# Compute the area of Quadrilateral $ABCD$

As title suggests, the question is to solve for the area of the given convex quadrilateral, with two equal sides, a side length of 2 units and some angles:

I have solved the problem with a synthetic geometric approach involving some angle chasing. However, I believe my solution (which I will post in a comment below since I don't want to clutter the question) is a little messy and not efficient. Are there any better ways to do this? Geometric and/or trigonometric approaches are all welcomed!

EDIT: I have posted my solution below!

• Oh well trig kills it. Law of sines immediately gives all the side lengths, then use whatever area method you would like (heron, sine area, etc.) Sep 23 at 1:18
• You're more than welcome post a trigonometric approach if you like!
– Goku
Sep 23 at 2:25
• See my answer.. Sep 23 at 4:09

This will be my approach to this problem. I shall add a brief explanation as well:

So this is how I go about it:

Please note that I forgot to mark the point of intersection of the two diagonals, therefore I will be referring to it as point $$X$$ throughout my explanation.

1.) Locate point $$A'$$ by extending segment $$AC$$ such that $$A'A$$=$$AC$$ and $$\angle A'AD$$=75. Connect point $$D$$ and $$A$$. Notice that $$\triangle A'AD$$ and $$\triangle ABC$$ are congruent via the SAS property.

2.) In $$\triangle A'DC$$, because the base $$A'C$$ is divided into two equal segments by a median, we can conclude that $$\triangle A'AD$$ and $$\triangle ADC$$ have the same area via a well-known lemma (the proof of which is trivial). By extension, via congruency that we proved earlier, we can also conclude that $$\triangle ABC$$ and $$\triangle ADC$$ also have the same area. This information will be helpful, as knowing the area of just one triangle will help is reach our desired result.

3.) Locate point $$F$$ on segment $$A'A$$ and connect it to point $$D$$ such that $$\angle DFA$$=45, $$\angle ADF$$=60 and note than $$\angle DXA$$=45. This implies that $$\triangle FDX$$ is an isosceles right triangle where segment $$DF$$=$$DX$$. Note further that $$\triangle FAD$$ is congruent to $$\triangle XBC$$ via the ASA property. Hence, we can conclude that segment $$FD$$=$$DX$$=$$XB$$. This proves that point $$X$$ is the midpoint of segment $$BD$$.

4.) Extend segment $$DA$$ to point $$E$$ and connect it to point $$B$$ such that segment $$EB$$ is perpendicular to segment $$AE$$. Notice that this forms a right triangle of the type 30-60-90, where $$\angle EBD$$=60. It is trivially known that, in such a triangle, the side opposite to the angle measure of 30 is half of the largest side hypotenuse (this can be proven via trigonometry as well). However, half of segment $$BD$$ would be equivalent to $$DX$$ and $$XB$$ as point $$X$$ was found to be a midpoint. Thus we can conclude that segment $$DX$$=$$XB$$=$$EB$$. Connect point $$E$$ and $$B$$, because $$\angle EBD$$=60, $$\triangle EXB$$ is equilateral, therefore $$EX$$=$$XB$$=$$EB$$.

5.) Above implies that $$\angle AEX$$=30, and $$\angle AXE$$=75, therefore $$\angle EAX$$ must be 75, thus $$\triangle AEX$$ is isosceles and segment $$AE$$=$$BE$$=$$BX$$=$$EX$$. This proves that $$\triangle AEB$$ is also an isosceles right triangle, therefore $$\angle ABX$$=60-45=15. However, this implies that $$\angle ABC$$=$$\angle ACB$$=75, and $$\angle BAC$$=30. Therefore $$\triangle ABC$$ is an isosceles triangle as well, thus line segment $$AC$$=$$AB$$= 2 units. We drop a perpendicular from point $$B$$ on segment $$AC$$ to meet at point $$G$$. $$\triangle ABG$$ is a right triangle of type 30-60-90, and as established above, the base opposite to angle measure 30 must be half of the hypotenuse, thus we can conclude that segment $$BG$$ is 1 unit.

6.) Since we now have a height and a base, we can compute the area of $$\triangle ABC$$, which is 1 unit^2. However as we established earlier, $$\triangle ABC$$ and $$\triangle ADC$$ have equal areas, therefore the area of $$\triangle ADC$$ is also 1 unit^2 and thus, the total area of the quadrilateral, our answer, is 2 unit^2

Alternative approach:

Area$$(\triangle ADC) = \dfrac{1}{2} \times \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)].$$

Area$$(\triangle ABC) = \dfrac{1}{2} \times \overline{BC} \times [ ~\overline{AC} \sin(75^\circ)].$$

Therefore, $$~\text{Area}(\triangle ADC) ~=~ \text{Area}(\triangle ABC).$$

Therefore

$$\text{Area(quadrilateral)} ~=~ \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)]. \tag1$$

Plan of Attack:

• Let $$X$$ denote the intersection of the two diagonals.

• Use the Law of Sines and the Law of Cosines to compute $$~\overline{AX}, ~\overline{BX}, ~\overline{CX}, ~\overline{DX}.$$

• Compute $$~\overline{AD}.$$

• Apply the formula in (1) above.

$$\displaystyle \frac{\overline{DX}}{\sin(105^\circ)} = \frac{\overline{AD}}{\sin(45^\circ)} = \frac{\overline{BC}}{\sin(45^\circ)} = \frac{\overline{BX}}{\sin(75^\circ)} \implies$$

$$\overline{DX} = \overline{BX}. \tag2$$

$$\displaystyle \frac{\overline{AX}}{\sin(30^\circ)} = \frac{\overline{AD}}{\sin(45^\circ)} = \frac{\overline{BC}}{\sin(45^\circ)} = \frac{\overline{CX}}{\sin(60^\circ)} \implies$$

$$\displaystyle \frac{1}{2} \times \overline{CX} = \frac{\sqrt{3}}{{2}} \times \overline{AX} \implies$$

$$\overline{CX} = \sqrt{3} \times \overline{AX}. \tag3$$

$$\overline{DX}^2 + \overline{AX}^2 - \left[2\overline{AX}~~\overline{DX} \times \frac{1}{\sqrt{2}}\right]$$

$$= ~\overline{CX}^2 + \overline{BX}^2 - \left[2\overline{CX}~~\overline{BX} \times \frac{1}{\sqrt{2}}\right] \implies$$

$$\overline{DX}^2 + \overline{AX}^2 - \left[\sqrt{2} ~\overline{AX}~~\overline{DX}\right]$$

$$= ~3\overline{AX}^2 + \overline{DX}^2 - \left[\sqrt{2} ~\left(\sqrt{3} ~\overline{AX}\right)~~\overline{DX}\right] \implies$$

$$= ~2\overline{AX}^2 + \left[ ~\sqrt{2} ~\overline{AX}~~\overline{DX} ~~\left(1 - \sqrt{3}\right) ~\right] = 0 \implies$$

$$= ~2\overline{AX} + \left[ ~\sqrt{2} ~~\overline{DX} ~~\left(1 - \sqrt{3}\right) ~\right] = 0 \implies$$

$$\overline{AX} ~=~ \overline{DX} ~~\left[\frac{\sqrt{3} - 1}{\sqrt{2}}\right] ~=~ \overline{BX} ~~\left[\frac{\sqrt{3} - 1}{\sqrt{2}}\right]. \tag4$$

Note $$\displaystyle ~: ~\cos(135^\circ) = \frac{-1}{\sqrt{2}}.$$

\begin{alignat*}{2} 4 ~ &=~ \overline{AX}^2 + \overline{BX}^2 + \sqrt{2} ~\overline{AX} ~~\overline{BX} \\ \\ &=~ \overline{BX}^2\left(\frac{\sqrt{3} - 1}{\sqrt{2}}\right)^2 ~+~ \overline{BX}^2 ~+~ \sqrt{2} ~\left(\frac{\sqrt{3} - 1}{\sqrt{2}}\right) ~\overline{BX}^2 \\ \\ &=~ \overline{BX}^2 ~\left[ ~\left( ~\frac{4 - 2\sqrt{3}}{2} ~\right) ~+~ 1 ~+~ \sqrt{3} - 1 ~\right] \\ \\ &=~ \overline{BX}^2 ~\left[ ~\frac{4 - 2\sqrt{3}}{2} ~+~ \frac{2\sqrt{3}}{2} ~\right] \\ \\ &=~ 2 ~\overline{BX}^2 \implies \end{alignat*}

$$\overline{BX} = \sqrt{2}. \tag5$$

Therefore

• $$~\overline{AX} = \sqrt{3} - 1.$$

• $$~\overline{BX} = \sqrt{2}.$$

• $$~\overline{CX} = 3 - \sqrt{3}.$$

• $$~\overline{DX} = \sqrt{2}.$$

$$\underline{\text{Computation of} ~\overline{AD}}$$

\begin{alignat*}{2} \overline{AD}^2 ~ &=~ \overline{AX}^2 + \overline{DX}^2 - \sqrt{2} ~\overline{AX} ~~\overline{DX} \\ \\ &=~ 4 - 2\sqrt{3} + 2 - 2\sqrt{3} + 2 \\ \\ &= 8 - 4\sqrt{3} \implies \end{alignat*}

$$\overline{AD} = \sqrt{8 - 4\sqrt{3}}. \tag6$$

$$\underline{\text{Final Computations}}$$

In general,

$$\cos(2\theta) = 2\cos^2(\theta) - 1 \implies$$

$$\displaystyle \frac{\sqrt{3}}{2} = \cos(30^\circ) = 2\cos^2\left(15^\circ\right) - 1 \implies$$

$$\displaystyle \cos\left(15^\circ\right) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{3}} = \sin\left(75^\circ\right) = \sin\left(105^\circ\right).$$

$$\overline{AC} = \overline{AX} + \overline{CX} = 2.$$

Therefore, using (1) above,

$$\text{Area(quadrilateral)} ~=~ \sqrt{8 - 4\sqrt{3}} \times 2 \times \frac{1}{2} \sqrt{2 + \sqrt{3}}$$

$$=~ 2 \times \sqrt{2 - \sqrt{3}} \times \sqrt{2 + \sqrt{3}} = 2.$$

Let the intersection of the two diagonals be $$O$$.

And let $$AO = x , OC = y, AD = z$$

Then it follows from the law of sines that

$$DO = a x \hspace{25pt}$$ where $$a = \dfrac{\sin(105^\circ)}{\sin(30^\circ) }$$

and

$$OB = b y \hspace{25pt}$$ where $$b = \dfrac{\sin(75^\circ)}{\sin(60^\circ)}$$

Apply the law of cosines to $$\triangle AOD$$, $$\triangle OBD$$, $$\triangle ABC$$ gives us the following three quadratic equations:

$$z^2 = x^2 (1 + a^2 - 2 a \cos(45^\circ) )$$

$$z^2 = y^2 (1 + b^2 - 2 b (\cos(45^\circ) )$$

$$z^2 + (x + y)^2 - 2 z (x + y) \cos(75^\circ) = 2^2 = 4$$

Using a quadratic system solver (for example from Wolframalpha.com), we get:

$$x = \sqrt{3} - 1$$.

$$y = 3 - \sqrt{3}$$.

$$z = \sqrt{2} x = \sqrt{6} - \sqrt{2}$$

Now the area is given by

$$\text{Area} = \dfrac{1}{2} \sin(45^\circ) \left( a x^2 + b y^2 + xy (a + b) \right)$$

Evaluating the above expression, yields

$$\text{Area} = 2$$

Let $$O$$ denote the intersection of the two diagonals, and $$\ell$$ denote the two equal sides $$AB=DC$$.

As suggested in the comments by TheBestMagician, let us make an extensive use of the law of sines:

• $$OA=\frac{\sin(30°)}{\sin(45°)}\ell=\frac1{\sqrt2}\ell$$,
• $$OD=\frac{\sin(105°)}{\sin(45°)}\ell=\frac{1+\sqrt3}2\ell$$,
• $$OC=\frac{\sin(60°)}{\sin(45°)}\ell=\frac{\sqrt3}{\sqrt2}\ell$$,
• $$OB=\frac{\sin(75°)}{\sin(45°)}\ell=\frac{1+\sqrt3}2\ell$$.

Knowing that $$AB=2$$, the law of cosines allows us to compute $$\ell^2$$: \begin{align}4&=OA^2+OB^2-2\,OA\,OB\,\cos(135°)\\&=\ell^2\left(\frac12+\frac{2+\sqrt3}2+\frac{1+\sqrt3}2\right)\\&=\ell^2(2+\sqrt3)\end{align}hence $$\ell^2=\frac4{2+\sqrt3}=8-4\sqrt3$$, and\begin{align}\text{Area}&=\frac12\,AC\,BD\,\sin(45°)\\&=\frac{\ell^2}2\,\frac{1+\sqrt3}{\sqrt2}\,(1+\sqrt3)\,\frac1{\sqrt2}\\&=\frac{8-4\sqrt3}2\,(2+\sqrt3)\\&=2.\end{align}