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

Question

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

Sadie Allen · Accepted Answer

≈ 8.68 square units

A rhombus has 4 equal sides and is constructed from 2 congruent isosceles triangles. The area of 1 triangle #=1/2 a.asintheta = 1/2 a^2 sintheta # where a is the length of side and #theta" the angle between them"# now the area of 2 congruent triangles ( area of rhombus) is area # = 2xx1/2 a^2 sintheta = a^2 sintheta # hence area of 1st rhombus #= 6^2 sin(pi/12) ≈ 9.32# and area of 2nd rhombus #= 6^2 sin(pi/6) = 18 # Difference in area = 18 - 9.32 = 8.68 square units

Autumn Armstrong · Answer

undefined The area of a rhombus can be calculated using the formula: Area = (diagonal1 * diagonal2) / 2. The diagonals of a rhombus bisect each other at right angles, forming four right-angled triangles. Given that the lengths of the sides of both rhombuses are 6 units, the diagonals can be calculated using trigonometric properties.

For the rhombus with a corner angle of π/12, the diagonals can be calculated as follows:
Diagonal1 = 6 / sin(π/12)
Diagonal2 = 6 / sin(π/2 - π/12)

For the rhombus with a corner angle of π/6, the diagonals can be calculated as follows:
Diagonal1 = 6 / sin(π/6)
Diagonal2 = 6 / sin(π/2 - π/6)

Once the diagonals are calculated for both rhombuses, you can use the area formula to find the areas of each rhombus. Then, subtract the smaller area from the larger area to find the difference.