Difference between revisions of "1965 AHSME Problems/Problem 12"
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==See Also== | ==See Also== | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Latest revision as of 17:02, 18 July 2024
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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