Two rhombuses have sides with lengths of #10 #. If one rhombus has a corner with an angle of #(11pi)/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 #10 #.  If one rhombus has a corner with an angle of #(11pi)/12  # and the other has a corner with an angle of #(pi)/4  #, what is the difference between the areas of the rhombuses?

Sophia Alfred · Accepted Answer

The difference between the areas of the rhombuses is #44.83 sq.units (2dp)# Angle of corner of first rhombus is #(11pi)/12= 165^0# Area of first rhombus is #10*10*sin165 =25.88 sq.unit# Angle of corner of second rhombus is #(pi)/4=45^0# Area of second rhombus is #10*10*sin45 =70.71 sq.unit# The difference between the areas of the rhombuses is #70.71-25.88=44.83 sq.units (2dp)#[Ans]

Theodore Abad · Answer

undefined The area of a rhombus can be calculated using the formula $ \text{Area} = \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 $.

For a rhombus with side length $ s $, the diagonals are $ s \times \sqrt{2} $.

Given that the side length of both rhombuses is 10:

- For the rhombus with a corner angle of $ \frac{11\pi}{12} $, the diagonals are $ 10\sqrt{2} $ each.
- For the rhombus with a corner angle of $ \frac{\pi}{4} $, the diagonals are also $ 10\sqrt{2} $ each.

Therefore, the areas of both rhombuses are equal.

Hence, the difference in their areas is $ 0 $.