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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