We are given in rhombus that height is equal $\displaystyle h = 2\sqrt{3} cm$ and diagonal $\displaystyle AC = 4\sqrt{3} cm$. Find length of a rhombus side $\displaystyle a$, other diagonal $\displaystyle BD$ all inner angles.
We are given in rhombus that height is equal $\displaystyle h = 2\sqrt{3} cm$ and diagonal $\displaystyle AC = 4\sqrt{3} cm$. Find length of a rhombus side $\displaystyle a$, other diagonal $\displaystyle BD$ all inner angles.
Nice picture! Calling the vertex at (0,0) "A" and going around counter clockwise, labeling vertices A, B. C, D, then $\displaystyle a= h cos(\theta)= 2\sqrt{3}cos(\theta)$ where $\displaystyle \theta$ is angle BAD. Further, triangle ABC is an isosceles triangle with base angles $\displaystyle \theta/2$ and two sides of length a. By the "cosine law", the length of the third side, the diagonal AC is given by $\displaystyle |AC|^2= (4\sqrt{3})^2= 48= a^2+ a^2+ 2a^2 cos(\theta/2)$. That gives two equations,
$\displaystyle a= 2\sqrt{3}cos(\theta)$ and $\displaystyle (4\sqrt{3})^2= 48= 2a^2(1+ a^2 cos(\theta/2))$ to solve for a and $\displaystyle \theta$.
@HallsofIvy,
I have a question. How can be the side $\displaystyle a = h\cos{\theta}$, if $\displaystyle a$ is a hypotenusis? Isn't it $\displaystyle \sin{\theta} = \frac{h}{a}$?
And cosine-law, I got: $\displaystyle a^{2} = a^{2} + e^{2} - 2\cdot{a}\cdot{e}\cdot\cos{\frac{\theta}{2}}$? Made I mistake, where $\displaystyle e = \left|AC\right|$?