Math (geometry problem)

[Deerslayersh]

Your equation 3 is wrong. The two triangles are not similar.

Hmm, maybe. I just used similarity for that equation. I better not spend any more of my client's $$$ thinking about why they are similar. They look like similar triangles to me.

[meltingfeather]

Math (geometry problem)

Hmm, maybe. I just used similarity for that equation. I better not spend any more of my client's $$$ thinking about why they are similar. They look like similar triangles to me.

They are not— you can see an angle difference between the point-to-point of the rectangle and the long sides of the parallelogram.
That’s why you got the wrong answer.

[JustinJ]

Got 'er did fellas.

Looks good to me (I think ). Been awhile since college. What do the lines over the variables mean? Are you just indicating that its Line AB, etc?

[JustinJ]

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Don't know how you get to .5/8.25=b/4

edit: just see where you used similarity. Triangles aren't the same.

[jb5001]

Got 'er did fellas.

Looks good to me. Nice work.


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[Tyler's Outdoor]

Just to be a smart*ss, the shape is actually a parallelogram.

[SAC]

a2+b2= c2.

a2(half of c2) +b2 (.25) =c2. parenthesis for description only

sohcahota to find the angle in the corner.

45- that answer. gives you the corner to shortside angle

That angle is the same as the small angle in triangle the top left corner. essentially your cut angle. from there the math should get a lil easier.


I have no idea if that's right, or makes sense. Just how I think I would solve it

Edit: yeah what they said

[JTCowpoke]

Got 'er did fellas.

Nice

[Hardware]

Don't have 2 sides in the original problem

You gave the 4" and the 8.25 in you r first post along with .5 width. That is enough to back into the answer.

I was in a hurry and assumed .5 for a measurement but once I looked again that would be greater since its cut on angle but the basic math formula given in link was correct.

[JustinJ]

You gave the 4" and the 8.25 in you r first post along with .5 width. That is enough to back into the answer.

agreed. just not as simple as plugging those numbers into that calc

[JustinJ]

Got 'er did fellas.

THANK YOU!!!!


There's the answer boys

[hpdrifter]

Hmm, maybe. I just used similarity for that equation. I better not spend any more of my client's $$$ thinking about why they are similar. They look like similar triangles to me.

Should be a geometry theorem for that; parallel lines intersect a ray creates similar angles.

[175gr7.62]

Looked at this over lunch. Realized you don’t need the angles. The diagonal lines being parallel allow you to use similarity. Then you can just use Pythagorean’s Theorem.


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[175gr7.62]

NVM, looks like Deerslayer wins the prize.


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[meltingfeather]

NVM, looks like Deerslayer wins the prize.


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Nope-- the parallel lines are not parallel to the diagonal of the 4" x 8.5" rectangle. The triangles are not similar. If they were you would have gotten the same answer as the CAD.
I have to admit it was a good thought. When I first saw Deerslarersh's work I thought, "how did I miss that?" So simple-- and incorrect.