This is a very nice problem I came across on Instagram, so I’ve decided to post it here. In the diagram below, we have a square $ABCD$ with some triangles in it including a small equilateral triangle, if the area of the orange triangle is $24$, the goal is to find the area of the blue triangle labeled $x$:
I’ll admit, this was a bit of a tricky problem, I first attempted to the length measures of the orange triangle, but that didn’t leave anywhere. I’m going to post my successful approach as an answer below, please let me know, if my answer is accurate or if my method is correct. Furthermore, please also share your own solutions as there are likely many different ways to solve this.
Best Answer
This is going to be my approach, I’ll explain my steps below as well:
1.) Rotate $\triangle FEA$ by $180^\circ$ counterclockwise about point $F$ such that $\triangle FGD$ is formed. Notice that $\angle GFI=60^\circ$, $FG=FE$, and $FI=FH$, this means that $\triangle FGI \cong \triangle FGH$. Since it has the same area, I’ll also label it as $x$.
2.) Since we know that $\angle FEH=75^\circ$, this means $\angle BEH=30^\circ$. We drop a perpendicular from $H$ onto $AB$ and $K$ is the foot of the perpendicular. It’s easy to see that $\angle KBH=\angle KHB=45^\circ$. Let’s label $BH$ as $a\sqrt2$, via Pythagorean we know that $KH=a$. Notice that $\triangle KEH$ is a “$30-60-90$” triangle, therefore $EH=2a$.
3.) Finally, we drop a perpendicular from $E$ onto $FH$, which has the length $a\sqrt2$, and from $H$ onto $FI$. Notice that the area of $\triangle BHI=(\frac{a\sqrt2}{2})(\frac{FI}{2}$) while Area $x$=$(\frac{a\sqrt2}{2})(FH)$, since we know $FI=FH$, we can see that Area $x$ is twice the area of $\triangle BHI$, therefore $x=48$