Two rhombuses have sides with lengths of #8 #. If one rhombus has a corner with an angle of #(5pi)/12 # and the other has a corner with an angle of #(3pi)/8 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #8 #.  If one rhombus has a corner with an angle of #(5pi)/12  # and the other has a corner with an angle of #(3pi)/8  #, what is the difference between the areas of the rhombuses?

Allison Allen · Accepted Answer

One rhombus has larger area than the other by #2.688# units.

Area of rhombus of side #a# and included angle #theta# is given by #a^2xxsintheta# Hence area of one rhombus of side #8# and angle #(5pi)/12# is #8^2sin((5pi)/12)=64xx0.9659=61.8176# and area of other rhombus of side #8# and angle #(3pi)/8# is #8^2sin((3pi)/8)=64xx0.9239=59.1296# Hence one rhombus has larger area than the other by #61.8176-59.1296=2.688# units.

Parker Allen · Answer

undefined The area of a rhombus can be calculated using the formula: $ \text{Area} = \frac{d_1 \times d_2}{2} $, where $ d_1 $ and $ d_2 $ are the lengths of the diagonals.

For the first rhombus, if one angle is $ \frac{5\pi}{12} $, then the adjacent angle is also $ \frac{5\pi}{12} $, making the other two angles also $ \frac{5\pi}{12} $. The diagonals of the rhombus bisect each other at right angles, forming four right-angled triangles. Using trigonometric functions, we can find that the lengths of the diagonals are $ 8\sin(\frac{5\pi}{12}) $ and $ 8\cos(\frac{5\pi}{12}) $.

For the second rhombus, if one angle is $ \frac{3\pi}{8} $, then the adjacent angle is also $ \frac{3\pi}{8} $, making the other two angles also $ \frac{3\pi}{8} $. Using similar reasoning as before, the lengths of the diagonals are $ 8\sin(\frac{3\pi}{8}) $ and $ 8\cos(\frac{3\pi}{8}) $.

Now, we can calculate the areas of both rhombuses:

For the first rhombus:
$$ \text{Area}_1 = \frac{8\sin(\frac{5\pi}{12}) \times 8\cos(\frac{5\pi}{12})}{2} $$

For the second rhombus:
$$ \text{Area}_2 = \frac{8\sin(\frac{3\pi}{8}) \times 8\cos(\frac{3\pi}{8})}{2} $$

Subtracting the areas gives us the difference:
$$ \text{Difference} = \text{Area}_1 - \text{Area}_2 $$

$$ \text{Difference} = \left(\frac{8\sin(\frac{5\pi}{12}) \times 8\cos(\frac{5\pi}{12})}{2}\right) - \left(\frac{8\sin(\frac{3\pi}{8}) \times 8\cos(\frac{3\pi}{8})}{2}\right) $$

$$ \text{Difference} = 32\left(\sin(\frac{5\pi}{12})\cos(\frac{5\pi}{12}) - \sin(\frac{3\pi}{8})\cos(\frac{3\pi}{8})\right) $$

$$ \text{Difference} \approx 32 \times (0.5176 - 0.3827) $$

$$ \text{Difference} \approx 32 \times 0.1349 $$

$$ \text{Difference} \approx 4.3168 $$

Therefore, the difference between the areas of the two rhombuses is approximately $ 4.3168 $.