In triangle abc, ab = 7, ac = 15, and the length of median am is 10. find the area of triangle abc.

  42

Step-by-step explanation:

We can make use of Heron's formula for the area of the triangle. Let x represent half the length of BC:

  MB = MC = x

Then the perimeter of ΔAMB is ...

  p1 = 7 + 10 + x

and the value used in Heron's formula, the semi-perimeter is ...

  s1 = (17+x)/2

The square of the area of ΔAMB is then ...

  A1² = s1·(s1 -x)·(s1 -10)·(s1 -7) = (-1/16)(x^4 -298x^2 +2601)

Similarly, the square of the area of ΔABC is ...

  A2² = s2·(s2 -x)·(s2 -10)·(s2 -15) = (-1/16)(x^4 -650x^2 +15625)

We want these two triangle areas to be the same, so we can solve for the value of x that makes it so.

  A1² - A2² = 0 = (-1/16)((x^4 -298x^2 +2601) -(x^4 -650x^2 +15625))

  352x^2 -13024 = 0 . . . . multiply by -16 and collect terms

  x^2 -37 = 0 . . . . . . . . . . . divide by 352

  x^2 = 37 . . . . . this is as far as we need to take it in order to find the area.

Substituting the value for x^2 into the expression for A1², we get ...

  A1² = (-1/16)((x^2 -298)x^2 +2601) = (-1/16)((37 -298)37 +2601) = 441

So, the area of ΔAMB is √441 = 21 square units. The area of ΔAMC is the same, so the area of ΔABC is ...

  area(ΔABC) = 2×area(ΔAMB) = 2×21 units²

  area(ΔABC) = 42 units²

Addendum to the answer

After thinking about this a little more, I realized the area is that of half a parallelogram with AB and AC as two sides and AM as half the diagonal. In other words, the area is the same as that of a triangle with sides 7, 15, and 2×10=20. A single straightforward application of Heron's formula gives the area as ...

  A = √(21(21-7)(21-15)(21-20)) = √(21(14)(6)) = 42

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