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!
52
I see where –– comes from, that's obvious. Where did the 1 come from? Thanks.
2
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?
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