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 $h$.
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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, $\frac{a}b = \frac{y}{y+h} \implies y = \frac{ah}{b-a}$
Therefore, the two areas are:
Smaller triangle: $\frac{a^2h}{2(b-a)}$
Larger triangle: $\frac{hb}{2b(b-a)}$
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, $\frac{a}b = \frac{y}{y+h} \implies y = \frac{ah}{b-a}$
Therefore, the two areas are:
Smaller triangle: $\frac{a^2h}{2(b-a)}$
Larger triangle: $\frac{hb}{2b(b-a)}$
for the smaller triangle your OK but the larger triangle is $\frac{hb^{2}}{2(b-a)}$ 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, $\frac{a}b = \frac{y}{y+h} \implies y = \frac{ah}{b-a}$
Therefore, the two areas are:
Smaller triangle: $\frac{a^2h}{2(b-a)}$
Larger triangle: $\frac{hb}{2b(b-a)}$
for the smaller triangle your OK but the larger triangle is $\frac{hb^{2}}{2(b-a)}$ if im wrong tell me Oh yea... opps, but the $b$ on the wrong side.