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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:

enter image description here

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!

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    $\begingroup$ 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.) $\endgroup$ Sep 23 at 1:18
  • $\begingroup$ You're more than welcome post a trigonometric approach if you like! $\endgroup$
    – Goku
    Sep 23 at 2:25
  • $\begingroup$ See my answer.. $\endgroup$ Sep 23 at 4:09

4 Answers 4

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

click here

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

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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,

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

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

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enter image description here

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

$\text{Area} = 2$

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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$$\begin{align}\text{Area}&=\frac12\,AC\,BD\,\sin(45°)\\&=\frac{\ell^2}2\,\frac{1+\sqrt3}{\sqrt2}\,(1+\sqrt3)\,\frac1{\sqrt2}\\&=\frac{8-4\sqrt3}2\,(2+\sqrt3)\\&=2.\end{align}$$

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