# Calculate area of a parallelogram given its sides and the angle between diagonals

The parallelogram ABCD has sides AB = 80 cm and BC = 60 cm. Let X be the intersection of its diagonals. How to calculate the area of the parallelogram, when given the angle between diagonals BXC = 60°.

I have calculated the angle AXB = 120° and written two equations based on the cosine law, but it has started to be complicated and I hope there is more elegant way.

Picture of the Parallelogram with the given values

• With cosine law you are on the right track: Give a name a=XB=XD, b=XA=XC; find two equations with the 2 unknowns $a$ and $b$ then solve the resulting system. – Jean Marie Nov 11 '16 at 16:22
• Thanks, actually that's what I did and got this result: link to WolframAlpha. It results in system of 2 quadratic equations which is quite complex and time consuming to solve. Is that the only possible way? – Štefan Schindler Nov 11 '16 at 17:07

You can use the law of cosines. You write the following system: $$\begin{cases} 80^2=x^2+y^2-2xy\cos120°\\60^2=x^2+y^2-2xy\cos60° \end{cases}$$ where $x$=$AX$ and $y$=$BX$.

Subtracting the equations: $$80^2-60^2=x^2+y^2-2xy\cos120°-(x^2+y^2-2xy\cos60°)$$ we have:

$2800=-2xy(\cos120°-\cos60°)$

$2800=-2xy(-\frac{1}{2}-\frac{1}{2})$

$2800=2xy$

so $xy=1400$

Now you can just calculate the area of the two triangles ABX and BCX using:

$$Area=\frac{1}{2}xy\sin\alpha,$$

where $\alpha$ is the angle between $x$ and $y$ (120° in ABX and 60° in BCX). Then it is easy.

• Thanks for great answer :) – Štefan Schindler Nov 11 '16 at 20:20
• Good short circuit compared to my solution. – Jean Marie Nov 11 '16 at 20:28

Let us work in meters in order to work with small numbers.

Setting $a=XB=XD, b=XA=XC,$ the law of cosines gives:

• In triangle ABX :

$$\tag{1}a^2+b^2+ab=0.8^2 \ \ \text{because} \ \ \cos(2\pi/3)=-1/2.$$

• In triangle BCX :

$$\tag{2}a^2+b^2-ab=0.6^2 \ \ \text{because} \ \ \cos(\pi/3)=1/2.$$

The by adding and subtracting (1) and (2), we obtain the following system:

$$\cases{a^2+b^2=0.5\\ab=0.14}.$$

By squaring both sides of the second equation, setting $A=a^2,B=b^2$, we obtain:

$$\cases{A+B=0.5\\AB=0.0196}.$$

Thus $A$ and $B$ are solutions of quadratic equation $X^2- 0.5 X + 0.0196=0$ that will not be difficult to solve: $A=0.0428768$ and $B=0.457123$. From which $a=\sqrt{A}=0.207067$ and $b=\sqrt{B}=0.676109$. It it the values (divided by 100) found through Wolfram, throwing away negatives ones.)

Then you proceed by computing areas of triangles XBC and XCD using formula: length of a side $\times$ length of another side $\times \sin$(angle) (angle between these sides).

• Thanks for answer, one thing not clear to me is the a^2 + b^2 = 1 part, shouldn't it be more like this: a^2 + b^2 = 0.5? – Štefan Schindler Nov 11 '16 at 20:22
• Sorry, you are right, I have forgotten to divide by 2... I correct it. – Jean Marie Nov 11 '16 at 20:31
• Ok, thanks once more, now its clear :) – Štefan Schindler Nov 11 '16 at 20:32