Difference between revisions of "1971 AHSME Problems/Problem 28"
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− | <math>\boxed{\textbf{(C) }200}</math>. | + | Let the triangle be <math>\triangle ABC</math> with base <math>\overline{BC}</math> and longest parallel segment <math>\overline{DE}</math> with <math>D</math> on <math>\overline{AB}</math> and <math>E</math> on <math>\overline{AC}</math>, as in the diagram. |
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+ | By the properties of [[transversals]], we have <math>\measuredangle ABC = \measuredangle ADE</math>. Thus, by [[AA Similarity]], we have <math>\triangle ADE \sim \triangle ABC</math> (because they share <math>\angle BAC</math>). From the problem, we know that <math>\tfrac{AE}{AC}=\tfrac9{10}</math>, so, by similarity, <math>\tfrac{DE}{BC}=\tfrac9{10}</math>, and so <math>DE=\tfrac9{10}BC</math>. | ||
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+ | Now, let <math>[\triangle ABC] = A</math>. Because <math>\tfrac{DE}{BE}=\tfrac9{10}</math>, we know that <math>\tfrac{[\triangle ADE]}{[\triangle ABC]}=(\tfrac9{10})^2=\tfrac{81}{100}</math>. From the problem, <math>[BCED]=38</math>, so <math>A=[\triangle ADE]+38=\tfrac9{10}A+38</math>. Solving for <math>A</math> yields <math>A=\boxed{\textbf{(C) }200}</math>. | ||
== See Also == | == See Also == | ||
{{AHSME 35p box|year=1971|num-b=27|num-a=29}} | {{AHSME 35p box|year=1971|num-b=27|num-a=29}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 12:02, 7 August 2024
Problem
Nine lines parallel to the base of a triangle divide the other sides each into equal segments and the area into
distinct parts.
If the area of the largest of these parts is
, then the area of the original triangle is
Solution
Let the triangle be with base
and longest parallel segment
with
on
and
on
, as in the diagram.
By the properties of transversals, we have . Thus, by AA Similarity, we have
(because they share
). From the problem, we know that
, so, by similarity,
, and so
.
Now, let . Because
, we know that
. From the problem,
, so
. Solving for
yields
.
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.