Two rhombuses have sides with lengths of #8 #. If one rhombus has a corner with an angle of #pi/12 # and the other has a corner with an angle of #pi/4 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #8 #.  If one rhombus has a corner with an angle of #pi/12  # and the other has a corner with an angle of #pi/4  #, what is the difference between the areas of the rhombuses?

James Atkins · Accepted Answer

Area of the rhombus with angle #pi/4# exceeds the rhombus with angle #pi/12# by #28.672#.

Area of rhombus of side #a# and one angle #theta# is given by #a^2xxsintheta#. Hence area of rhombus of side #8# and one angle #pi/12# is #8^2xxsin(pi/12)=64xx0.259=16.576# and area of rhombus of side #8# and one angle #pi/4# is #8^2xxsin(pi/4)=64xx0.707=45.248# Hence, the area of the latter's exceeds former by #45.248-16.576=28.672#

Sadie Allen · Answer

undefined To find the area of a rhombus, you can use the formula: Area = (diagonal1 * diagonal2) / 2.

Given that the sides of both rhombuses are 8 units long, the diagonals can be found using trigonometric relationships.

For the rhombus with an angle of π/12, the diagonals can be calculated as follows:
Diagonal1 = 2 * side * sin(π/12)
Diagonal2 = 2 * side * sin(π - π/12) = 2 * side * sin(11π/12)

For the rhombus with an angle of π/4, the diagonals can be calculated as follows:
Diagonal1 = 2 * side * sin(π/4)
Diagonal2 = 2 * side * sin(π - π/4) = 2 * side * sin(3π/4)

Once you have the diagonals for both rhombuses, plug them into the area formula to find the areas. Then, subtract the smaller area from the larger area to find the difference.