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

Ella Armstrong · Accepted Answer

≈ 8.33 square units

A rhombus has 4 equal sides and is constructed from 2 congruent isosceles triangles. The area of 1 triangle = # 1/2 a.a sin theta = 1/2a^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 # = 2xx1/2a^2 sintheta = a^2 sintheta # hence area of 1st rhombus #= 12^2sin(pi/3) ≈ 124.71# and area of 2nd rhombus #= 12^2sin((5pi)/8) ≈ 133.04# difference in area = 133.04 - 124.71 = 8.33 square units

Sadie Allen · Answer

undefined To find the areas of the rhombuses, you can use the formula: Area = (side length)^2 * sin(angle).

For the first rhombus with an angle of π/3, its area is (12)^2 * sin(π/3) = 72√3 square units.

For the second rhombus with an angle of (5π)/8, its area is (12)^2 * sin((5π)/8) ≈ 120.95 square units.

The difference between the areas of the two rhombuses is approximately 48.95 square units.

James Atkins · Answer

undefined The difference between the areas of the two rhombuses is $12^2 \times \left(\frac{\pi}{3} - \frac{5\pi}{8}\right)$.