since this is an isosceles triangle we could split and have a perpendicular bisector. since it is a bisector we know the bottom part of it would be 5 (half of 10) and the hypothenuse would be 13. this is a right triangle since the line we got was a perpendicular bisector.
using the pythagorean theorem we get height perpendicular bisector is 12.
13^2 - 5^2 = 144 = 12^2 so side is 12 and it would be the altitude to BC
now we find the area which is 1/2 ( b ) ( h ) we got the height/altitude which is 12 when base is 10
so area will be
1/2 ( 12 ) ( 10 ) = 60
using AB as the base the perpendicular which is the height would be solved by
2 A / b = h
120 / 13 would be the altitude to BC
you could use heron's formula/ hero's formula if you want but it would be too tedious