Question

Four parallel lines are drawn parallel to one side of an equilateral triangle such that it cuts the other sides at equal intervals. the area of the largest segment thus formed is 27m^2. find the area of the triangle.

Answer 1

Four parallel lines are drawn parallel to one of the sides of an equilateral triangle.

The area of the blue region PQCB in the figure is given as 27 m²

Now

ΔABC ~ ΔAPQ

$\frac{ABC}{APQ}=(\frac{5x}{4x})^2=\frac{25}{16}$

Let area of ΔABC = 25y

Area of ΔAPQ = 16y

Area of PQCB = area of ΔABC - area of ΔAPQ

Area of PQCB = 25y-19y = 6y

But 6y = 27

Dividing by 6 on both sides

y = 27/6 = 4.5

Hence area of ΔABC = 25y = 25(4.5) = 112.5 cm²

Answer 2

This is the same as taking an equilateral triangle and subtracting the area of a smaller equilateral triangle with 4/5 of the linear size, so you are subtracting 16/25 of the area.  

Therefore 27 m^2 is 9/25 of the area of the original triangle, and the area of the original triangle is 75 m^2.

I hope it helps you. If there is still any confusion. Please leave a comment below.