How do you solve, "A rectangular piece of paper,10cm by 24cm, is folded so that a pair of diagonally opposite corners coincide. Find the length of the crease."

Question

Vocaloid · Answer

@AZ when you get a chance

surjithayer · Answer

you have to find the diagonal.
$$d=\sqrt{24^2+10^2}$$
$$=\sqrt{576+100}$$
$$=\sqrt{676}$$
$$=26$$

AZ · Answer

The wording of this question confuses me. At first glance, it seems like they're just asking you to calculate the diagonal of the rectangle (and the numbers work perfectly too since it's a Pythagorean triple) and this would be the simplest solution and this interpretation makes most sense However, they don't need to be two diagonals from the opposite end as shown in the images on here: https://questioncove.com/study#/updates/52568d3be4b0ede9e446db69 http://quora.com/A-rectangular-paper-of-4-cm-in-length-and-3-cm-in-width-is-folded-so-that-a-pair-of-diagonally-opposite-vertices-coincide-How-long-is-the-crease-fold-mark-formed But then this diagram just confuses me https://math.stackexchange.com/questions/984177/solve-pythagoras-problem-solving But the question listed here seems more clear and the responses are very clear https://web2.0calc.com/questions/pythagorean-theorem-help_3 But my head is still trying to wrap around what they're exactly asking for haha

surjithayer · Answer

|dw:1616521458384:dw|

surjithayer · Answer

d=AC=26
Join CQ
Let BQ=x
AQ=24-x
PC=24-x
CQ=PC=24-x
$$in~\triangle~CBQ$$
$$(24-x)^2=x^2+10^2$$
$$576+x^2-48x=x^2+100$$
$$48x=576-100=476$$
$$x=\frac{476}{48}=\frac{119}{12}$$
$$24-x=24-\frac{119}{12}=\frac{24 \times 12 -119}{12}=\frac{288-119}{12}=\frac{169}{12}$$
$$y^2=(\frac{ 169 }{ 12})^2-13^2=\frac{ 169^2-169 \times 144 }{ 144}=169(\frac{ 169-144 }{ 144})=\frac{ 169 \times 25 }{ 144 }$$
$$y=\sqrt{\frac{ 169 \times 25 }{ 144 }}=\frac{ 13\times 5 }{ 12 }=\frac{ 65 }{ 12 }$$
$$2y=2\times \frac{ 65 }{ 12 }=\frac{ 65 }{ 6 }\approx 10.833$$
so reqd. length of crease 10.833...