1971 AHSME Problems/Problem 28

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Problem

Nine lines parallel to the base of a triangle divide the other sides each into $10$ equal segments and the area into $10$ distinct parts. If the area of the largest of these parts is $38$, then the area of the original triangle is

$\textbf{(A) }180\qquad \textbf{(B) }190\qquad \textbf{(C) }200\qquad \textbf{(D) }210\qquad  \textbf{(E) }240$

Solution

[asy]  import geometry;  point B = origin; point A = (3,5); point C = (7,0); triangle t = triangle(A,B,C);  point D = B*9/10 + A/10; point E;  // Defining point E pair[] e = intersectionpoints(parallel(D,line(B,C)),A--C); E = e[0];  // Triangle ABC and Parallel Segment draw(t); draw(D--E);  // Point Labels dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(D); label("D",D,NW); dot(E); label("E",E,NE);  [/asy]

Let the triangle be $\triangle ABC$ with base $\overline{BC}$ and longest parallel segment $\overline{DE}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$, as in the diagram.

By the properties of transversals, we have $\measuredangle ABC = \measuredangle ADE$. Thus, by AA Similarity, we have $\triangle ADE \sim \triangle ABC$ (because they share $\angle BAC$). From the problem, we know that $\tfrac{AE}{AC}=\tfrac9{10}$, so, by similarity, $\tfrac{DE}{BC}=\tfrac9{10}$, and so $DE=\tfrac9{10}BC$.

Now, let $[\triangle ABC] = A$. Because $\tfrac{DE}{BE}=\tfrac9{10}$, we know that $\tfrac{[\triangle ADE]}{[\triangle ABC]}=(\tfrac9{10})^2=\tfrac{81}{100}$. From the problem, $[BCED]=38$, so $A=[\triangle ADE]+38=\tfrac9{10}A+38$. Solving for $A$ yields $A=\boxed{\textbf{(C) }200}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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