Two rhombuses have sides with lengths of #2 #. If one rhombus has a corner with an angle of #pi/12 # and the other has a corner with an angle of #pi/4 #, what is the difference between the areas of the rhombuses?

A rhombus has 4 equal sides and is constructed from 2 congruent isosceles triangles.

Now the area of 2 congruent triangles ( area of rhombus ) is

Difference in area = 2.828 - 1.035 = 1.793 square units

The area of a rhombus can be calculated using the formula ( A = \frac{1}{2} d_1 d_2 ), where ( d_1 ) and ( d_2 ) are the lengths of the diagonals of the rhombus. Since the diagonals of a rhombus are perpendicular bisectors of each other, the lengths of the diagonals can be calculated using trigonometry.

For the rhombus with an angle of ( \frac{\pi}{12} ), we can calculate the length of its diagonals as follows: Let ( x ) be half the length of one diagonal, then ( \tan(\frac{\pi}{12}) = \frac{2}{x} ) which gives ( x = \frac{2}{\tan(\frac{\pi}{12})} ). Therefore, the length of one diagonal is ( 2x = \frac{4}{\tan(\frac{\pi}{12})} ). Since the diagonals are perpendicular bisectors of each other, the other diagonal is the same length.

For the rhombus with an angle of ( \frac{\pi}{4} ), both diagonals are equal in length and can be calculated using the Pythagorean theorem. Let ( y ) be half the length of one diagonal, then ( y = \frac{2}{\sqrt{2}} = \sqrt{2} ). Therefore, the length of one diagonal is ( 2y = 2\sqrt{2} ).

Now, the areas of the rhombuses can be calculated as follows: For the rhombus with angle ( \frac{\pi}{12} ), the area is ( A_1 = \frac{1}{2} \cdot \frac{4}{\tan(\frac{\pi}{12})} \cdot \frac{4}{\tan(\frac{\pi}{12})} ). For the rhombus with angle ( \frac{\pi}{4} ), the area is ( A_2 = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2\sqrt{2} ).

The difference in areas is ( A_2 - A_1 ).