Question

Find the area of a parallelogram with sides 6 and 12 and an angle of 60

Answer 1

Hello,

Let's assume h the heigth of the parallelogram

h/6=sin 60°==>h=√3/2*6=3√3
Area=3√3 * 12=36√3

Answer 2

Explanation:

Consider ABCD is a parallelogram and AB=CD=6 and BC=DA=12, Let AE be the height of the parallelogram, then from ΔAED, we have

`(AE)/(AD)=sin60^(\circ)`

⇒
`(AE)/(12)=(√(3))/(2)`

⇒
`AE=6√(3)`

Then, the area of parallelogram=
`Base{*}height`

=
`CD{*}AE`

=
`6{*}6√(3)`

=
`36√(3)sq units`

Therefore, the area of parallelogram is
`36√(3)sq units`.