Second Proof: Area of a Trapezium (Trapezoid US)

I saw an interesting animation on twitter recently and decided to have a go at a formal proof (formal ish).

In essence, we’re going to chop a bit off the top of a trapezium and turn the whole thing into a triangle with the same height, and a base of (a+b). I stupidly called b ‘x’ in my scribblings, but you get the idea:

IMG_0983

Now extend AD by length ‘x’ and call the new point ‘E’, and create line BE:

IMG_0987

Angles CFB and DFE are clearly equal. Now box in BE*:

 

IMG_0988

We now have a rectangle, with diagonal BE. Hence we have two (green) triangles that are congruent, and hence angles EBH and BEG are equal. If two angles of these slim (red) triangles are equal, then the third must also be equal, and seeing as they both have an equal length side, they must be congruent.

IMG_0990

And if they’re congruent, then the area of the trapezium ABCD must be equal to the area of the triangle ABE.

The base of triangle ABE is ‘a’ + ‘x’ (imagine x is b… doh!), and the height of the triangle is the height of the trapezium, hence the area of the trapezium has the formula ((a+b)/2)*h

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One thought on “Second Proof: Area of a Trapezium (Trapezoid US)

  1. Pingback: The Emergency Maths Kit Strikes Again! | Solve My Maths

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