Difference between revisions of "1959 AHSME Problems/Problem 10"
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In <math>\triangle ABC</math> with <math>\overline{AB}=\overline{AC}=3.6</math>, a point <math>D</math> is taken on <math>AB</math> at a distance <math>1.2</math> from <math>A</math>. Point <math>D</math> is joined to <math>E</math> in the prolongation of <math>AC</math> so that <math>\triangle AED</math> is equal in area to <math>ABC</math>. Then <math>\overline{AE}</math> is: <math>\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6</math> | In <math>\triangle ABC</math> with <math>\overline{AB}=\overline{AC}=3.6</math>, a point <math>D</math> is taken on <math>AB</math> at a distance <math>1.2</math> from <math>A</math>. Point <math>D</math> is joined to <math>E</math> in the prolongation of <math>AC</math> so that <math>\triangle AED</math> is equal in area to <math>ABC</math>. Then <math>\overline{AE}</math> is: <math>\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6</math> | ||
Revision as of 13:58, 16 July 2024
Problem
In with , a point is taken on at a distance from . Point is joined to in the prolongation of so that is equal in area to . Then is:
Solution
Note that . Since , we have , so that . Therefore, . Thusly, our answer is , and we are done.