Two rhombuses have sides with lengths of #2 #. If one rhombus has a corner with an angle of #pi/3 # and the other has a corner with an angle of #(5pi)/12 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #2 #.  If one rhombus has a corner with an angle of #pi/3  # and the other has a corner with an angle of #(5pi)/12  #, what is the difference between the areas of the rhombuses?

Sadie Allen · Accepted Answer

Area of 1st rhombus #A_1= 2xx2xxsin(pi/3)=3.46squnit#
Area of 1st rhombus #A_2= 2xx2xxsin(5pi/12)=3.86squnit#
So difference in area =#3.86-3.46=0.4squnit#

Aurora Ayala · Answer

undefined To find the difference between the areas of the rhombuses, we need to calculate the area of each rhombus and then find the difference between them.

The formula to find the area of a rhombus is: Area = (diagonal1 * diagonal2) / 2.

Given that both rhombuses have sides with lengths of 2, we can calculate the diagonals using the given angles.

For the first rhombus with an angle of π/3 (60 degrees), each diagonal will form two 30-60-90 triangles. Using trigonometric ratios, we find that the length of each diagonal is 2 * √3.

For the second rhombus with an angle of (5π)/12 (75 degrees), we can use trigonometric ratios to find the length of each diagonal. After some calculations, we find that each diagonal is approximately 2.6131.

Now, we can calculate the areas of the rhombuses:

For the first rhombus: Area1 = (2 * √3) * (2 * √3) / 2 = 6.

For the second rhombus: Area2 = (2.6131) * (2.6131) / 2 ≈ 3.409.

The difference between the areas of the rhombuses is: Difference = Area1 - Area2 = 6 - 3.409 ≈ 2.591.

Therefore, the difference between the areas of the rhombuses is approximately 2.591 square units.