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

Question

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

Autumn Armstrong · Accepted Answer

#color(purple)("Difference in Areas " = A_d = A_s - A_r = 144 - 37.27 = 106.73 "sq units"# #Rhombus 1 : a = 12, theta = pi/12# #"Area of Rhombus " A_(r) = a^2 sin theta = 12^2 sin (pi/12) = 37.27# #Rhombus 2 : a = 12, theta = pi/2, " it's a square as "theta = pi/2# #"Area of Square " A_(s) = a^2 = 12^2 = 144# #color(purple)("Difference in Areas " = A_d = A_s - A_r = 144 - 37.27 = 106.73 "sq units"#

Noah Atkinson · 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 its diagonals.

Given that both rhombuses have sides with lengths of 12, and we are provided with the angles at one of their corners, we can calculate the lengths of their diagonals using trigonometry.

For the rhombus with a corner angle of $ \frac{\pi}{12} $:
- Since the angle at one corner is $ \frac{\pi}{12} $, the opposite angle is also $ \frac{\pi}{12} $.
- Using trigonometry, we can find the length of the diagonals. Let $ x $ be the length of one diagonal. Then, $ \tan\left(\frac{\pi}{12}\right) = \frac{6}{x} $.
- Solving for $ x $, we get $ x = \frac{6}{\tan\left(\frac{\pi}{12}\right)} $.

For the rhombus with a corner angle of $ \frac{\pi}{2} $:
- In a rhombus, the diagonals are perpendicular to each other and bisect each other at their intersection point, forming right-angled triangles.
- Since one angle of the right triangle is $ \frac{\pi}{2} $, the other angle will be $ \frac{\pi}{4} $ (because the sum of angles in a triangle is $ \pi $).
- Given that one leg of the right triangle (half the length of a diagonal) is 6 (half the length of a side), we can find the length of the other leg (half the length of the other diagonal) using trigonometry. Let $ y $ be the length of the other diagonal. Then, $ \tan\left(\frac{\pi}{4}\right) = \frac{6}{y} $.
- Solving for $ y $, we get $ y = 6 $.

Now, we have the lengths of the diagonals for both rhombuses. We can calculate their areas and find the difference.

For the rhombus with angle $ \frac{\pi}{12} $:
- Diagonals: $ x = \frac{6}{\tan\left(\frac{\pi}{12}\right)} $ and $ 12 $ (given)
- Area: $ \frac{6}{\tan\left(\frac{\pi}{12}\right)} \times 12 $

For the rhombus with angle $ \frac{\pi}{2} $:
- Diagonals: $ 6 $ (given) and $ y = 6 $
- Area: $ 6 \times 6 $

Now, compute the areas and find their difference.