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A rectangular sheet of paper measures 12 inches by 9 inches. One corner is folded onto the diagonally opposite corner and the paper is creased. What is the length in inches of the crease? Express your answer as a mixed number.

michaelcai  Jul 18, 2017
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6+0 Answers

 #1
avatar+76929 
+3

 

 

I did this with a sheet of paper [ pretending that it was 9  x 12 ] .......I orient the sheet with the smaller side at the top and fold from the top left corner to the bottom right corner

 

First.....fold as instructed......at the "bottom" of the sheet we will have a right triangle whose leg lengths are 9 and x  and whose hypotenuse is 12 - x

 

Using the Pythagorean Theorem to solve for x, we have

 

sqrt ( 9^2 + x^2)  = 12 - x     square both sides

9^2 + x^2  = x^2 - 24x + 144   subtract x^2 from both sides

81 = -24x + 144  rearrange

24x  = 63

x = 63/24  =  21 / 8

 

Now.....unfold the paper......if we draw a line from the bottom left fold of the crease  parallel to the bottom edge of the paper all the way across to the other side, we will have a right triangle whose sides are 9  and [12 - 2(21/8)] = [ 12 - 42/8] = [96 - 42] / 8 = ]54/8] =  27/4........and the crease will form the hypotenuse of this right triangle

 

So......using the Pythagorean Theorem again the length of the crease is given by

 

sqrt [  9^2  + (27/4)^2 ] =  sqrt [ 81 + 729/16] =

sqrt [ 1296 + 729] / 4 =  sqrt [ 2025]/ 4 =  45/4  inches = 11 + 1/4 inches

 

 

cool cool cool

CPhill  Jul 18, 2017
edited by CPhill  Jul 18, 2017
 #2
avatar+4716 
+2

I didn't understand at first so I tried doing it, I'll share some pictures in case it helps smiley

 

Since I didn't have any paper that was big enough to make it 12 inches by 9 inches,

I made it 12 centimeters by 9 centimeters.

 

 

 

 

Then, fold from the top right corner to the bottom left right corner ( I did it backwards..oops! ).

 

 

At that point, you can find what  x  is.

 

This picture shows why the side length is  x  and the hypotenuse is  12 - x  .

 

 

Then unfold it and draw a the line parallel to the bottom edge like this...

 

 

Now you're left with a right triangle where the legs are  9  and  12 - 2x  cm long, and the hypotenuse is the crease.  And looky here...it is right about 11 and a quarter centimeters long! laugh

 

 

I dont think I ever would have figured that one out on my own!!

hectictar  Jul 19, 2017
 #3
avatar+76929 
+1

 

 

Thanks for those pics, hectictar....I tried to explain what I did as well as possible, but I see how it can be difficult to follow....this one might be a good candidate for a "YouTube"  presentation....!!!!!

You really deserve more than a single point for all that effort...!!!

 

BTW.....I really liked this one....I still wonder how someone figured out that we would get a rational number as an answer....it seems, at first blush, that we would end up with some nasty radical....LOL!!!!

 

 

cool cool cool

CPhill  Jul 19, 2017
 #4
avatar+4716 
+1

Yeah, I was thinking a video might be helpful..

 

I also wouldn't have expected a clean answer either...it beats me!!

hectictar  Jul 19, 2017
 #5
avatar+18612 
+2

A rectangular sheet of paper measures 12 inches by 9 inches.

One corner is folded onto the diagonally opposite corner and the paper is creased.

What is the length in inches of the crease? Express your answer as a mixed number.

 

Let a = 12 inches
Let b = 9 inches
Let x = crease = ?


\(\begin{array}{|rcll|} \hline x &=& \frac{b}{a}\cdot \sqrt{a^2+b^2} \\ &=& \frac{9}{12}\cdot \sqrt{12^2+9^2} \\ &=& \frac{3}{4}\cdot \sqrt{225} \\ &=& \frac{3}{4}\cdot 15 \\ &=& \frac{45}{4} \\ &=& 11 \frac14\ \text{ inches}\\ \hline \end{array}\)

 

 

\(\begin{array}{|lrcll|} \hline (1) & b^2 &=& p\cdot a \\ & p &=& \frac{b^2}{a} \\\\ (2) & x^2 &=& p\cdot(p+a) \\ & x^2 &=& \frac{b^2}{a} \cdot(\frac{b^2}{a}+a) \\ & x^2 &=& \frac{b^2}{a} \cdot \frac{a}{a} \cdot(\frac{b^2}{a}+a) \\ & x^2 &=& \frac{b^2}{a^2} \cdot a \cdot(\frac{b^2}{a}+a) \\ & x^2 &=& \frac{b^2}{a^2} \cdot(b^2+a^2) \\ & x^2 &=& \frac{b^2}{a^2} \cdot(a^2+b^2) \\ & \mathbf{x} & \mathbf{=} & \mathbf{ \frac{b}{a} \cdot \sqrt{a^2+b^2} } \\ \hline \end{array}\)

 

 

laugh

heureka  Jul 19, 2017
edited by heureka  Jul 19, 2017
edited by heureka  Jul 19, 2017
 #6
avatar+76929 
+1

 

Thanks, heureka......your answer is more elegant.....but hectictar and I had more fun folding paper........!!!!!!

 

 

cool cool cool

CPhill  Jul 20, 2017

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