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)/12 #, 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)/12  #, what is the difference between the areas of the rhombuses?

Sadie Adams · Accepted Answer

Difference in Area#=11.31372" "#square units

To compute the area of a rhombus Use the formula Area#=s^2*sin theta" "#where #s=#side of the rhombus and #theta=# angle between two sides Compute the area of rhombus #1.# Area#=4*4*sin ((5pi)/12)=16*sin 75^@=15.45482# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Compute the area of rhombus #2.# Area#=4*4*sin ((pi)/12)=16*sin 15^@=4.14110# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Compute the difference in Area#=15.45482-4.14110=11.31372# God bless....I hope the explanation is useful.

Theodore Abad · Answer

undefined The area of a rhombus can be calculated using the formula $ A = s^2 \sin(\theta) $, where $ s $ is the length of a side and $ \theta $ is the measure of one of the angles.

For the first rhombus with an angle of $ \pi/12 $:
$$ A_1 = 4^2 \sin\left(\frac{\pi}{12}\right) = 16\sin\left(\frac{\pi}{12}\right) $$

For the second rhombus with an angle of $ \frac{5\pi}{12} $:
$$ A_2 = 4^2 \sin\left(\frac{5\pi}{12}\right) = 16\sin\left(\frac{5\pi}{12}\right) $$

The difference between the areas of the two rhombuses is:
$$ \Delta A = A_2 - A_1 = 16\sin\left(\frac{5\pi}{12}\right) - 16\sin\left(\frac{\pi}{12}\right) $$

Using trigonometric identities or a calculator:
- $ \sin\left(\frac{\pi}{12}\right) \approx 0.2588 $
- $ \sin\left(\frac{5\pi}{12}\right) \approx 0.9659 $

Thus,
$$ \Delta A = 16(0.9659) - 16(0.2588) = 15.4144 - 4.1408 = 11.2736 $$

Therefore, the difference between the areas of the two rhombuses is approximately $ 11.27 $ square units.