The area of a parallelogram is the product of the lengths of any two of its adjacent sides by the sine of the angle between them.
= = = 2*5*3 = 30 square units.
Answer by amalm06(224) (Show Source): You can put this solution on YOUR website! Consider parallelogram ABCD, where A is the vertex on the bottom left, B is the vertex on the top left, C is the vertex on the top right, and D is the vertex on the bottom right.
Let AB=4 and AD=5sqrt (3).
Draw a perpendicular from B to AD and call the point where B touches AD as F.
Now, triangle ABF has angles 60, 30, and 90.
The height of the parallelogram is simply (4)(cos 30)=3.464
We are given that the length of the base is 5*sqrt(3).
Since the area of the parallelogram is the product of the length and width,
we have that A=5*sqrt(3)*3.464=30 (Answer) Answer by TeachMath(96) (Show Source): You can put this solution on YOUR website! Altitude: sin ∡A = O/H
sin 60^o = x/4
x, or altitude = 4 sin 60^o = 4 * sqrt(3)/2 = 2sqrt(3)
Area = Base * Height = 5sqrt(3) * 2sqrt(3) = 10(3) = 30 sq units