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

#1**+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

CPhill
Jul 18, 2017

#2**+2 **

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

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!

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

hectictar
Jul 19, 2017

#3**+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!!!!

CPhill
Jul 19, 2017

#5**+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** = ?

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

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

heureka
Jul 19, 2017