# Area of an equilateral triangle divided by three lines

An equilateral triangle is divided by three straight lines into seven regions whose areas are shown in the image below. Find the area of the triangle.

How to solve this problem ?

Let the big triangle be $ABC$, the central $DEF$ and the remainig points be $P,Q,R$, so that we have lines $ADEP$, $BEFQ$, and $CFDR$. Let $x$ be the area of $\Delta DEF$.

Then $$AE:AP = \Delta ABE . \Delta ABP = (4+20):(4+20+4)=6:7$$ and $$AQ:AC = \Delta ABQ:\Delta ABC = (20+4+20+x):(4+20+4+20+4+20+x)=(44+x):(72+x).$$ Therefore $$(6\cdot(44+x)):(7\cdot(72+x))=\Delta AEQ:\Delta APC = (20+x):(20+4+20+x)$$ This should give you a quadratic equation in $x$ iwth only one positive solution.

By symmetry, the inner triangle is equilateral. Then the triangle marked with area $4$ is similar to one with area $28$ (on the smaller side of a cevian), so the similarity ratio is $7$.

We can then apply Menelaus to one of the triangles composed of the inner equilateral triangle, two $20$ area triangles, and a small $4$ area triangle. The transversal is a cevian. If we let the smaller part of a side be $1$ arbitrary unit and the larger be $x$, we get

$$\frac{x}{1}\frac{1+x}{1}\frac{1}{6} = 1.$$

We can now solve for $x$ and find the ratio that the cevian cuts the side into, which gives us the area ratio of the two triangles that the cevian cuts the triangle into, and hence the area of the small triangle.

• So $x=2$ and area of the triangle is $84$. – Xoff Dec 23 '13 at 12:07