Two rhombuses have sides with lengths of #1 #. If one rhombus has a corner with an angle of #pi/6 # 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 #1 #.  If one rhombus has a corner with an angle of #pi/6  # and the other has a corner with an angle of #(3pi)/4  #, what is the difference between the areas of the rhombuses?

Mason Adams · Accepted Answer

See explanation.

The area of a rhombus can be calculated using: where #a# is the side length and #alpha# is any of the rhombus' angles. Here both rhombuses have #a=1# so their areas are just and #A_2=sin((3pi)/4)=sin(pi-pi/4)=sin(pi/4)=sqrt(2)/2# The difference is then:

James Atkins · Answer

undefined The area of a rhombus can be calculated using 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 a rhombus with an angle of $\frac{\pi}{6}$ at one of its corners, the diagonals will be perpendicular bisectors of each other. Let's denote the lengths of the diagonals of this rhombus as $d_1$ and $d_2$. Since the angle at one of the corners is $\frac{\pi}{6}$, we have a $30^\circ$-$60^\circ$-$90^\circ$ triangle, and the ratio of the sides is $1:\sqrt{3}:2$. Therefore, one diagonal ($d_1$) is twice the length of the side of the rhombus, i.e., $2 \times 1 = 2$, and the other diagonal ($d_2$) is $\sqrt{3}$ times the length of the side, i.e., $\sqrt{3} \times 1 = \sqrt{3}$.

So, the area of the rhombus with an angle of $\frac{\pi}{6}$ is $A_1 = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}$.

For the other rhombus with an angle of $\frac{3\pi}{4}$ at one of its corners, the diagonals will bisect each other at right angles. In this case, we can use the property that the diagonals of a rhombus are perpendicular bisectors of each other to find the lengths of the diagonals. Since the sides of the rhombus are 1, by symmetry, the diagonals will be $\sqrt{2}$.

So, the area of the rhombus with an angle of $\frac{3\pi}{4}$ is $A_2 = \frac{1}{2} \times \sqrt{2} \times \sqrt{2} = 1$.

The difference between the areas of the two rhombuses is $A_1 - A_2 = \sqrt{3} - 1$.