1) An isosceles triangle and an isosceles trapezoid
share a common base (solid line) and have the same
area. However the trapezoid has 1/3 the height of
the larger triangle. What fraction of the longer base
of the trapezoid extends outside the triangle?
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Let B = length of the common base and
    Bt =
length of the trapezoid's second base. (the long one)
    H = the triangle's height and
    Ht = the trapezoid's height = H/3
- The formula for the area of a triangle is       (B H) / 2
- The formula for the area of the trapezoid is
    (B + Bt) Ht / 2
- Set these two areas equal to each other and solve for Bt =
BH/2 = (B+Bt) (H/3)/2
BH = (B+Bt) H/3
B = (B+Bt)/3
3B = B + Bt
2B = Bt therefore Bt = 2B.
In other words, the second base of the trapezoid (Bt) is twice the length of the other base (B).
- The 2 triangles are similar, so the length of the second base that is inside the triangle is 2⁄3 B because it's height is 2⁄3 of the big triangle's height.
- The portion that is outside the base is
2B - 2⁄3B = 4⁄3B
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The fraction of the long trapezoid base that is outside the triangle is 4⁄3B / 2B =
2⁄3
So, this means the longer base of the trapezoid, which is twice the length of the other base, B, is cut into 3 equal parts, each of which is 2⁄3B in length, so 2 of those 3 equal parts are outside the triangle!
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