Two rhombuses have sides with lengths of #1 #. If one rhombus has a corner with an angle of #(pi)/2 # 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 #1 #.  If one rhombus has a corner with an angle of #(pi)/2  # and the other has a corner with an angle of #(3pi)/8  #, what is the difference between the areas of the rhombuses?

James Atkins · Accepted Answer

The difference between areas of two rhombuses is #0.076# Formula for area of a parallelogram with sides #a# and #b# and included angle #theta# is given by #axxbxxsintheta# In a rhombus two sides are equal and as included angle is #theta# its area will be #a^2sintheta#. Hence area of first rhombus is #1^2xxsin(pi/2)=1xx1=1# Area of other rhombus is #1^2xxsin(3pi)/8=1xx0.924=0.924# Hence the difference between areas of two rhombuses is #1-0.924=0.076#

Caroline Aucoin · Answer

undefined The area of a rhombus is given by the formula $ A = \frac{1}{2} \times d_1 \times d_2 $, where $ d_1 $ and $ d_2 $ are the lengths of its diagonals.

For the rhombus with a corner angle of $ \frac{\pi}{2} $, the diagonals are equal in length and each diagonal is $ \sqrt{2} $ (since it's a square). So, the area of this rhombus is $ A_1 = \frac{1}{2} \times \sqrt{2} \times \sqrt{2} = 1 $.

For the rhombus with a corner angle of $ \frac{3\pi}{8} $, the diagonals are not equal. We can use trigonometry to find the lengths of the diagonals. Let $ d_1 $ be the longer diagonal and $ d_2 $ be the shorter diagonal. Using the given angle, we can calculate $ d_1 $ and $ d_2 $ as $ d_1 = 1 $ and $ d_2 = 2\sin\left(\frac{3\pi}{16}\right) $.

Then, the area of this rhombus is $ A_2 = \frac{1}{2} \times 1 \times 2\sin\left(\frac{3\pi}{16}\right) $.

The difference in area between the two rhombuses is $ A_1 - A_2 $. Therefore, the difference between the areas is $ 1 - \sin\left(\frac{3\pi}{16}\right) $.