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

Answer:

Step-by-step 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

$\frac{AE}{AD}=sin60^{\circ}$

⇒$\frac{AE}{12}=\frac{\sqrt{3}}{2}$

⇒$AE=6\sqrt{3}$

Then, the area of parallelogram=$Base{\times}height$

=$CD{\times}AE$

=$6{\times}6\sqrt{3}$

=$36\sqrt{3}sq units$

Therefore, the area of parallelogram is $36\sqrt{3}sq units$.