Two rhombuses have sides with lengths of #7 #. If one rhombus has a corner with an angle of #pi/4 # and the other has a corner with an angle of #(3pi)/4 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #7 #.  If one rhombus has a corner with an angle of #pi/4  # and the other has a corner with an angle of #(3pi)/4  #, what is the difference between the areas of the rhombuses?

Sadie Allen · Accepted Answer

Both rhombuses are identical and hence no difference in areas.

Given #a = 7, theta_1 = pi/4, theta_2 = ((3pi)/4)#

Area of Rhombus #A_r = a^2 sin theta#

We know, sin theta = sin pi - theta)#

#:. sin (3pi)/4 = sin (pi - (3pi)/4) = sin (pi/4)#

In both the rhombuses, theta is #(pi/4)#. Hence, their areas are the same

#A_r = 7^2 sin (pi/4) = 49 / sqrt2 = 34.6482#

Sadie Adams · Answer

undefined The area of a rhombus can be calculated using the formula $ A = \frac{1}{2} d_1 d_2 $, where $ d_1 $ and $ d_2 $ are the lengths of the diagonals.

For a rhombus with side length $ s $ and a corner angle $ \theta $, the diagonals have lengths $ s $ and $ s\sqrt{2} $.

Using this information, we can calculate the areas of the two rhombuses:

1. For the rhombus with an angle of $ \frac{\pi}{4} $:
   $$ A_1 = \frac{1}{2} \cdot 7 \cdot 7\sqrt{2} = 49\sqrt{2} $$

2. For the rhombus with an angle of $ \frac{3\pi}{4} $:
   $$ A_2 = \frac{1}{2} \cdot 7 \cdot 7\sqrt{2} = 49\sqrt{2} $$

The difference between the areas is:
$$ \text{Difference} = |A_1 - A_2| = |49\sqrt{2} - 49\sqrt{2}| = 0 $$

So, the difference between the areas of the two rhombuses is 0.