Question

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

Answer 1

Refer to the attached image.

Consider triangle EHC,

$\tan 60^{\circ} = \frac{EP}{PC}$

$\sqrt3= \frac{h}{x}$

So,$x = \frac{h}{\sqrt 3}$

Now, Let side BC = 'a' units.

Therefore, DE = BC - x - x

DE = a-2x units

DE = $a -(2 \times \frac{h}{\sqrt 3})$

$DE = a - \frac{2h}{\sqrt 3}$

Now, consider the area of trapezium BDEC =

$\frac{1}{2}$ $\times$ (sum of parallel sides) $\times$ height

$\frac{1}{2}h(BC+ED)=27$

$\frac{1}{2}h(a+a-\frac{2h}{\sqrt{3}})=27$

$\frac{1}{2}h(2a-\frac{2h}{\sqrt{3}})=27$

$h({\sqrt{3}}a-h)=27\sqrt{3}$ (Equation 1)

Since, area of triangle = $\frac{1}{2}\times Base \times height$

Area of triangle ABC = $\frac{1}{2}\times BC \times (h+h+h+h+h)$

= $\frac{1}{2}\times a \times (5h)$

= $\frac{5ah}{2}$

Since, area of triangle is also equal to $\frac{\sqrt 3}{4} \times (Side)^2$

So, area of triangle ABC=$\frac{\sqrt{3}}{4}a^{2}$

Therefore,$\frac{5ah}{2} =\frac{\sqrt{3}}{4}a^{2}$

So, $a = \frac{10h}{\sqrt 3}$

Substituting the value of a in equation 1, we get

$h((\sqrt{3}\times \frac{10h}{\sqrt{3}})-h)=27\sqrt{3}$

$h(9h)=27\sqrt{3}$

$(9h)^{2}=27\sqrt{3}$

$(h)^{2}=3\sqrt{3}$

Since , area of triangle ABC =$\frac{\sqrt{3}}{4}a^{2}$

= $\frac{\sqrt{3}}{4}(\frac{10h}{\sqrt{3}})^{2}$

By substituting the value of h from above, we get

$=\frac{\sqrt{3}}{4}(\frac{100}{{3}})\times 3\sqrt{3}$

= $\frac{300}{4}$

= 75 square meters.

Therefore, the area of triangle is 75 square meters.

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.