Undoing a pentagon

Geometry Level 3

A white rectangular paper $ABCD$ with an area of $20\text{ cm}^2$ is bent as shown, forming a pentagon $BCD'EF$ with an area of $14 \text{ cm}^2$ . If we paint both sides of this pentagon in blue and undo the fold, the rectangle will have an unpainted area. What is the area of this region?

18\text{ cm}^2

10\text{ cm}^2

12\text{ cm}^2

16\text{ cm}^2

14\text{ cm}^2

2 solutions

Jessica Wang
Jul 1, 2015

Moderator note:

Great solution! It's just a composite figure problem after all!

For completeness, how would one go about showing that $[ECBF ] = \frac{1}{2} [ABCD]$ ?

Thanks for the note!

For a very completed solution, I shall add the following content:

$\because \textrm{In }Rt\bigtriangleup AED\; \: and\;\: Rt\bigtriangleup CED',$

$\left\{\begin{matrix} AD=CD' & & \\ DE=D'E& & \end{matrix}\right.$

$\therefore Rt\bigtriangleup AED\cong Rt\bigtriangleup CED'.$

$\therefore AE=CE,\: \: \textrm{and}\; \angle DEA=\angle D'EC.$

$\textrm{Also},\: \: \because \angle DEF=\angle D'EF,$

$\therefore \angle AEF=\angle CEF.$

$\because CD\parallel AB,$

$\therefore \angle CEF=\angle EFA,\: \: \angle DEF=\angle EFB,$

$\therefore \angle AEF=\angle AFE=\angle CEF=\angle CFE,$

$\therefore AE=AF=CE=CF.$

$\textrm{Also, in }Rt\bigtriangleup ADE\; \: and\; \: Rt\bigtriangleup CBF,$

$\because \left\{\begin{matrix} AE=CF & & \\ AD=CB & & \end{matrix}\right.$

$\therefore Rt\bigtriangleup ADE\cong Rt\bigtriangleup CBF,$

$\therefore DE=FB.$

$\therefore [ECBF]=\frac{1}{2}(EC+BF)\times CB$

$\: \: \: \: \: \: \: \: \: \: \: \:=\frac{1}{2}(AF+DE)\times AD=[ADEF].$

$\because [ECBF]+[ADEF]=[ABCD],$

$\therefore [ECBF]=\frac{1}{2}[ABCD].$

Jessica Wang - 5 years, 11 months ago

Potsawee Manakul
Jul 5, 2015

Nice solution Punpun :)

Jessica Wang - 5 years, 11 months ago

Undoing a pentagon

2 solutions

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