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and LB = 90�, Which group cleaned more area and by how much? Find the total area cleaned by the students (neglecting the width of the lanes).

Answer 1

Answer:

Given, <ABC is a right-angle triangle

Therefore, AC^2 = AB^2+BC^2...

[Using Pythagoras Theorem]

= AC^2 = 9^2+40^2

= AC = root 81+1600 = root 1681 = 41m

Area of triangle ABC = 1/2×base×height

= 1/2×40×9 = 180m^2

Now, in triangle ADC, we have s = 28+15+41/2 = 84/2 = 42m

= Area of triangle DAC = root 42×(42-28)×(42-15)×(42-41)m^2

= root 42×14×27×1m^2 = 126m^2

So, group 1 covered area triangle ABC

= 180m^2 and group 2 covered area

= triangle DAC = 126m^2

Hence, group 1 covers more area by group 2 which is 54m^2 = (180m^2-126m^2) more.

Now, area covered by both the groups = Area ABC+Area DAC = 180m^2+126m^2 = 306m^2.