Question 1109507: What is the area of a parallelogram ABCD if AB=4 units,AD=5root 3 and measure ofangle A=60?
Found 3 solutions by ikleyn, amalm06, TeachMath:
Answer by ikleyn(52765)   (Show Source):
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
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