1975 AHSME Problems/Problem 23
Problem 23
In the adjoining figure and are adjacent sides of square ; is the midpoint of ; is the midpoint of ; and and intersect at . The ratio of the area of to the area of is
Solution
First, let's draw a few auxiliary lines. Drop altitudes from to and from to . We can label the points as and , respectively. This forms square . Connect .
Without loss of generality, set the side length of the square equal to . Let , and since is the midpoint of , would be . With the same reasoning, and
We can also see that is similar to . That means .
Plugging in the values, we get: . Solving, we find that . Then, . The area of and together would be . Subtract this area from the total area of to get the area of .
So, . The question asks for the ratio of the area of to the area of , which is .
The answer is . ~jiang147369
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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