Given a Square $ABCD$, find triangle Area $x$ if the area of the orange triangle is $24$

Question

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.

Answer 1

I am not too good with modifying images so I will refer to the triangle that OP used in their solution.

I begin by finding the side length of the square, or really half of that. Let $DI$ have length $t$, then we have the following: $\text{Area}(\triangle {FDI})=t^2/2$, $\text{Area}(\triangle {FHI})=\sqrt{3}t^2/2$ and $\text{Area}(\triangle {BCI})=t^2$. Now letting $y:=\text{Area}(ABHF)$, where $ABHF$ is a quadrilateral, we compare the areas of the regions listed thus far with the area of the square to find $$y=\left(\frac{5-\sqrt3}{2}\right)t^2-24.$$

Furthermore noticing that the line $BHD$ is the diagonal of our square we gather that the area of $\triangle {ABD}$ leaves us with $$y=\left(\frac{7-\sqrt3}{4}\right)t^2.$$ Hence we have $t=4\sqrt{3+\sqrt3}$. Now we need only notice that $\angle {AFE}=\pi/12$, therefore the length of $FE$ equals $8\sqrt{3-\sqrt3}$, using the properties of cosine.

Finally, we can use the following formula for the area of a triangle $$\text{Area}(\triangle {EFH})=\frac{1}{2}(\overline{FE})(\overline{FH})\sin{(\angle {EFH})}=\frac{1}{2}\left(8\sqrt{3-\sqrt3}\right)\left(4\sqrt{2}\sqrt{3+\sqrt{3}}\right)\sin\left(\frac{\pi}{3}\right)=48.$$

Answer 2

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 $x=48$

Answer 3

Consider the diagram

Since $\angle AEF=\frac\pi4$ and $\angle FEG=\frac\pi3$ and $\angle GEH=\frac\pi3$, we get $$ \angle DEH=\frac\pi{12}\tag1 $$ and $\sec\left(\frac\pi{12}\right)=\sqrt6-\sqrt2$. Thus, if the side of the square is $s$, we have $$ \begin{align} EH &=ED\left(\sqrt6-\sqrt2\right)\tag{2a}\\ &=\frac12\left(\sqrt6-\sqrt2\right)s\tag{2b} \end{align} $$ Furthermore, $$ \begin{align} GC &=AC-AK-KG\tag{3a}\\[6pt] &=\left(\sqrt2-\frac{\sqrt2}4-\frac{\sqrt6}4\right)s\tag{3b}\\ &=\frac{\sqrt3}4\left(\sqrt6-\sqrt2\right)s\tag{3c} \end{align} $$ We are given that $\angle HEG=\frac\pi3$. Since $\angle KGF=\frac\pi6$, we get $\angle FGC=\frac{5\pi}6$, $$ \begin{align} \frac{x}{24} &=\frac{\frac12\color{#C00}{EH}\cdot EG\color{#090}{\sin\left(\frac\pi3\right)}}{\frac12\color{#C00}{GC}\cdot GF\color{#090}{\sin\left(\frac{5\pi}6\right)}}\tag{6a}\\ &=\color{#C00}{\frac2{\sqrt3}}\color{#090}{\sqrt3}\tag{6b}\\[6pt] &=2\tag{6c} \end{align} $$ Thus, $x=48$.

Answer 4

Since $\angle GFB=\angle DHE=75^o$, then $GF\parallel HE$, and joining $HF$,$$\triangle HFE=\triangle HGE$$Extend $CG$ to $A$ through $K$, the midpoint of $EF$, and join $HK$.

Now $\triangle CGF$ has base $CG$ and height $KF$, while triangles $HKF$, $HKE$ have base $KF=KE$ and height $HL$. And since parallels $HL=CG$, then$$\triangle HGE=\triangle HFE=2\triangle CGF=48$$