Two rhombuses have sides with lengths of #3 #. If one rhombus has a corner with an angle of #pi/12 # and the other has a corner with an angle of #pi/4 #, what is the difference between the areas of the rhombuses?

Question

Two rhombuses have sides with lengths of #3 #.  If one rhombus has a corner with an angle of #pi/12  # and the other has a corner with an angle of #pi/4  #, what is the difference between the areas of the rhombuses?

Aaliyah Anderson · Accepted Answer

Difference in areas between the two rhombuses is 4.0344

Area of rhombus #= (1/2) * d_1 * d_2 or a * h#
Where #d_1 , d_2 # are the diagonals, a is the side and h is the altitude.

In this case we will use the formula Area = a * h.

Jayden Jackson · Answer

undefined The area of a rhombus can be calculated 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 first rhombus with an angle of $ \frac{\pi}{12} $, the diagonals are $ 3 $ and $ 3 $, because all sides are equal in a rhombus.

For the second rhombus with an angle of $ \frac{\pi}{4} $, one diagonal is $ 3 $, and to find the other diagonal, we can use trigonometry. Since the angle is $ \frac{\pi}{4} $, which is $ 45^\circ $, and the sides adjacent to the angle are equal, we can use the cosine rule:

$$ \text{Adjacent side} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} $$

So, both diagonals of the second rhombus are $ \frac{3\sqrt{2}}{2} $.

Now, we can calculate the areas of both rhombuses.

For the first rhombus:
$$ Area_1 = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} $$

For the second rhombus:
$$ Area_2 = \frac{1}{2} \times 3 \times \frac{3\sqrt{2}}{2} = \frac{9\sqrt{2}}{4} $$

The difference in their areas:
$$ Difference = Area_2 - Area_1 $$
$$ Difference = \frac{9\sqrt{2}}{4} - \frac{9}{2} $$

$$ Difference = \frac{9\sqrt{2}}{4} - \frac{18}{4} $$

$$ Difference = \frac{9\sqrt{2} - 18}{4} $$

So, the difference between the areas of the two rhombuses is $ \frac{9\sqrt{2} - 18}{4} $.