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In a trapezium the lengths of the parallel sides are 10 and 24. $$P$$, $$Q$$ are the mid points of the non parallel sides. If the line $$PQ$$ intersects the diagonals in $$R$$ and $$S$$ then $$RS$$ is equal to
Let the trapezium be $$ABCD$$ with $$AB = 24$$, $$CD = 10$$ and $$P$$ the mid point of $$AD$$. The line $$PQ$$ is parallel to the bases. In $$\triangle ABD$$ it meets $$BD$$ at $$R$$ with $$PR = \frac{AB}{2} = 12$$, and in $$\triangle ACD$$ it meets $$AC$$ at $$S$$ with $$PS = \frac{CD}{2} = 5$$. Hence $$RS = 12 - 5 = 7$$.
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