Trapezoid from perpendiculars in an equilateral triangle

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From the foot $H$ of the altitude $CH$ of an equilateral triangle $A\hat{B}C$ of area $1$, drop the perpendicular to side $AC$, meeting it at $K$, and the perpendicular to $BC$, meeting it at $R$. Determine the perimeter and area of the trapezoid $ABRK$.

Solution

Let $a$ be the side. From the area $\dfrac{\sqrt3}{4}a^2=1$ we get $a^2=\dfrac{4}{\sqrt3}$, i.e. $a=\dfrac{2}{\sqrt[4]{3}}\approx1{,}520$. $H$ is the midpoint of $AB$, so $AH=\dfrac a2$. In the right triangle $AHK$ (right-angled at $K$) the angle at $A$ is $60^\circ$: $AK=AH\cos 60^\circ=\frac a4,\qquad HK=AH\sin 60^\circ=\frac{a\sqrt3}{4}.$ By symmetry $BR=\dfrac a4$ and $HR=\dfrac{a\sqrt3}{4}$; moreover $K$ and $R$ are at the same height, with $KR\parallel AB$ and $KR=\dfrac{3a}{4}$. Perimeter of $ABRK$: $\;a+\dfrac a4+\dfrac{3a}{4}+\dfrac a4=\dfrac{9a}{4}\approx3{,}42$. Area. The trapezoid has bases $AB=a$ and $KR=\dfrac{3a}{4}$, height $\dfrac{a\sqrt3}{4}$: $\mathcal A=\frac12\!\left(a+\frac{3a}{4}\right)\frac{a\sqrt3}{4}=\frac{7\sqrt3}{64}a^2=\frac{7}{16}=0{,}4375.$ $\boxed{\mathcal A_{ABRK}=\frac{7}{16}=0{,}4375,\qquad 2p_{ABRK}=\frac{9a}{4}\approx3{,}42.}$