Difference between revisions of "1972 AHSME Problems/Problem 18"
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<math>\textbf{(A) }3\textstyle\frac{2}{3}\qquad \textbf{(B) }3\frac{3}{4}\qquad \textbf{(C) }4\qquad \textbf{(D) }3\frac{1}{2}\qquad \textbf{(E) }3</math> | <math>\textbf{(A) }3\textstyle\frac{2}{3}\qquad \textbf{(B) }3\frac{3}{4}\qquad \textbf{(C) }4\qquad \textbf{(D) }3\frac{1}{2}\qquad \textbf{(E) }3</math> | ||
==Solution== | ==Solution== | ||
− | <math>\ | + | We begin with a diagram: |
+ | |||
+ | <asy> | ||
+ | pair A, B, C, D, E; | ||
+ | |||
+ | A = (0, 0); | ||
+ | B = (8, 0); | ||
+ | C = (7, 4); | ||
+ | D = (3, 4); | ||
+ | E = intersectionpoint(A--C, B--D); | ||
+ | |||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--C); | ||
+ | draw(B--D); | ||
+ | |||
+ | label("$A$", A, W); | ||
+ | label("$B$", B, SE); | ||
+ | label("$C$", C, NE); | ||
+ | label("$D$", D, NW); | ||
+ | label("$E$", E, 3W); | ||
+ | label("$11$", midpoint(A--C), 2S); | ||
+ | </asy> | ||
+ | |||
+ | The bases of a trapezoid are parallel by definition, so <math>\angle EDC</math> and <math>\angle EBA</math> are alternate interior angles, and therefore equal. We have the same setup with <math>\angle ECD</math> and <math>\angle EAB</math>, meaning that <math>\triangle ABE \sim \triangle CDE</math> by AA Similarity. We could've also used the fact that <math>\angle BEA</math> and <math>\angle DEC</math> are vertical angles. | ||
+ | |||
+ | With this information, we can setup a ratio of corresponding sides: | ||
+ | <cmath>\frac{AB}{CD} = \frac{AE}{CE} \implies \frac{2CD}{CD} = \frac{11 - CE}{CE}.</cmath> | ||
+ | And simplify from there: | ||
+ | <cmath> \begin{align*} | ||
+ | \frac{2CD}{CD} &= \frac{11 - CE}{CE} \ | ||
+ | 2 &= \frac{11 - CE}{CE} \ | ||
+ | 2CE &= 11 - CE \ | ||
+ | 3CE &= 11 \ | ||
+ | CE &= \frac{11}{3} = 3 \frac{2}{3}. | ||
+ | \end{align*} </cmath> | ||
+ | Therefore, our answer is <math>\boxed{\textbf{(A) }3\textstyle\frac{2}{3}.}</math> |
Latest revision as of 03:54, 23 June 2022
Problem
Let be a trapezoid with the measure of base
twice that of base
, and let
be the point of intersection of the diagonals. If the measure of diagonal
is
, then that of segment
is equal to
Solution
We begin with a diagram:
The bases of a trapezoid are parallel by definition, so and
are alternate interior angles, and therefore equal. We have the same setup with
and
, meaning that
by AA Similarity. We could've also used the fact that
and
are vertical angles.
With this information, we can setup a ratio of corresponding sides:
And simplify from there:
Therefore, our answer is