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Question from Martin, a teacher:

Given the area of an isosceles trapezoid is 150, that the difference of its bases is 5 (that is b1 - b2 = 5), and that its equal sides are 10 greater than two-thirds of the sum of its bases (that is, s1 = s2 = 2/3(b1 +b2) + 10
Solve the problem using an incorrect Babylonian formula for the area of a trapezoid A = (b1 + b2)/2 x (s1 + s2)/2 Find the length of the sides of the isosceles trapezoid.

Hi Martin,

I think what is being asked is

Given that

b1 - b2 = 5
s1 = s2 = 2/3(b1 + b2) + 10 and
(b1 + b2)/2 × (s1 + s2)/2 = 150

find s1.

Since s1 = s2 the last equation is

(b1 + b2)/2 × s1 = 150            (*)

From the second equation

b1 + b2 = 3/2 (s1 - 10)

Substitution into equation (*) gives

3/4 (s1 - 10) × s1 = 150

which simplifies to

s12 - 10 s1 - 200 = 0

Factor and solve for s1.

Penny

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