1954 AHSME Problems/Problem 29

Problem

If the ratio of the legs of a right triangle is $1: 2$ , then the ratio of the corresponding segments of the hypotenuse made by a perpendicular upon it from the vertex is:

$\textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5$

Solution

Let $\triangle ABC$ be a right triangle with right angle at $B$ , $AB = 2$ , and $BC = 1$ . Let the altitude from $B$ to $AC$ intersect $AC$ at $D$ . From the Pythagorean Theorem, $AC = \sqrt{1^2 + 2^2} = \sqrt5$ , so $BD = \frac{1\cdot 2}{\sqrt5} = \frac{2\sqrt5}{5}$ . Since $\triangle ADB \sim \triangle BDC \sim \triangle ABC$ , $\frac{AD}{BD} = 2$ and $\frac{CD}{BD} = \frac12$ . Therefore, $AD = \frac{4\sqrt5}{5}$ and $CD = \frac{\sqrt5}{5}$ , so $CD:AD = \boxed{\textbf{(A) }1:4}$ .

1954 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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1954 AHSME Problems/Problem 29

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