Question

Given: △ABC, CM ⊥ AB , BC=5, AB=7, CA=4sqaureroot2 Find: CM.

Answer 1

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.

Answer 2

Remark

This is a nice problem. Thanks for posting it.

The general plan is to find BM and from that CM. You need 2 equations to do that.

Step One

Set up the two equations.

(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32

BM^2 + CM^2 = 5^2 = 25

Step Two

Subtract the two equations.

(7 - BM)^2 + CM^2  = 32

BM^2 + CM^2         = 25

(7 - BM)^2 - BM^2 = 7               (3)

Step three

Expand the left side of the new equation labeled (3)

49 - 14BM + BM^2 - BM^2 = 7    

Step 4

Simplify And Solve

49 - 14BM = 7              Subtract 49 from both sides.

-49 - 14BM = 7 - 49

- 14BM = - 42              Divide by - 14

BM = -42 / - 14

BM = 3

Step  Five

Find CM

CM^2 + BM^2 = 5^2

CM^2 + 3^2 = 5^2        Subtract 3^2 from both sides.

CM^2 = 25 - 9            

CM^2 = 16                     Take the square root of both sides.        

sqrt(CM^2) = sqrt(16)

CM = 4    < Answer