Question

The sides of a rhombus with angle of 60° are 6 inches. Find the area of the rhombus.

Answer 1

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, $DO= \sqrt{ DC^{2}- OC^{2} }= \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}=$ 
$=\sqrt{9}* \sqrt{3} =3\sqrt{3} (in)$

6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*$\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}$=$=2*9 \sqrt{3}=18 \sqrt{3}$ ($in^{2}$)