Area of trapezium

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how can you find the area of trapezium for which all area are given but no width?

is there a simple formula?

see the attached image.
 

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I dont know what width you are referring to. You could break it up into two triangles and a rectangle and sum up the areas, or use the general rule for trapezoids:

[tex] A = \frac{b_1+b_2}{2}h [/tex] where b's represent the parallel sides.
 

Integral

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I do not know of a simple formula, but you can solve for h by using properties of the right triangles formed by cutting out the center rectangle.
 

HallsofIvy

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Without knowing more than just the lengths of the sides (in order to use whozum and integral's ideas you would need to know at least some of the angles at which the sides join). Imagine this trapezoid made from sticks pinned together so that the sides can rotate on the pins (you can change the angles). You can flex the trapezoid to get a number of different area with the same side lengths.
 
112
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that's right.

i think so. the information given does not give a unique trapezium.
you can keep on shearing it until you get zero area.

that's why it looks odd.

thanks.
 
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no it is possible to do what he says.
Suppose that we "slide" the right side of the trapezium along the upper width until it meets the starting point of the other side. Now we have a triangle with sides(speaking for the numerical example)4,6,7. We can find this one's area by using

[tex]\sqrt u(u-a)(u-b)(u-c) [/tex]

where u is (a+b+c)/2 and a,b and c are the sides of the triangle. Now that we know the area of the triangle we can find its height. This is also the height of the trapezium so we have the area of the trapezium
 
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I have provided a jpeg file to make myself more clear on what I'm talking about
 

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mathelord

it actually is possible,but the formula is complicated
 
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it is definately possible and pretty easy but there is no such formula or it maybe be complicated.
 

Integral

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http://home.comcast.net/~Integral50/Math/trapezoid.pdf [Broken] is the bulk of the alegbra.
 
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NateTG

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As long as the parralel sides [itex]a[/itex] and [itex]c[/itex] are of different lengths, you can use Heron's formula to find the height, and from there it's easy.
If the parralel sides are of the same length, then the area is not determined by the side lenghts.
 
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stupidkid said:
it is definately possible and pretty easy but there is no such formula or it maybe be complicated.
If that was an answer to my post, there exists such a formula and it is called Heron's Formula. It is used to find the area of a triangle only by using the sides. What I did is clear and correct. Integral provided just another way to find the area.

If you have a look at my attachment in my previous post you will understand what I mean.
 
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thanks a lot for the help.

actually my friend asked me this question, and i said to him
it is not possible because the trapezium is not unique.

heh he. i am wrooong.....

thanks.
 

NateTG

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sniffer said:
thanks a lot for the help.

actually my friend asked me this question, and i said to him
it is not possible because the trapezium is not unique.
The trapezoid is not necessarily unique. You are correct:

For example, consider a trapezod with side lenghts 1,1,1, and 1. It's quite easy to see that any quadrilateral with side lengths like that is going to be a trapezoid . However, since this trapezoid happens to be a parralelogram we know that the area of that trapezoid is going to be [itex]\sin\theta[/itex] where [itex]\theta[/itex] is the one of the corner angles.

Since the corner angle can vary, so can the sine of it.

There is also an IMHO pedantic argument that given just four numbers it's not necessarily possible to identify which pair of sides is parralel.
 

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