Find area of a quadrilateral inside a rectangle

Question

I am asked to find the area of the shaded portion of the rectangle ABCD, whre $AD=5$, $AB=4$ and $E$ is the midpoint of $BC$.

I know that if I subtract $\triangle ABC+\triangle AFD - \triangle AFG$ to the total area of the rectangle I get the answer but I am having troubles figuring out the area of $\triangle AFG$. Can someone point me in the right direction or a simpler way to solve it?

a quadrilateral in rectangle

where is $F$ situated? – Dr. Sonnhard Graubner Sep 26 '17 at 17:01 — Dr. Sonnhard Graubner, Sep 26 '17 at 17:01
Is $F$ the midpoint of $AB$? – Jack D'Aurizio Sep 26 '17 at 17:14 — Jack D'Aurizio, Sep 26 '17 at 17:14

Michael Rozenberg · Answer 1 · 2017-09-27T01:32:00.317

0

I hope you mean that $F$ is a midpoint of $AB$ and we need to calculate $S_{DGEC}$.

Let $AE\cap DC=\{K\}$.

Thus, since $DC=CK=4$, we obtain

$$FG:GD=AG:GK=AF:DK=2:8=1:4.$$

Hence, $$S_{\Delta ADG}=\frac{4}{5}S_{\Delta ADF}=\frac{4}{5}\cdot\frac{4\cdot5}{4}=4.$$ Also, $$S_{\Delta ABE}=\frac{1}{4}S_{ABCD}=5.$$ Id est, $$S_{DGEC}=S_{ABCD}-S_{\Delta ABE}-S_{\Delta AEG}=20-5-4=11.$$

edited Sep 27 '17 at 01:32

answered Sep 26 '17 at 17:16

Michael Rozenberg

194,933

Thanks, adding $K$ clarifies the solution. Do you have any recommendations for a good book on geometry with practice problems? – Rafa U U Sep 26 '17 at 20:58
@Rafa U U What do you think about my solution? Thank you! – Michael Rozenberg Sep 27 '17 at 01:32

score 0 · Answer 2 · answered Sep 26 '17 at 20:42

\begin{equation} \text{You need to subtract } (\triangle ABE +\triangle AFD -\triangle AFG)\text{ to the total area of the rectangle.} \end{equation}

let T be the total area of the rectangle: \begin{equation} T=AB\times BC=20\end{equation}

\begin{equation} \triangle ABE=(AB\times BE) /2=AB\times BC /4=T/4=5\\ \triangle AFD=(AD\times AF) /2=AD\times AB /4=T/4=5 \end{equation} \begin{equation} \triangle AFG=(base\triangle AFG\times height\triangle AFG) /2=AF\times GH/2 \\ \text{where H is the projection of G on AB}\\ GH=BC/5 \text{ (to be proved)}\\ \text{so }\triangle AFG=AB\times BC/20=T/20=1\\ \end{equation}

\begin{equation} \text{then: } \triangle ABE +\triangle AFD -\triangle AFG=5+5-1=9 \\ \text{and the area of the shaded part is: } T-9=20-9=11 \end{equation}

\begin{equation} \text{Why } GH=BC/5 \text{ ??} \end{equation}

let K be the intersection of DF With CB \begin{equation} \end{equation} to under stud the following steps you must help your self by drawing the figure!

\begin{equation} \text{we have: } CK=2AB \text{ and } CE=AB/2\\ \text{that gives } EK=3AB/2\\ \text{then } GE=3AG/2\\ \text{then } AE=5AG/2\\ \text{then } EB=5GH/2\\ \text{then } CB/2=5GH/2\\ \text{then } CB/5=GH\\ \text{so } GH=CB/5 \end{equation}

score 0 · Answer 3 · answered Apr 03 '21 at 16:24

Let the area $I=[ABCD]= 20$ and evaluate $$\frac{GE}{AE} = \frac{[ADF]}{[ADEF]}=\frac{[ADF]}{[ABCD]-[CDE]-[BEF]} =\frac{\frac14I}{I-\frac14I-\frac18I} = \frac35 $$ Then $[DEG]=\frac{GE}{AE}[ADE] = \frac35\cdot \frac12I = \frac3{10}I$ and $$[CDGE] = [DER]+[CDE] = (\frac3{10}+\frac14)I = 11$$

Find area of a quadrilateral inside a rectangle

3 Answers3