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)/8`, what is the difference between the areas of the rhombuses?

Answer 1

Difference between the areas of the rhombuses is `5.32` sq.units

Explanation:

Area of a parallelogram with sides `a` and `b` and included angle `theta` is given by `1/2xxaxxbxxsintheta`. As it is a rhombus, two sides are equal area will be `1/2xxa^2xxsintheta`.

Hence area of rhombus with side `4` and angle `pi/12` is

`1/2xx4^2xxsin(pi/12)=1/2xx16xx0.259=2.072`

Hence area of rhombus with side `4` and angle `5pi/8` is

`1/2xx4^2xxsin(5pi/8)=1/2xx16xx0.924=7.392`

Difference between the areas of the rhombuses is `7.392-2.072=5.32`

Answer 2

Difference between the areas`=10.641" "`square units

Explanation:

The formula for computing the area of the rhombus when side length `s` and corner angle `theta` are given

Area `=s^2*sin theta`

Therefore to compute for the difference:

Difference between the areas
`=s^2sin theta_1-s^2*sin theta_2`

`=s^2*(sin theta_1-sin theta_2)`

`=4^2*(sin ((5pi)/8)-sin(pi/12))`

`=10.641" "`square units

God bless....I hope the explanation is useful.