Two rhombuses have sides with lengths of #8 #. If one rhombus has a corner with an angle of #pi/6 # and the other has a corner with an angle of #(3pi)/4 #, what is the difference between the areas of the rhombuses?

The area of a rhombus can be calculated using the formula: ( A = \frac{1}{2} \times d_1 \times d_2 ), where ( d_1 ) and ( d_2 ) are the lengths of the diagonals.

For the rhombus with an angle of ( \frac{\pi}{6} ), we can calculate the area using trigonometry to find the diagonals: ( d_1 = 2 \times 8 \times \sin\left(\frac{\pi}{6}\right) ) and ( d_2 = 2 \times 8 \times \sin\left(\frac{5\pi}{6}\right) ).

For the rhombus with an angle of ( \frac{3\pi}{4} ), we can similarly calculate the diagonals: ( d_1 = 2 \times 8 \times \sin\left(\frac{3\pi}{4}\right) ) and ( d_2 = 2 \times 8 \times \sin\left(\frac{\pi}{4}\right) ).

Once we have the diagonals for both rhombuses, we can find their areas and then calculate the difference.

Let's proceed with the calculations.For the rhombus with an angle of ( \frac{\pi}{6} ), the diagonals are:

( d_1 = 2 \times 8 \times \sin\left(\frac{\pi}{6}\right) = 16 \times \frac{1}{2} = 8 )

( d_2 = 2 \times 8 \times \sin\left(\frac{5\pi}{6}\right) = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3} )

Area of this rhombus: ( A_1 = \frac{1}{2} \times 8 \times 8\sqrt{3} = 32\sqrt{3} )

For the rhombus with an angle of ( \frac{3\pi}{4} ), the diagonals are:

( d_1 = 2 \times 8 \times \sin\left(\frac{3\pi}{4}\right) = 16 \times \frac{\sqrt{2}}{2} = 8\sqrt{2} )

( d_2 = 2 \times 8 \times \sin\left(\frac{\pi}{4}\right) = 16 \times \frac{\sqrt{2}}{2} = 8\sqrt{2} )

Area of this rhombus: ( A_2 = \frac{1}{2} \times 8\sqrt{2} \times 8\sqrt{2} = 64 )

The difference between the areas of the two rhombuses is:

( A_2 - A_1 = 64 - 32\sqrt{3} )