Two rhombuses have sides with lengths of #7 #. If one rhombus has a corner with an angle of #pi/6 # 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 #7 #.  If one rhombus has a corner with an angle of #pi/6  # and the other has a corner with an angle of #(pi)/4  #, what is the difference between the areas of the rhombuses?

Savannah Aguilar · Accepted Answer

The difference between the areas of the rhombuses is #10.15(2dp) sq.unit # We know the area of rhombus is #A_r= s^2*sin theta# where #s# is the length of sides and #theta# is the angle of a corner. So the Area of first rhombus is#A_(r1)=7^2*sin(pi/6)=49sin30=49*1/2=24.5 sq.unit# Area of second rhombus is#A_(r2)=7^2*sin(pi/4)=49sin45=49*1/sqrt2=34.65 sq.unit :.#The difference between the areas of the rhombuses is #=A_(r2)-A_(r1)=34.65-24.5=10.15(2dp) sq.unit #[Ans]

Jayden Jackson · Answer

undefined The area of a rhombus can be calculated using the formula: Area = (diagonal1 * diagonal2) / 2.

For the first rhombus with an angle of π/6, if one side has length 7, the diagonals will have lengths of 7 and 14.

For the second rhombus with an angle of π/4, if one side has length 7, the diagonals will have lengths of 7 and 7√2.

The areas of the rhombuses are: 
- Rhombus 1: (7 * 14) / 2 = 49 
- Rhombus 2: (7 * 7√2) / 2 = 24.5√2

The difference in areas is: 
49 - 24.5√2 ≈ 15.97 square units.