1. Check the drawing of the rhombus ABCD in the picture attached.
2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.
3. The diagonals:
i) bisect the angles so m(ODC)=60°/2=30°
ii) are perpendicular to each other, so m(DOC)=90°
4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.
5. By the pythagorean theorem, [tex]DO= \sqrt{ DC^{2}- OC^{2} }= \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}= [/tex]
[tex]=\sqrt{9}* \sqrt{3} =3\sqrt{3} (in)[/tex]
6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*[tex] \frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2} [/tex]=[tex]=2*9 \sqrt{3}=18 \sqrt{3} [/tex] ([tex] in^{2} [/tex])
