Difference between revisions of "2018 AMC 10A Problems/Problem 9"
Ishankhare (talk | contribs) (Created page with "All of the triangles in the diagram below are similar to iscoceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each of the 7 smallest triangles has area 1, and <ma...") |
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<math>\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 </math> | <math>\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | Let <math>x</math> be the area of <math>ADE</math>. Note that <math>x</math> is comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area ratio of <math>9:16</math>). Thus, we have <cmath>x=7+\dfrac{9}{16}x</cmath> This gives <math>x=16</math>, so the area of <math>DBCE=40-x=\boxed{24}</math>. |
Revision as of 15:27, 8 February 2018
All of the triangles in the diagram below are similar to iscoceles triangle , in which
. Each of the 7 smallest triangles has area 1, and
has area 40. What is the area of trapezoid
?
Solution
Let be the area of
. Note that
is comprised of the
small isosceles triangles and a triangle similar to
with side length ratio
(so an area ratio of
). Thus, we have
This gives
, so the area of
.