1959 AHSME Problems/Problem 26

Problem 26

The base of an isosceles triangle is $\sqrt 2$ . The medians to the leg intersect each other at right angles. The area of the triangle is: $\textbf{(A)}\ 1.5 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.5\qquad\textbf{(E)}\ 4$

Solution

Drop the altitude $AD$ from the vertex $A$ to base $BC$ , and note that it intersects the centroid $M$ . Then $BCM$ is an isosceles right triangle, so $MB=1$ , so $MD=\frac{\sqrt{2}}{2}$ . By centroid properties, $AD=3MD=\frac{3\sqrt2}{2}$ . Then the area of $ABC$ is $\frac12\sqrt{2}\frac{3\sqrt2}{2}=1.5\rightarrow\boxed{\textbf{A}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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All AHSME Problems and Solutions

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1959 AHSME Problems/Problem 26

Problem 26

Solution

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