Find the area of a rhombus ABCD with Side length of 3 and measure of angle ABC of 120.

Question

This is a case of SAS. Could you do it without using the Laws of Sin. My friend showed me how he could use the area formula for triangle 1/2 ac sinB for this. I don't understand. Could you attach a diagram?

Answer 1

Hint Draw the diagonals. They intersect each other perpendicularly, thus the result will be four congruent $30-60-90$ right triangles because $m \angle ABC = 120^\circ$.

Answer 2

Draw the rhombus. Now draw the two diagonals. They divide the rhombus into $4$ congruent right-angled triangles with hypotenuse $3$.

We find the area of one of these triangles, and multiply by $4$.

In our case, the triangles are the familiar $30$-$60$-$90$ triangles. So the legs of any triangle are $3\cdot \frac{1}{2}$ and $3\cdot \frac{\sqrt{3}}{2}$. Thus each triangle has area $\frac{9\sqrt{3}}{8}$. Finally, multiply by $4$.