Question

The area of a trapezium is 6y⁵.

The sum of its parallel sides is 4y² .

Derive an expression for the perpendicular distance between the parallel sides.

Answer 1

Answer: 3y^3

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Work Shown:

A = area = 6y^5

b1+b2 = sum of the parallel bases = 4y^2

h = unknown height, i.e. distance between the parallel sides

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A = 0.5*h*(b1+b2)

6y^5 = 0.5*h*4y^2

6y^5 = 0.5*4y^2*h

6y^5 = 2y^2*h

2y^2*h = 6y^5

h = (6y^5)/(2y^2)

h = (6/2)*y^(5-2)

h = 3y^3

The parallel sides are separated by a distance of 3y^3 units.