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A rectangular piece of paper $ABCD$ is folded so that edge $CD$ lies along edge $AD,$ making a crease $DP.$ It is unfolded, and then folded again so that edge $AB$ lies along edge $AD,$ making a second crease $AQ.$ The two creases meet at $R,$ forming triangles $PQR$ and $ADR$. If $AB=5\mbox{ cm}$ and $AD=8\mbox{ cm},$ what is the area of quadrilateral $DRQC,$ in $\mbox{cm}^2?$

 

 

Thank you!

 Dec 6, 2020
 #1
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Area of DRQC = 52/2 - 1

 Dec 6, 2020
 #2
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                     52

I see where  ––  comes from, that's obvious.  Where did the 1 come from?  Thanks. 

                     2  

Guest Dec 6, 2020
 #3
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PQ = 2             22 / 4 = 1           area of PQR = 1

Guest Dec 6, 2020
 #4
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Okay, with your help, I understand it.  Here's what I did:   

 

When you fold the corner up, that leaves          BP  =  8 – 5  =  3  

And when you fold the other corner down         CQ  =  8 – 5  =  3  

So                                                                PQ  =  BC – (BP + CQ) 

                                                                    PQ  =  8 – 6  =  2  

 

Once you have PQ = 2 you can calcularate QR  

                                                         

PQ is a diagonal of a square,               2              2 • sqrt(2)  

so the side of the square is               –––––   =   –––––––––  

                                                       sqrt(2)              2  

 

The 2's cancel out leaving        QR = sqrt(2) 

And PR is the same                PR = sqrt(2)  

 

                                                        1

Thus the area of triangle PQR is         ––– • sqrt(2) • sqrt(2)  =  1  

                                                        2    

Thanks.

Ron  

 

I had to edit it several times to get stuff to align properly.  What happened to WYSIWYG?  

.

Guest Dec 6, 2020
edited by Guest  Dec 6, 2020
edited by Guest  Dec 6, 2020
edited by Guest  Dec 6, 2020
edited by Guest  Dec 6, 2020
edited by Guest  Dec 6, 2020

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