# Find the area of the figure below: http://www.flickr.com/photos/93084714@N07/8615304582/in/photostream

*print*Print*list*Cite

### 1 Answer

We are given a rhombus with side 8 and interior angle of `60^@` . We are asked to find the area of the rhombus. Here are three good methods:

I. The area of a rhombus is given by `A=1/2d_1d_2` where `d_1,d_2` are the lengths of the diagonals.

The diagonals bisect the angles of the rhombus and bisect each other; also the diagonals are perpendicular. Thus the diagonals form 4 congruent triangles. The triangles are 30-60-90 right triangles with hypotenuse 8.

So `d_1=2(4)=8,d_2=2(4sqrt(3))=8sqrt(3)` .

Then `A=1/2(8)(8sqrt(3))=32sqrt(3)`

II. The area of a rhombus is `A=bh` where b is the length of a base, and h the height to that base.

Label the rhombus ABCD. Drop an altitude from A to E on CD. Then triangle AEC is a 30-60-90 right triangle. If AC is the shorter diagonal then AC=8. ( Triangle ABC is an equilateral triangle.) Then AE=`4sqrt(3)`

So `A=8(4sqrt(3))=32sqrt(3)`

III The area of a rhombus is given by `A=s^2sinalpha` where s is the side length and `alpha` is any of the angles. ( Opposite angles are congruent so they have the same sine, and adjacent angles are supplementary and will have the same sine.)

Then `A=8^2sin60^@=64*sqrt(3)/2=32sqrt(3)`