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Home » Quadrangles » Trapezoids » Line Parallel to the Base of a Trapezoid

Line Parallel to the Base of a Trapezoid

Last updated: Sep 30, 2019 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

There are a number of interesting features of lines drawn parallel to the bases of a trapezoid. We saw one of these in this problem- a line parallel to the bases and connecting the midpoints of the legs.

If the bases of the trapezoid are a and b, that line's length is equal to ½(a+b). We will later deal with another such interesting line - a line parallel to the bases of the trapezoid which partitions it into two similar trapezoids.

But here, we'll consider the following line- drawn parallel to the bases of the trapezoid, and partitioning the trapezoid into 2 equal areas. This line is sometimes called the area bisector.

Problem

ABCD is a trapezoid, AB||CD. Ef is parallel to the bases (EF||AB, EF||AB), and divides ABCD into two equal areas. Find a formula for the length of EF using AB and CD.

trapezoid with two equal parts

Strategy

Let's call the top base 'a', the bottom base 'b', and the middle line 'c'.

Line c creates two trapezoids within the original trapezoid. The areas of the two small trapezoids are equal, so let's call that area 'S', and the area of the big trapezoid is 2S.

We'll write out the formulas for the areas of the two small trapezoids using the bases, a, b, and c, and then use some basic algebra to solve for c, making use if the fact that the areas are equal, and that the combined heights of the smaller trapezoids is equal to the height of the large one: h1+h2=h

trapezoids with heights

Solution

For the top small trapezoid:
(1) S=(a+c) · h1 / 2 → h1 = 2S/(a+c)

For the bottom small trapezoid:
(2) S=(b+c) · h2 / 2 → h2 = 2S/(b+c)

For the large trapezoid, h= h1+h2 and the area is 2S, so
(3) 2S = (a+b)· h /2 = (a+b)·[2S/(a+c)+2S/(b+c)]/2
(4) 1 = (a+b)/[1/(a+c) +1/(b+c)]/2 //divide both sides by 2S
(5) 2 = (a+b)/(a+c) +(a+b)/(b+c) //rearrange
(6) 2·(a+c)·(b+c) = (a+b)(b+c)+(a+b)(a+c) //common denominator and cross multiply
(7) 2ab+2ac+2cb+2c2=ab+ac+b2+bc+a2+ac+ba+bc
(8) 2c2=b2+a2
(9) c=√[(b2+a2)/2]

So c is the quadratic mean (or Root Mean Square - RMS) of the two bases of the trapezoid.

« Trapezoid Midsegment Theorem
Area of Semicircle »

About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

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