Given: △ABC, CM ⊥ AB , BC=5, AB=7, CA=4sqaureroot2 Find: CM.
2 answers:
We assume M is on AB, so that CM is the altitude to AB.
The triangle altitude can be found different ways. We choose to find the angle opposite the irrational side length using the Law of Cosines:
... b² = a² + c² - 2ac·cos(B)
... (4√2)² = 5² + 7² - 2·5·7·cos(B)
... 32 = 74 -70·cos(B)
... cos(B) = (74 -32)/70 = 3/5
Then
... BM = BC·cos(B) = 5·(3/5) = 3
By the Pythagorean theorem, ...
... BC² = BM² + CM²
... 5² = 3² + CM²
... CM = √(25 -9) = 4
The length of CM is 4.
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A diagram helps.
In the first diagram, we marked angle A to be 45° after realizing that AMC is an isosceles right triangle. This let us use a triangle solver with whole numbers to check the answer (second attachment). The area being 14 tells us the altitude to the side of length 7 is (2·14)/7 = 4, as we computed above.
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