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The diagram shows a rectangle with side lenths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. If the area of the shaded region inside the square and the rectangle is $A$, then find $\dfrac85A$.

dia

My method:

dia

By the Pythagorean theorem, $EG=\sqrt{5^2-4^2}=3$ .

By trigonometry:

$$\tan{\theta}=\frac{4-x}{z}=\frac34 \tag{1}$$

$$\tan{\theta}=\frac34 \implies \cos{\theta}=\frac45 , \sin{\theta}=\frac35$$

$$BG=z=5\cos{\theta} \implies AE=5-z=5(1-\cos{\theta})=10\sin^2{\frac{\theta}{2}}$$

Solving for $\sin^2{\frac{\theta}{2}}$ from $\sin{\theta}=\dfrac35$:

$$\sin^2{\frac{\theta}{2}}=\frac{1}{10}$$

$$\therefore 5-z=AE=10 \cdot \frac{1}{10} =1 \implies z=4$$

Substituting this into $({1}):$

$$\frac{4-x}{4}=\frac34 \implies x=1$$

$$\therefore \text{ Shaded Area}=[GFH]+[FHC]-[JCH]$$

The red and blue angles are same; hence $JC=\frac34$.

Hence:

$$\text{Shaded Area}=[GFH]+[FHC]-[JCH]=\frac{25+1}{2}-\frac12 \cdot \frac34 =\frac{125}{8}$$

Hence $\dfrac{8A}{5}=25.$

What are other ways to determine the area?

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4 Answers 4

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In this generalized figure ...

enter image description here

$\triangle ABC\sim\triangle AB'C'$ with scale factor $\dfrac{c}{b}$, so that the desired area ("$A$" in OP, but I'll call $X$) is $$\begin{align} X = c^2 - |\triangle AB'C'| \;=\; c^2 - \left(\frac{c}{b}\right)^2\cdot|\triangle ABC| \;=\; c^2 - \frac{c^2}{b^2}\cdot\frac12ab \;=\; \frac{c^2}{2b}(2b-a) \end{align}$$ With $(a,b,c)=(3,4,5)$, this gives $\dfrac{125}{8}$, so that $\dfrac85X=25$. $\square$

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In my previous answer, I showed that the target value (eight-fifths of the shaded area) is $25$, which is the area of the square. It occurred to me that this could be shown directly, leveraging the evident $3$-$4$-$5$ triangles.

Update. An insight in @Neil's answer led me to this:

enter image description here

$$|\square ABCD| = \frac85\;|\square ABCE|$$

Done! $\square$

Note: Generalizing, with $O$ the bottom corner, and defining $a:=|OB|$ and $b:=|OA|$, the corresponding result is immediately $$|\square ABCD| = \frac{2b}{2b-a}\;|\square ABCE|$$ which agrees with the formula in my earlier answer.

Previously in this answer ...

Consider this figure, with $M$ the midpoint of $\overline{BC}$, and $\triangle MBB' \cong \triangle MCC'$.

enter image description here

Then $$\begin{align} |\square ABCD| &\;=\; |\triangle AC'D| \;+\; |\square ABCC'| \\[6pt] &\;=\; |\triangle AC'D| \;+\; |\triangle AC'B'| \\[4pt] &\;=\; \frac12\cdot 3 \cdot |AC'| \;+\; \frac12\cdot 5\cdot |AC'| \\[4pt] &\;=\; \frac{8}{5}\;|\square ABCC'| \end{align}$$ Done! $\square$

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    $\begingroup$ Both of your answers are great +2, but this one is just soooo beautiful! $\endgroup$ Commented Apr 17 at 17:23
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    $\begingroup$ If I would have been able to select two answers, I would have selected both... $\endgroup$ Commented Apr 18 at 17:32
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    $\begingroup$ I like your updated diagram, it really shows what I was failing to put into words. $\endgroup$ Commented Apr 19 at 6:12
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Firstly, in $\triangle EFG,\ \angle EFG=\vartheta$ which is your purple angle. That is, $\triangle EFG$ is the much better way to read off that $(\cos\vartheta,\sin\vartheta,\tan\vartheta)=(\dfrac45,\dfrac35,\dfrac34)$ than you started with.

The knowledge of that then allows you to just write down that $(BG,BH)=(4,3)$ and hence $\therefore\ CH=1$

Crucially, $JH=\dfrac{CH}{\cos\vartheta}=\dfrac54$

This is actually already sufficient, because $$ \begin{align}A &=\frac12FG\times GH+\frac12FH\times JH\sin\angle FHJ\\ &=\frac{25}2+\frac{5\sqrt2}2\cdot\frac54\sin\frac\perp2\\ &=\frac{25}2+\frac{25\sqrt2}8\cdot\frac1{\sqrt2}\\ &=\frac{125}8 \end {align} $$ and we get where you are, much faster.

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dia

$F$ is $4$ units above $\overline{AB}$ and $H$ is $3$ units above $\overline{AB}$ therefore $I$ is $7$ units above $\overline{AB}$. This means that $\overline{FC}$ divides $\overline{HI}$ in a $1:3$ ratio and $\triangle FHJ$ (where $J$ is the point of intersection of $\overline{FC}$ and $\overline{HI}$) is therefore $\frac14$ of $\triangle FHI$ or $\frac18$ of $\square FGHI$ meaning $A$ is $\frac58$ of $\square FGHI$ and therefore $\frac85A$ has the same area as $\square FGHI$ i.e. 25 square units.

Edit: @Blue's latest diagram shows this more clearly. Points $F, G, H, I, J$ on @DhairyaKumar's diagram map to $A, B, C, D, E$ below:

dia

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  • $\begingroup$ Your method too is concise and short. $\endgroup$ Commented Apr 19 at 2:58

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