Two rhombuses have sides with lengths of #15 #. If one rhombus has a corner with an angle of #pi/12 # and the other has a corner with an angle of #(5pi)/6 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #15 #.  If one rhombus has a corner with an angle of #pi/12  # and the other has a corner with an angle of #(5pi)/6  #, what is the difference between the areas of the rhombuses?

Wyatt Anderson · Accepted Answer

Difference in areas between the two rhombuses is 54.2655

Area of rhombus #= (1/2) * d_1 * d_2 or a * h#
Where #d_1 , d_2 # are the diagonals, a is the side and h is the altitude.

In this case we will use the formula Area = a * h.

Sophia Alfred · Answer

undefined The area of a rhombus can be calculated using the formula: $ \text{Area} = \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 $.

For a rhombus with side length $ s $ and an angle $ \theta $ between two adjacent sides, the diagonals are $ d_1 = s $ and $ d_2 = s \times \text{abs}(2\sin(\theta)) $.

Given:
- Both rhombuses have side lengths of 15.
- One rhombus has a corner angle of $ \frac{\pi}{12} $ radians.
- The other rhombus has a corner angle of $ \frac{5\pi}{6} $ radians.

Let's calculate the areas of both rhombuses using the formula mentioned above.

For the rhombus with angle $ \frac{\pi}{12} $:
- $ d_1 = 15 $
- $ d_2 = 15 \times \text{abs}(2\sin(\frac{\pi}{12})) $

For the rhombus with angle $ \frac{5\pi}{6} $:
- $ d_1 = 15 $
- $ d_2 = 15 \times \text{abs}(2\sin(\frac{5\pi}{6})) $

Calculate the areas of both rhombuses using these diagonal lengths and then find the difference.