Answer:
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
_____
<em>Addendum to the answer</em>
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
