document.write( "Question 697587: A parallelogram ABCD has all its sides measure 4, one of the diagonals also measures 4. What is its area?
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Algebra.Com's Answer #430132 by jsmallt9(3758) $\"\"$ $\"About$

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The formula for the area of a parallelogram is: A = b*h. The base in these area formulas can be any side of the polygon. Since all the sides of this polygon are 4's then we know our base will be 4.

\n" ); document.write( "The hard part is the height. If you draw a diagram of what you describe then you will see that an equilateral triangle with sides of 4 is formed by two sides of the parallelogram (rhombus actually) and the diagonal whose length is 4. The height of this triangle will also be the height of the parallelogram.

\n" ); document.write( "The angles in all equilateral triangles are always 60 degrees. If we draw in the height for the triangle we will have a 30-60-90 right triangle with a hypotenuse of 4. From Trigonometry or from remembering the relationships between the sides of 30-60-90 triangles we should be able to figure out that height is $\"2sqrt%283%29\"$

\n" ); document.write( "So the area of this parallelogram will be:
\n" ); document.write( " $\"A+=+b%2Ah+=+4%2A2sqrt%283%29+=+8sqrt%283%29\"$ \n" ); document.write( "

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