Two rhombuses have sides with lengths of #2 #. 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?

1 Answer
Mar 13, 2016

Answer:

#Difference=|Adiamond_1-Adiamond_2|= 2sqrt(3)-2sqrt(2+sqrt(2))#
#|Adiamond_1-Adiamond_2|= 2(sqrt(3)-sqrt(2+sqrt(2)))#
#|Adiamond_1-Adiamond_2| ~~ 2|1.732-1.848| = .231#

Explanation:

The easiest way to approach is using the following approach to finding the area of a parallelogram:
#Adiamond= vecs_1 xx vecs_2 = |s_1*s_2| sin theta#, where #theta# is the angle between the sides of the rhombuses. Since #s_1=s_2# we write:#Adiamond= |s^2| sin theta# thus,
#Adiamond_1= 2^2 sin (pi/3) = 2sqrt(3) #
#Adiamond_2= 2^2 sin (5pi/8) = 2sqrt(2+sqrt(2))#
#Difference=|Adiamond_1-Adiamond_2|= 2sqrt(3)-2sqrt(2+sqrt(2))#
#|Adiamond_1-Adiamond_2|= 2(sqrt(3)-sqrt(2+sqrt(2)))#
#|Adiamond_1-Adiamond_2| ~~ 2|1.732-1.848| = .231#