In triangle ABC, AB=4cm, AC=8cm and AM=3cm. AM is the median which meets BC and M. Then, BC is equal to
BM = MC = x which is half of BC
angle AMB = T
then angle AMC = 180 - T
cos angle AMB = = cos T
cos angle AMC = cos(180-T)=-cos T
now law of cosines
4^2 = x^2 + 3^2 - 2(3x)cos T
8^2 = 3^2 + x^2 - 2(3x)(-cos T)
16 = x^2 + 9 - 6 x cos T
64 = x^2 + 9 + 6 x cos T
------------------------add
80= 2 x^2 + 18
2 x^2 = 62
x^2 = 31
x = sqrt 31
2x = BC = 2 sqrt 31
about 11.1
To find the length of BC in triangle ABC, we can use the property of medians in a triangle. The property states that a median of a triangle divides the opposite side into two equal parts.
Let's label the midpoint of BC as D. Now, we have AD as another median of the triangle.
Since AM is a median, it divides BC into two equal parts, so we have BD = CD. Similarly, since AD is also a median, it divides BC into two equal parts, so we have BD = DC.
Therefore, BD = DC.
Now, let's consider triangle ABD. It is a right triangle with AB as the hypotenuse.
Using the Pythagorean theorem, we can find the length of BD.
AB^2 = AD^2 + BD^2.
Substituting the given values, we have:
4^2 = 3^2 + BD^2.
16 = 9 + BD^2.
BD^2 = 16 - 9 = 7.
Taking the square root of both sides, we get:
BD = sqrt(7).
Since BD = DC, BC can be found by doubling the length of BD:
BC = 2 * sqrt(7).
Therefore, BC is equal to 2 * sqrt(7) cm.