The Area of a Trapezoid
Date: 1/27/96 at 14:18:9 From: Anonymous Subject: Area of a Trapezoid My 7th grade daughter has been assigned a project to derive the area of a trapezoid, i.e. not simply look up the formula in CRC. My math was in prehistoric times and I have banged my head for a couple of days on this. Any direction on how to derive this from basic geometric principles would be appreciated i.e. not just 1/12 . l3 . sqrt(2). Thanks
Date: 2/5/96 at 9:38:43 From: Doctor Elise Subject: Re: Area of a Trapezoid Hi! Okay. Draw yourself a trapezoid - any old trapezoid. It has two parallel sides of different lengths, and two non-parallel sides of different lengths (unless it's an isosceles trapezoid, and then these two are the same length). Just so we can talk about this trapezoid, draw the two parallel lines horizontally with the longer side on the bottom. Now label the upper left corner "A", the upper right corner "B", the lower right "C", and the lower left "D". Yay! Now we have a trapezoid we can talk about. You don't do trigonometry yet in 7th grade, do you? I'll assume not. In order to find the area of this trapezoid, we have to know a few things: The length of the line segment AB (how long is the side between A and B) The length of the line segment DC The length of AD, and the length of BC, The height of the trapezoid, OR the angles of the corners. Okay. Look at your trapezoid again, and draw a line from "A" perpendicular to DC. Call the point where the new line hits DC point E. Then draw another line from B perpendicular to DC, and call that point F. Still with me? You should have a rectangle and two triangles, like this: A _____________________________________________ B /| |\ / | | \ / | | \ / | | \ / | | \ / | | \ /______|____________________________________________|______\ D E F C Since we know that AB and DC are parallel line segments, and we drew AE and BF perpendicular to DC, we know a bunch of things: AB = EF (because ABEF is a rectangle) DAE is a right triangle (because angle AED is a right angle) BCF is a right triangle (because angle BFC is a right angle). So, the area of the trapezoid is equal to the area of the rectangle plus the area of each of the triangles. Let's take a trapezoid where AB = 5, DC = 9, the height (which is AE AND BF) = 3, and lets say it's an isosceles trapezoid where AD = BC. You might want to label your trapezoid picture so you can follow along. Okay. We already know that AB = EF, because it's a rectangle, so they're both 5. And we know that the height is 3, because they told us. So the area of the rectangle is the base times the height, 5 * 3 = 15. Since this is an isosceles trapezoid, we know that the two triangles are exactly the same size and proportions, which we call "congruent" triangles. (If you cut two congruent triangles right out of the paper, turn them around the right way, and put them on top of each other, they are exactly the same.) So DE = FC. Since we know EF = 5, (because AB = 5), and DC = 9. DE has to be half of DC - AB so we know that DE and FC each have to be 2. The area of any right triangle is 1/2 base times height. We know the base is 2, and the height is 3 (because it's our old friend AE), so the area of one of these triangles is 1/2 * 2 * 3 = 3. And we have two of these triangles, so the total area of this trapezoid would be 15 + 3 + 3 = 21. If I were going to make a generic formula for an isoceles trapezoid, I'd do it this way: AREA = RECTANGLE + TRIANGLE + TRIANGLE AREA = base * height + 1/2 base1 * height + 1/2 base2 * height AREA = AB * AE + 1/2 (1/2(DC - AB))* AE + 1/2 (1/2(DC - AB)) * AE AREA = AB * AE + 1/2(DC - AB) * AE AREA = AE * (AB + 1/2(DC - AB)) AREA = AE * (AB + 1/2 * DC - 1/2 * AB) AREA = AE * (1/2 * AB + 1/2 * DC) AREA = 1/2 * AE * (AB + DC) If you check my work here, and plug in AE = 3, AB = 5, and DC = 9, do you still get 21? Hope this helps! -Doctor Elise, The Math Forum
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