# Parallelogram Area

Given a parallelogram with two diagonals 6 cm and 8 cm. And the angle between them is 120 degrees.

Find the length of the sides and the parallelogram area.

BD = 6 cm

AC = 8 cm

(AEB) = 120 degrees

AB = ?

Area of parallelogram = ?

• the area is $1/4\times d_1 d_2 \sin\theta$ – danimal May 1 '15 at 16:26
• For the side $AB$ use the Cosine Law. – André Nicolas May 1 '15 at 16:28
• What if I don't remember the Cosine Law in the test ? Is there any more logical solution ? – whyguy May 1 '15 at 16:32
• Drop a perpendicular from $A$ to $K$ on $BD$. – André Nicolas May 1 '15 at 16:34
• A interesting 90-60-30 triangle, what next? – whyguy May 1 '15 at 16:39

Without knowing the parallelogram law Bernard mentioned, you can use the fact that the diagonals bisect each other and solve the triangles with Law of Sines and Law of Cosines.

First, $AB^2 = (AC/2)^2 + (BD/2)^2 - 2(AC/2)(BD/2)\cos 120^{\circ}$, and similarly for $BC$.

Then,

$$\frac{AB}{\sin(\angle AEB)} = \frac{BE}{\sin(\angle BAE)}$$

and similarly for $\angle EAD$.

Then as Andre suggested in the comments, you can drop a perpendicular from $A$ down to the line defined by $CD$. The height will be $h = AD \cos (\angle EAD + \angle BAE)$ and then your area is $AB \times h$.

• Certainly. There are a number of ways to do it. To find the area you can just add the areas of the four small triangles and use Law of Cosines to find the sides. That would work, too, and wouldn't involve Law of Sines. – John May 1 '15 at 22:09

Area of the Parallelogram = $\frac{1}{2}.d1.d2.\sin \theta$ = $\frac{1}{2}.8.6.\sin120 = 4.6.\frac {\sqrt 3}{2} = 12\sqrt 3$ sq units.

Using cosine formula for finding out the third side, we get, $$AB^2 = AE^2 + EB^2 - 2.AE.EB.\cos \angle AEB$$ and $$AD^2 = AE^2 + ED^2 - 2.AE.EB.\cos \angle AED$$ Hence, $$AB^2 = 4^2 + 3^2 - 2.4.3.(-\frac{1}{2}) = 16 + 9 + 12 = 37$$ and $$AD^2 = 4^2 + 3^2 - 2.4.3.(\frac{1}{2}) = 16 + 9 - 12 = 13$$ Thus, length AB = $\sqrt 37$ and length AD = $\sqrt 13$