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

Question

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

Noah Atkinson · Accepted Answer

Difference in areas between the two rhombus = 4.1618

Area of a rhombus = #(d_1 * d_2) / 2# Where #d_1 , d_2 # are diagonals. #side = a# Diagonals bisect each other at right angles. #:. d_1/2 = (a/2 )* sin (theta /2)# #d_2 /2 = (a/2) * cos (theta/2)# Rhombus #1 : side= 5 and /_theta = pi/12# #d_1 = 5 * sin (pi/24) = 0.6526# #d_2 = 5 * cos (pi/24) = 4.9752# #Area = (d_1 * d_2) / 2 = (0.6526 * 4.9752)/2 = 1.6125# Rhombus #1 : side= 5 and /_theta = (3pi)/8# Rhombus #1 : side= 5 and /_theta = pi/12# #d_1 = 5 * sin ((3pi)/16) = 4.1573# #d_2 = 5 * cos ((3pi)/16)= 2.7779# #Area = (d_1 * d_2) / 2 = (4.1573 * 2.7779)/2 = 5.7743# Diff. in areas between the Rhombus = 5.7743 - 1.6125 = 4.1618#

Sophia Alfred · Answer

undefined The difference between the areas of the two rhombuses is $ \frac{45}{2} - \frac{5}{2}\sqrt{6} $.