Two rhombuses have sides with lengths of #4 #. If one rhombus has a corner with an angle of #pi/12 # 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 #4 #.  If one rhombus has a corner with an angle of #pi/12  # and the other has a corner with an angle of #(5pi)/8  #, what is the difference between the areas of the rhombuses?

Noah Atkinson · Accepted Answer

Difference between the areas#=10.641" "#square units

The formula for computing the area of the rhombus when side length #s# and corner angle #theta# are given Area #=s^2*sin theta# Therefore to compute for the difference: Difference between the areas #=s^2sin theta_1-s^2*sin theta_2# #=s^2*(sin theta_1-sin theta_2)# #=4^2*(sin ((5pi)/8)-sin(pi/12))# #=10.641" "#square units God bless....I hope the explanation is useful.

Sadie Allen · Answer

Difference between the areas of the rhombuses is #5.32# sq.units

Area of a parallelogram with sides #a# and #b# and included angle #theta# is given by #1/2xxaxxbxxsintheta#. As it is a rhombus, two sides are equal area will be #1/2xxa^2xxsintheta#. Hence area of rhombus with side #4# and angle #pi/12# is #1/2xx4^2xxsin(pi/12)=1/2xx16xx0.259=2.072# Hence area of rhombus with side #4# and angle #5pi/8# is #1/2xx4^2xxsin(5pi/8)=1/2xx16xx0.924=7.392# Difference between the areas of the rhombuses is #7.392-2.072=5.32#