The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.
[Math] Find side of an equilateral triangle inscribed in a rhombus
triangles
Related Solutions
Square both sides, and you end up with a quadratic equation in $a$.
Sorry for not typing! There is an easy way to find $k$ if we use some elementary trigonometry!
Best Answer
Given the picture, let $x$ be the side of the equilateral triangle. We have: $$ 6 = \frac{x}{2}\cot\arctan\frac{4}{3}+\frac{x}{2}\cot\frac{\pi}{6},$$ or: $$ 6 = \frac{3x}{8}+\frac{\sqrt{3}\,x}{2},$$ so: $$ 48 = x(4\sqrt{3}+3) $$ and: $$ 48(4\sqrt{3}-3) = 39 x,$$ so $k=\color{red}{\frac{16}{13}}$.