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Find the area of the figure below:...

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user6933212 | eNoter

Posted April 3, 2013 at 12:00 AM via web

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Find the area of the figure below:

http://www.flickr.com/photos/93084714@N07/8615304582/in/photostream

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted April 3, 2013 at 2:05 AM (Answer #1)

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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)`

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