Two rhombuses have sides with lengths of #13 #. If one rhombus has a corner with an angle of #(7pi)/8 # and the other has a corner with an angle of #(pi)/6 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #13 #.  If one rhombus has a corner with an angle of #(7pi)/8  # and the other has a corner with an angle of #(pi)/6  #, what is the difference between the areas of the rhombuses?

Mason Adams · Accepted Answer

Difference between the areas of the rhombuses is #9.887# 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 #13# and angle #7pi/8# is #1/2xx13^2xxsin7(pi/8)=1/2xx169xx0.383=32.363# Hence area of rhombus with side #13# and angle #pi/6# is #1/2xx13^2xxsin(pi/6)=1/2xx169xx0.5=42.25# Difference between the areas of the rhombuses is #42.25-32.363=9.887#

Claire Akin · Answer

undefined To find the difference between the areas of the two rhombuses, we first need to calculate the area of each rhombus.

The area of a rhombus can be found using the formula: $ Area = \frac{1}{2} \times d_1 \times d_2 $, where $ d_1 $ and $ d_2 $ are the lengths of the diagonals.

For the rhombus with an angle of $ \frac{7\pi}{8} $, we can calculate the lengths of its diagonals using trigonometry. Since the given angle is $ \frac{7\pi}{8} $, we know that the acute angle of this rhombus is $ \frac{\pi}{8} $ (angles in a rhombus are congruent). Using trigonometric ratios, we can find the lengths of the diagonals.

For the rhombus with an angle of $ \frac{\pi}{6} $, we can similarly calculate the lengths of its diagonals using trigonometry.

Once we have the lengths of the diagonals for each rhombus, we can plug them into the area formula to find the areas of the two rhombuses. Then, we subtract the smaller area from the larger area to find the difference between their areas.