What is the name of this shape with unequal sides?

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The shape in question is identified as a pentagon, specifically an irregular pentagon since not all sides are equal. The discussion revolves around finding the area of the shape, which is given as 3375 cm². Participants suggest subdividing the shape into simpler components, such as a trapezoid and a triangle, to calculate the area more easily. There is some confusion regarding the terminology, with one participant mistakenly referring to a parallelogram instead of a trapezoid. Ultimately, the correct value for X is determined to be 75 after addressing a calculation error.
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Hi FP! :smile:

A pentagon is a good name for this. A pentagon is just a thingy with 5 corners and 5 sides.
It isn't a regular pentagon though, for that we would require all sides and all angles to be equal.
 
Giant diamond.
 
Hi micromass! :) (I love this new "hi" thing lol!)

The reason I'm asking is that I'm trying to find X here

http://img10.imageshack.us/img10/3268/yjjjj.jpg

And I know the area is 3375 cm squared. I was hoping the relate the formula for area for this shape to X, but I'm unable to find its formula on my formula page (I can't find the shape, even)

Oh, and there's no "diamond" there :P
 
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Ah, I see. Well, I guess the area of a parallellogram is on your formula page?

Well, let's try to find a formula for the giant diamond (I love that name :biggrin:)

The trick is to subdivide the giant diamond into a parallellogram and a triangle. Just take the two mid-vertices (not the vertices on top and not the vertex on the bottom) and connect them with a line. Now the top is a parallellogram and the bottom is a triangle. Can you find the area of the parallellogram and the triangle?
 
micromass said:
Ah, I see. Well, I guess the area of a parallellogram is on your formula page?

Well, let's try to find a formula for the giant diamond (I love that name :biggrin:)

The trick is to subdivide the giant diamond into a parallellogram and a triangle. Just take the two mid-vertices (not the vertices on top and not the vertex on the bottom) and connect them with a line. Now the top is a parallellogram and the bottom is a triangle. Can you find the area of the parallellogram and the triangle?

I think you mean a trapezoid, not a parallelogram.
 
eumyang said:
I think you mean a trapezoid, not a parallelogram.

Yes, indeed. Sorry for the confusion! :blushing:
 
Femme_physics said:
The reason I'm asking is that I'm trying to find X here

The figure is symmetrical about that dotted centre line, so concentrate on finding the area of just one of those halves and make it easier for yourself.

Divide the shape up into simpler shapes whose area you can find, viz., rectangles and triangles. Looks like the height of the lowest triangle will be (X - 30). Add the areas together, and equate them to half of 3375 cm2

You end up with one equation and one unknown. Solve for X. :smile:
 
micromass said:
Ah, I see. Well, I guess the area of a parallellogram is on your formula page?

Well, let's try to find a formula for the giant diamond (I love that name :biggrin:)

The trick is to subdivide the giant diamond into a parallellogram and a triangle. Just take the two mid-vertices (not the vertices on top and not the vertex on the bottom) and connect them with a line. Now the top is a parallellogram and the bottom is a triangle. Can you find the area of the parallellogram and the triangle?
Hmm.. I see what you're saying! That's brilliant.

I can just take a trapezoid area and add the triangle's area, like thathttp://img695.imageshack.us/img695/5277/triy.jpg

Though from some reason I've still gotten the wrong X. Is this the right equation?
 
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  • #10
Yes, that seems to be the correct formula. Weird that you find the incorrect X now...
 
  • #11
My late Uncle Jor-el says it's our old family crest. He says it should have a big "S" in it.
 
  • #12
NascentOxygen said:
Divide the shape up into simpler shapes whose area you can find, viz., rectangles and triangles. Looks like the height of the lowest triangle will be (X - 30). Add the areas together, and equate them to half of 3375 cm2
That's a bit more work than Micromass' method, IMO.

Femme_physics said:
Though from some reason I've still gotten the wrong X. Is this the right equation?
What did you get for X, and what was the answer supposed to be?
 
  • #13
It's X = 75 which I actually got -- I forgot to multiply the other side by 2 *oops*

Thanks, everyone!
 

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