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Novosadov [1.4K]
3 years ago
14

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

Mathematics
2 answers:
KiRa [710]3 years ago
8 0
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

stepladder [879]3 years ago
8 0

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.

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