How can I approach finding quadrilateral area in a triangle with 2 segments drawn?

Question

Sorry for the lack of "work", I just genuinely don't know what I'm supposed to notice in a problem like this. What I got was $AD=13$, $AC=5\sqrt{2}$, $AE= \frac{65}{17}$ from angle bisector, and other obvious segments. I haven't found $BF$ or $FE$ or the length of $BE$ or any of its segments.

Overall what I'm asking is what can I notice about this problem to be able to answer it. Should I find $\Delta {AEF}$ because I know the area of $\Delta{ABC}$? Is there an easier way? Any help is appreciated.

Edite: Answer is $14.93$

Answer 1

Here's another solution. Shamelessly cribbing the image from another answer (thanks g.kov! :) ), the strategy is to find the area of $\triangle AEF$, since the area of $\triangle ACD$ is trivially found ($\frac 12 \cdot 5 \cdot 7 = \frac{35}{2}$).

$BF = \frac{5}{\sqrt 2}$.

$BE$ can be found most easily by a slightly less used corollary of the angle bisector theorem (other than this, I'm not sure if this theorem has a specific name), which gives that $BE^2 = AB\cdot BD - AE \cdot ED$. We can easily determine $AE$ and $ED$ by the more usual statement of the angle bisector theorem as $\frac{65}{17}$ and $13 - \frac{65}{17}$, respectively. So $BE = \sqrt{60 - \frac{65}{17}(13 - \frac{65}{17})}$

$FE$ is of course $BE - BF = \sqrt{60 - \frac{65}{17}(13 - \frac{65}{17})} - \frac{5}{\sqrt 2}$

$AF = BF = \frac{5}{\sqrt 2}$ and $AE$ is already known to be $\frac{65}{17}$.

We now have three sides of a triangle and can apply Heron's formula:

Area of $\triangle AEF = \sqrt{s(s-AF)(s-FE)(s-AE)}$, where $s$ (the semiperimeter) is given by $s = \frac{AF+FE+AE}{2}$. Working this out and subtracting it away from $\frac{35}{2}$, we get the expected value of $14.93$ to the nearest hundreth, as required.

Answer 2

Let me try. We have $S_{ABD} = 30$ and $S_{ABC} = \frac{25}{2}$. Then $S_{ACD} = \frac{35}{2}$.

Now, we have $\frac{S_{ABE}}{S_{BED}} = \frac{AE}{ED} = \frac{AB}{BD}$, then $\frac{S_{ABE}}{S_{ABD}} = \frac{AB}{AB+BD} = \frac{5}{17}$, or $S_{ABE} = \frac{150}{17}$.

On the other hand, we have $S_{ABF} = \frac{S_{ABC}}{2} = \frac{25}{4}$.

Thus, we get $S_{AEF} = S_{ABE} - S_{ABF} = \frac{175}{68}$.

Summing up, we have $S_{EFCD} = S_{ACD} - S_{AEF}$.

Answer 3

You can readily find the area of triangle ACD:

$$ Area_{ACD} = \frac{1}{2} \cdot 7 \cdot 5 = 17.5 $$

Now if you can find the area of AEF, you can subtract that from the area of ACD to get the area of EFCD and you are done.

For line AD: $ y = -\frac{5}{12} x + 5 $

For line BE: $ y = x $

$$ y = -\frac{5}{12} y + 5 $$

$$ \frac{17}{12} y = 5 $$

$$ y = 5 \frac{12}{17} = \frac{60}{17} $$

So point E is at $ \left( \frac{60}{17}, \frac{60}{17} \right) $

Point F is at $ \left( \frac{5}{2}, \frac{5}{2} \right) $

The length of EF is $ \left( \frac{60}{17} - \frac{5}{2} \right) \sqrt{2} = \frac{120-85}{34} \sqrt{2} = \frac{35}{34} \sqrt{2} $

The length of AF is $ \frac{5}{2} \sqrt{2} $

The area of AEF = $ \frac{1}{2} \cdot \frac{35}{34} \sqrt{2} \cdot \frac{5}{2} \sqrt{2} = \frac{175}{68} \approx 2.57 $

The area of EFCD $ \approx 17.5 - 2.57 = 14.93 $

Tada. No trig required, although trigons were exploited.

Ced

P.S. Lawyers doing math?

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Hindsight:

Once E is known the area of BED $ = \frac{1}{2} \cdot 12 \cdot \frac{60}{17} \approx 21.18 $

Area of BCF is a quarter of 5 x 5 $ = \frac{25}{4} = 6.25 $

Area of EFCD = BED - BCF $ \approx 21.18 - 6.25 = 14.93 $

Answer 4

\begin{align} Area(\triangle ACD) &= Area(\triangle ABD) - Area(\triangle ABC) \\ &= \frac12(5)(12) - \frac12(5)(5) \\ &= \frac{35}{2}. \end{align}

Dropping a perpendicular from $D$ to the extended line $BE$ and taking similar triangles, we have $AE:ED = AB:BD = 5:12,$ so $AE = \frac{5}{17}AD,$ $$Area(\triangle ACE) = \frac{5}{17}Area(\triangle ACD),$$ and $$Area(\triangle AEF) = \frac12 Area(\triangle ACE) = \frac{5}{34}Area(\triangle ACD).$$

Therefore \begin{align} Area(EFCD) &= Area(\triangle ACD) - Area(\triangle AEF) \\ &= Area(\triangle ACD) - \frac{5}{34}Area(\triangle ACD) \\ &= \frac{29}{34}Area(\triangle ACD) \\ &= \frac{1015}{68} \\ &\approx 14.92647. \end{align}

Answer 5

\begin{align} [EFCD]&=[EBD]-[BCF] . \end{align}

Note that four $\triangle BCF$ make a square with the side $BC$, hence the area
\begin{align} [BCF]&=\tfrac14\cdot|BC|^2=\frac{25}4 . \end{align}

The area of $\triangle EBD$ can be found from known side $|BD|$ and $\angle EBD$, $\angle BDE$ as follows.

\begin{align} [EBD]&=\frac{|BD|^2}2\cdot \frac{\sin{\angle EBD}\sin{\angle BDE}}{\sin(\angle EBD+\angle BDE)} \\ &=\frac{|BD|^2}2\cdot \frac{\sin{\angle EBD}\sin{\angle BDE}} {\sin\angle EBD\cos\angle BDE +\cos\angle EBD\sin\angle BDE } \tag{1}\label{1} . \end{align}

Since \begin{align} |AD|&=\sqrt{|AB|^2+|BD|^2}=13 ,\\ \sin\angle BDE&=\frac{|AB|}{|AD|}=\frac5{13} ,\\ \cos\angle BDE&=\frac{|BD|}{|AD|}=\frac{12}{13} ,\\ \sin{\angle EBD}&=\cos{\angle EBD}=\frac{\sqrt2}2 , \end{align}

\eqref{1} becomes

\begin{align} [EBD]&=\frac{12^2}2\cdot \frac{\sin{\angle BDE}} {\cos\angle BDE+\sin\angle BDE} =\frac{6\cdot6\cdot2\cdot5}{5+12} =\frac{360}{17} . \end{align}

The answer is \begin{align} [EFCD]&=[EBD]-[BCF] =\frac{360}{17} -\frac{25}4 =\frac{1015}{68} \approx14.926 . \end{align}