Area of Trapezoid Given Only the Side LengthsDate: 12/15/2008 at 11:08:48 From: Jeromy Subject: sides are 1, 2, 3, and 4. find the area of the trapezoid How would you find the area of a trapezoid without the height, but when all sides are given and they don't tell you which sides are parallel? Ex. given trapezoid has side lengths of 1,2,3, and 4. Find the area. How would you find the area without the height? Date: 12/15/2008 at 11:51:54 From: Doctor Peterson Subject: Re: sides are 1, 2, 3, and 4. find the area of the trapezoid Hi, Jeromy. Interesting question! You can find the area without knowing the height ahead of time, but you have to know which sides are parallel. Different pairs of parallel sides may yield different areas. To find the area given only the sides AND which are parallel, you can draw a parallelogram within the trapezoid: +---------+ / / \ / / \ / / \ +---------+-----------+ You will now know all three sides of the triangle, from which you can find its height. Let's try doing that with the obvious first choice of 2 and 4 for the parallel sides: 2 +---------+ / / \ 1/ 1/ \3 / 2 / 2 \ +---------+-----------+ 4 Looking closely at the triangle, we see that this is impossible! The Triangle Inequality Theorem says that the sum of any two sides must be greater than the other side, but 1+2 = 3. So the 2 and 4 sides can't be parallel. Can we come up with a rule to figure out which pairs of parallel sides are possible? Let's use variables this time: a +---------+ / / \ c/ c/ \d / a / b-a \ +---------+-----------+ b In order for the triangle to be possible (assuming, as shown, that b>a), we must have (by the Triangle Inequality Theorem) c + d > b - a c + (b-a) > d, so d-c < b-a d + (b-a) > c, so c-d < b-a The last two facts can be combined, and the assumption that b>a eliminated, by writing |d-c| < |b-a| The first inequality can be written similarly as |b-a| < c+d So we're looking for a pair of sides whose difference is between the sum and the difference of the other two sides: |d-c| < |b-a| < c+d There are 6 ways to choose a pair of parallel sides; let's see which fit this requirement: bases sides ------- ------- a b c d d-c b-a c+d --- --- --- --- --- --- --- 1 2 3 4 1 = 1 < 7 no 1 3 2 4 2 = 2 < 6 no 1 4 2 3 1 < 3 < 5 yes 2 3 1 4 3 > 1 < 5 no 2 4 1 3 2 = 2 < 4 no 3 4 1 2 1 = 1 < 3 no So there's only one way to place the sides that will work. (That may not always be true!) 1 +----+ / / \ 2/ 2/ \3 / 1 / 3 \ +----+-----------+ 4 Now we need to find the height: + /: \ 2/ :h \3 / : \ +---+-------+ 3 We can use the Pythagorean theorem to find the two parts of the base in terms of h, and then solve for h: sqrt(2^2 - h^2) + sqrt(3^2 - h^2) = 3 Try solving this; you'll get a fairly nice radical expression after some pretty hard work. Then you can use this in the formula for the area of a trapezoid to find the area you want. An alternative way to get the area is to find the area of the triangle using Heron's formula (see our FAQ on Formulas) and then add on the area of the parallelogram. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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