1965 AHSME Problems/Problem 12
Problem
A rhombus is inscribed in in such a way that one of its vertices is and two of its sides lie along and . If inches, inches, and inches, the side of the rhombus, in inches, is:
Solution
As in the diagram, suppose the rhombus is inscribed in with on , on , and on . Let the side length of the rhombus be . Because a rhombus is a parallelogram, its opposite sides are parallel. Thus, by AA similarity, , and so . Because and are sides of the rhombus, they both have length . Furthermore, , and , because, combined with sides of the rhombus, they form sides of the triangle. Thus, by substituting into the proportion derived above, we see that: \begin{align*} \frac{FE}{FB}&=\frac{DC}{DE} \\ \frac{x}{12-x}&=\frac{6-x}{x} \\ x^2&=72-12x-6x+x^2 \\ 18x&=72 \\ x&=4 \\ \end{align*}
Thus, the side of the rhombus has length .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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