# To find Area of rectangular with given 3 parameters $a,b,c$ are given parameters . I would like to find Area of (ABCD) rectangular.

I can find $d$ from $a,b,c$.

$$(x-m)^2+(y-n)^2=a^2$$ $$(x-m)^2+n^2=b^2$$ $$m^2+(y-n)^2=c^2$$ $$m^2+n^2=d^2$$

$$m^2+n^2+(x-m)^2+(y-n)^2=a^2+d^2=b^2+c^2$$

$$d=\sqrt {b^2+c^2-a^2}$$

Let's define

$\angle AEB =\alpha$, $\angle DEC =\beta$ ,$\angle AED =\gamma$ , $\angle BEC =\phi$

$$x^2=a^2+c^2-2ac \cos (\alpha)$$ $$x^2=b^2+d^2-2bd \cos (\beta)$$ $$y^2=c^2+d^2-2cd \cos (\gamma)$$ $$y^2=a^2+b^2-2ab \cos (\phi)$$

$$b^2+d^2-2bd \cos (\beta)=a^2+c^2-2ac \cos (\alpha)$$ $$a^2+b^2-2ab \cos (\phi)=c^2+d^2-2cd \cos (\gamma)$$

And also we know that

$$\alpha + \beta + \phi + \gamma = 2 \pi$$ $$\cos (\alpha + \beta + \phi + \gamma)= \cos (2 \pi)=1$$

Area of $ABCD =\frac{1}{2} [ac \sin (\alpha) + bd \sin (\beta) + cd \sin (\gamma))+ ab \sin (\phi)]=xy$

I am stuck to solve the equations and find the area by given $a,b,c$, Is it possible to find area of ABCD rectangular via given 3 parameters $a,b,c$ ? Thanks for hints and answers.

UPDATE: Nov, 14th 2014:

I proved that Area of ABCD does not depend on only $a,b,c$ $$y=a.\sin P +b \sin Q$$

$$|EF|=a \cos P=b \cos Q$$

$$x=a.\cos P +\sqrt{c^2-a^2 \sin^2 P}$$

$$y=a.\sin P +b \sin Q=a.\sin P +b \sqrt{1-\frac{a^2 \cos^2 P}{b^2}}$$

$$y=a.\sin P +b \sin Q=a.\sin P + \sqrt{b^2-a^2 \cos^2 P}$$

Area of $ABCD=x.y=(a.\cos P +\sqrt{c^2-a^2 \sin^2 P})(a.\sin P + \sqrt{b^2-a^2 \cos^2 P})$

The formula shows that The Area also depends on an angle not only $a,b,c$

• another +1 for showing your work.from georgia – dato datuashvili Apr 5 '13 at 6:42
• $a,b,c,d$ are length right?is $m,n$ given? – dato datuashvili Apr 5 '13 at 6:48
• if m,n is known, then d is known. but if m,n are not known, it is imposible to find the area as there is unlimit possible. – chenbai Apr 5 '13 at 8:56 for m,n is fixed, $d=\sqrt{m^2+n^2}$ .
• Thanks a lot for answer and the graph. Yellow area does not follow the rule. Red and green lengths must be on opposite corner. Another important thing that $d=\sqrt {b^2+c^2-a^2}$. Thus you can draw another circle for $d$. I still suspect that $x.y$ can be constant. I need to check more some formulas. – Mathlover Apr 8 '13 at 10:39