A trapezium has parallel sides of length equal to a and b (a<b), and the distance between the parallel sides is the altitude h. The extensions of the non-parallel lines intersect at a point that is a vertex of two triangles that have as sides the parallel sides of the trapezium. Express the areas of the triangles as functions of a,b and h.
Problem
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Tags: geometry, trapezoid, function, similar triangles
ssilwa
17.03.2013 20:42
Say that the two non parallel lines intersect at point X. Let the height of the smaller triangle with base a be y.
From similar triangles, ab=yy+h⟹y=ahb−a
Therefore, the two areas are:
Smaller triangle: a2h2(b−a)
Larger triangle: hb2b(b−a)
rahman
17.03.2013 22:28
ssilwa wrote:
Say that the two non parallel lines intersect at point X. Let the height of the smaller triangle with base a be y.
From similar triangles, ab=yy+h⟹y=ahb−a
Therefore, the two areas are:
Smaller triangle: a2h2(b−a)
Larger triangle: hb2b(b−a)
for the smaller triangle your OK but the larger triangle is hb22(b−a) if im wrong tell me
ssilwa
17.03.2013 22:44
rahman wrote: ssilwa wrote:
Say that the two non parallel lines intersect at point X. Let the height of the smaller triangle with base a be y.
From similar triangles, ab=yy+h⟹y=ahb−a
Therefore, the two areas are:
Smaller triangle: a2h2(b−a)
Larger triangle: hb2b(b−a)
for the smaller triangle your OK but the larger triangle is hb22(b−a) if im wrong tell me Oh yea... opps, but the b on the wrong side.