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Posted: 2/7/2006 5:35:19 PM EDT
[#1]
is a trapezoid a four sided figure?
if so, I don't understand how only one end could be angled. |
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Posted: 2/7/2006 5:37:34 PM EDT
[#2]
What are you talking about
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Posted: 2/7/2006 5:40:12 PM EDT
[#3]
rhombus
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Posted: 2/7/2006 5:42:20 PM EDT
[#4]
A parallelogram.
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Posted: 2/7/2006 5:42:34 PM EDT
[#5]
a trapezoid is defined as :
A quadrilateral with exactly one pair of parallel sides, called bases. The other sides are legs. so your question doesn't make sense. |
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Posted: 2/7/2006 5:44:00 PM EDT
[#6]
Nope. Imagine a trapezoid with 2 90 degree corners only one end is angled. |
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Posted: 2/7/2006 5:44:11 PM EDT
[#7]
a parallelogram has a point on the opposite corner from the other "pointy" corner |
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Posted: 2/7/2006 5:45:01 PM EDT
[#8]
I'll guess:
a "right" trapezoid considering the angle between the base and one "leg" is 90 degrees |
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Posted: 2/7/2006 5:45:12 PM EDT
[#9]
if a trapezoid has 90 degree corners it would be a square. or rectangle |
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Posted: 2/7/2006 5:45:40 PM EDT
[#10]
WTF all of the corners are angled |
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Posted: 2/7/2006 5:46:49 PM EDT
[#11]
If you guys don't stop this I am gonna start over with a new thread!
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Posted: 2/7/2006 5:50:11 PM EDT
[#12]
Where did ya go?
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Posted: 2/7/2006 5:50:35 PM EDT
[#13]
It can be approximated by a combination rectangle and right triangle, and the area found from there, and that's all you need to know.
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Posted: 2/7/2006 5:51:15 PM EDT
[#14]
The correct answer is a duck.
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Posted: 2/7/2006 5:51:38 PM EDT
[#15]
n/m |
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Posted: 2/7/2006 5:53:07 PM EDT
[#16]
Here is a VERY BAD graphic representation of that he is talking about using only the keyboard
_____ |_____\ Fill in the gaps and you have what he is discribing 
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Posted: 2/7/2006 5:53:12 PM EDT
[#17]
Quadrilateral?
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Posted: 2/7/2006 5:54:06 PM EDT
[#18]
Irregular quadrilateral.
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Posted: 2/7/2006 5:54:15 PM EDT
[#19]
 THANK YOU! Does this shape have a name? |
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Posted: 2/7/2006 5:57:15 PM EDT
[#20]
What you say is only true is the quadrlateral has all side of equal lenth. Having one angle be 90° does not mean all the other angles have to be 90° also. 2 sides are parellel not the other 2, if they were we could start to figure out the angles. |
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Posted: 2/7/2006 5:58:18 PM EDT
[#21]
"I'll take Quadrilaterals for $100 Alex."
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Posted: 2/7/2006 5:58:26 PM EDT
[#22]
I can't recall any 'name' for it other than trapizoid. |
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Posted: 2/7/2006 5:59:39 PM EDT
[#23]
a kite ? nah, isosceles trapezoid |
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Posted: 2/7/2006 6:03:25 PM EDT
[#24]
Please note that I got it right six posts up, before the title was edited.
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Posted: 2/7/2006 6:03:28 PM EDT
[#25]
That shape is a trapezoid. The shape usually associated with the word trapezoid is in fact an isosceles trapezoid.
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Posted: 2/7/2006 6:03:45 PM EDT
[#26]
Ok the vertical height A=30", the length of the angle B=42" With that info can you tell me what C&D are?
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Posted: 2/7/2006 6:03:50 PM EDT
[#27]
It is a trapezoid.
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Posted: 2/7/2006 6:04:38 PM EDT
[#28]
No matter what else it is, it's a quadrilateral, which is any enclosed shape with four sides.
Here are some examples - regentsprep.org/Regents/math/quad/LQuad.htm mathforum.org/dr.math/faq/formulas/faq.quad.html |
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Posted: 2/7/2006 6:09:26 PM EDT
[#29]
well you can get the other leg of the triangle (sqrt864=29.4), but to get D i don't know. oops, dumb error |
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Posted: 2/7/2006 6:09:29 PM EDT
[#30]
C = D+29.4 in.
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Posted: 2/7/2006 6:12:22 PM EDT
[#31]
No, insufficient information. However, the minimum that C can be is zero. So, D = C + [42^2 - 30^2]^(1/2) = C + 29.4 as someone else already posted above. |
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Posted: 2/7/2006 6:16:36 PM EDT
[#32]
Ok here is what I get by rough measuring A=30" B=42" C=77.375" D=47.25" Does that compute? |
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Posted: 2/7/2006 6:18:23 PM EDT
[#33]
C and D can be whatever you want (larger than 0)- there are no proportions or other info to check them. but close enough, since 47.25 + 29.4 = 76.65 which if someone was kinda rough on the tape measure would be close enough to 77 3/8 for gubmint work |
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Posted: 2/7/2006 6:20:40 PM EDT
[#34]
Indiana, of course.
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Posted: 2/7/2006 6:22:21 PM EDT
[#35]
I'll take "who cares" for $200 alex
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Posted: 2/7/2006 6:23:44 PM EDT
[#36]
With A,C, and D can you verify that B would be correct?
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Posted: 2/7/2006 6:25:11 PM EDT
[#37]
B = squareroot( A^2 + (C-D)^2)
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Posted: 2/7/2006 6:34:39 PM EDT
[#38]
Yes. For a basic quadrilateral you'll need three measurements to determine #4. And, FWIW, it's not an Irregular Quadrilateral, it's just a Quadrilateral. The "shape" does not have a geometrical name. regentsprep.org/Regents/math/quad/LQuad.htm So, where's the plot of land you're trying to determine acreage for?
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Posted: 2/7/2006 6:51:04 PM EDT
[#39]
Mirror for my new mini-bar.
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Posted: 2/7/2006 6:57:28 PM EDT
[#40]
Tin Foil is shiny and you don't really have to measure it. Just staple it up and use a razor blade to trim it. |
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