Standard letter paper is 8.5 inches wide and 11 inches long. To the nearest tenth of an inch, what is the diagonal distance d across the paper?
Standard letter paper is 8.5 inches wide and 11 inches long. To the nearest tenth of an inch, what is the diagonal distance d across the paper?
Can somebody else than Eliz be more willing to show some work so I can see if i did it correctly. Eliz. Thanks for the "tip" but you didn't really help much.
SaeLK, if you're unfamiliar with the Pythagorean Theorem, then you need classroom help...or at least (if you're not too busy or tired) look it up: use Google
Have a nice day.
I'm just an imagination of your figment !
If you've done the exercise, then please show your work. We'll be glad to check to see if you did this correctly.Originally Posted by SaeLk
Note: Since you used some method other than the Pythagorean Theorem (or, which is the same thing, the Distance Formula), please state what method you used when you reply.
Thank you.
Eliz.
[tex]a^{2}+b^{2}=c^{2}[/tex]
I'll let you write the letters on the paper and maybe figure out how to fold it just right.
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
8.5^2 in + 11^2 in.
a^2 + b^2 = d ?
d = 193.25 or d = 13.901.....
72.25 + 121 = 193.25
This method I used to get the answer.
but my first one was 11 in - 8.5 in. = 2.5 in
then 11 + 2.5 = 13.5 in was my answer.
I'm not sure which answer is the correct one. or even if i have a correct answer.
Hey, well done Saelk!
The math gurus like to see:
let d = diagonal; then:
d^2 = 8.5^2 + 11^2
d^2 = 72.25 + 121
d^2 = 193.25
d = sqrt(193.25)
d = ~13.90143....
d = 13.9 (nearest 10th)
Now take a sheet of 8.5 by 11, draw the diagonal,
then measure it with your ruler: 13.9 looks good ?
And if you CUT along the diagonal, you're left with 2 triangles,
same size and with right angle, right?
I'm just an imagination of your figment !
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