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

Maverick Smith · Accepted Answer

#=3.43# Area of rhombus with angle #theta=a^2 sintheta# where #a# is the side. So Area #A1#of rhombus with side-length #2# and angle #pi/12# can be written as #A1=(2^2) sin(pi/12)# or #A1=4sin(pi/12)# or #A1=1.03# And Area #A2#of rhombus with side-length #2# and angle #pi/3# can be written as #A2=(2^2) sin(pi/3)# or #A2=4 sin(pi/3)# or #A2=3.46# The difference #=A2-A1# #=3.46-1.03# #=3.43#

James Atkins · Answer

undefined The area of a rhombus can be calculated using the formula: Area = (diagonal_1 * diagonal_2) / 2.

For both rhombuses, since the length of each side is 2, the diagonals can be calculated using trigonometry. The diagonals of a rhombus are twice the length of the side times the sine of the angle formed by the sides.

For the rhombus with an angle of π/12, the diagonal is: $2 \times 2 \times \sin(\pi/12)$.

For the rhombus with an angle of π/3, the diagonal is: $2 \times 2 \times \sin(\pi/3)$.

Using these diagonals, you can find the areas of both rhombuses and then find the difference between their areas.