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I know there is already a question about resolving a quadrilateral from three sides and two angles, but I want to ask about a special case. Firstly, two of the sides are known to be of equal size. Secondly, I'm only interested in the area, not in the remaining angles or lengths. Can anyone suggest a simple formula?

enter image description here

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  • $\begingroup$ Are you looking for this here en.wikipedia.org/wiki/Bretschneider%27s_formula $\endgroup$ – Dr. Sonnhard Graubner Sep 10 '18 at 17:40
  • $\begingroup$ No. I don't know the perimeter, and hence don't know the semiperimeter, nor do I know two opposite angles. Yes, I could figure them out with trigonometry, but the point of the question is whether there is a simple formula that doesn't involve (eg) the cosine rule. $\endgroup$ – Tom Sep 10 '18 at 17:51
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Here's a brute-force derivation ...

enter image description here

$$\begin{align} |\square PQRS| &= |\triangle PQR| + |\triangle PRS| \\[4pt] &=\frac12 p q \sin \angle Q + \frac12 d r \sin(\angle R-\angle PRQ) \\[4pt] &=\frac12 \left(\;p q \sin \angle Q + r \left(\;\sin\angle R\cdot d \cos\angle PRQ - \cos\angle R\cdot d\sin\angle PRQ\;\right) \;\right)\\[4pt] &=\frac12 \left(\;p q \sin \angle Q + r \left(\;\sin\angle R\cdot(q-p\cos\angle Q) - \cos\angle R\cdot p\sin\angle Q\;\right) \;\right)\\[4pt] &=\frac12 \left(\;p q \sin \angle Q +qr\sin\angle R - pr \left(\;\sin\angle R \cos\angle Q + \cos\angle R\sin\angle Q\;\right) \;\right)\\[16pt] &=\frac12 \left(\;p q \sin \angle Q +qr\sin\angle R - pr \sin(\angle Q+\angle R)\;\right)\\ \end{align}$$


In the specific case of $p=r$, the formula can be manipulated thusly: $$\begin{align} |\square PQRS| &= \frac{p}{2}\;\left(\;q(\sin Q + \sin R) - p \sin(Q+R) \;\right) \\[4pt] &= p\sin\frac{Q+R}{2}\;\left(\;q\cos\frac{Q-R}{2}- p \cos\frac{Q+R}{2} \;\right) \end{align}$$

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  • $\begingroup$ Yep, I was afraid of that. Thanks anyway. $\endgroup$ – Tom Sep 11 '18 at 7:11
  • $\begingroup$ @Tom: I'm not sure what there is to be afraid of. The final formula itself is simple, and uses only the three sides and two angles, even though its derivation got a little complicated and introduces other quantities. The simplicity of the final form makes me think that there's a slicker way to get there, I just haven't had time to give it much thought. $\endgroup$ – Blue Sep 11 '18 at 7:37
  • $\begingroup$ Well, this is part of a larger problem (which I might post here at some point). Actually, what you've posted can be simplified further because you haven't taken advantage of p = r. But this should be enough to get me going. $\endgroup$ – Tom Sep 11 '18 at 9:14
  • $\begingroup$ @Tom: p=r doesn't make the derivation I gave appreciably easier, so I thought I'd post the more-general form (which might be useful to future readers). Even so, I should've included the specific case. I'll make an edit. $\endgroup$ – Blue Sep 11 '18 at 9:30
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We can try using coordinate geometry and determinants to find the area.


WLOG, suppose $u=AB$ lie on the $x$-axis with $A=(0,0)$ and $B=(u,0)$. We can drop a line perpendicular to the $x$-axis through $E$, and we have a right triangle. Since $L=EA$, we can easily write points $E,F$ in $xy$-coordinate space. And thus we have the following vertices:

$$A=(0,0)\\B=(u,0)\\E=(L \cos a,L\sin a)\\ F=(u-L\cos b,L\sin b)$$ enter image description here

Let's split the quadrilateral into two triangles: $\triangle AEF$ and $\triangle ABF$. From here, we can write the area as: $$A_{ABEF}=\frac12\left| \begin{array}{ccc} 0 & 0 & 1 \\ u & 0 & 1 \\ u-L \cos (b) & L \sin (b) & 1 \\ \end{array} \right|-\frac{1}{2}\left| \begin{array}{ccc} 0 & 0 & 1 \\ L \cos (a) & L \sin (a) & 1 \\ u-L \cos (b) & L \sin (b) & 1 \\ \end{array} \right|$$

Which simplies into: $$A_{ABEF}=\frac{1}{2} L \,u (\sin (a)+\sin (b))-\frac{1}{2} L^2 \sin (a+b)$$


Note: it is important to consider the signs of the resulting $\cos$ of angles $a,b$ and of the determinants in the derivation of the area.

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  • $\begingroup$ This is the same result Blue posted, though an interesting way (to me, anyway) of getting there. $\endgroup$ – Tom Sep 13 '18 at 8:51
  • $\begingroup$ @Tom yes it is exactly the same, but I thought this was clever (-ish) enough to warrant to put out there. $\endgroup$ – John Glenn Sep 13 '18 at 14:04

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