A (regular) trapezoid with bases a and b is a quadrilateral with two sides paral

Question

A (regular) trapezoid with bases a and b is a quadrilateral with two sides parallel and of lengths a and b where the midpoints of the bases line up: The Pythagoreans had the following idea for a mean (an average) of two numbers: "Any line parallel to the bases between the other two sides is a mean of a and b .'' As there are an infinite number of parallel lines to the bases, there are an infinite number of means. This exercise will investigate 4 of them. Find algebraic expressions (i.e. formulas) in the variables a and b only for each of the following means (so they are independent of the height of the trapezoid that we drew): The mean (m) that bisects the sides. The mean (m) that creates similar trapezoids. Hint consider the 2 interior trapezoids. (This is much more general than ) The mean (m) that bisects the area. Hint: You may want to draw and label the heights of the two trapezoids that the mean makes. (Note: this part may be the most challenging of the 4 parts) The mean (m) that goes through the intersection of the two diagonals. Hint: Label segments and use similar triangles. Compute the four means from - when a= 2 and b= 6. Rank these from largest to smallest.

Explanation / Answer

1) Drop a perpendicular from two ends of side 'a' to 'b'
now b is divided into 3 parts and mean 'm' is also divided into 3 parts.
the middle parts of both will be equal to a.
and as mean is passing through mid points the other two parts of mean are equal to half of other two parts of 'b'
the other two parts of b = (b-a)/2
so mean m = a+([(b-a)/2]/2)+([(b-a)/2]/2) = (b+a)/2

2)as they are similar trapeziums a/m = m/b
so m = sqrt(ab) = √ab

3)let height of first trapezium be h1 and lower one be h2

h1+h2 = h height of big trapezium

areas are equal 

so (h1/2)*(a+m) = (h2/2)*(m+b) = ((h1+h2)/4)*(a+b) = k

h1 = 2k/(a+m) and h2= 2k/(m+b)

h1+h2 = 4k/(a+b)

so 2k/(a+m) + 2k/(b+m) = 4k/(a+b)

so 1/(a+m) + 1/(b+m) = 2/(a+b)

solving this we get m

m = sqrt[(a^2 + b^2)/2]

 

4) mark vertices top left one is A, top right one is B, bottom left one C, and bottom right one D

now take triangle ABD

the diagonals intersection point devides mean into two equal parts = m/2

and let mean devide AD at E

so now taking similar triangles, DE/AD = m/2a

and in triangle ACD AE/AD = m/2b

AE+DE = AD

AD(m/2a) + AD(m/2b) = AD

so m/2a + m/2b = 1

so m = 2ab/(a+b)

 

 

now taking a= 2 and b =6

mean 1 = 4

mean 2 = sqrt(12)

mean 3 = sqrt(20)

mean 4 = 3

 

largest to smallest mean 3 > mean 1 > mean 2 > mean 4

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