1971 AHSME Problems/Problem 28

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

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

$\boxed{\textbf{(C) }200}$ .

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.

Page

Toolbox

Search

1971 AHSME Problems/Problem 28

Problem

Solution

See Also