Question

find the area of triangle of its two side measure 6 in and 9 in and the bisector angle between sides id 4 square root of 3

Answer 1

Find the area of a triangle if its two sides measure 6in.6in. and 9in.9in., and the bisector of the angle between the sides is 43√43 in. I'm thinking of using the formula AA=12bh12bh I can't find the base or height, I used the Angle bisector formula which is l=l=ab[(a+b)2−c2]√a+bab[(a+b)2−c2]a+b So i found out C which i think is the base should i multiply it by 22? because I think it's the half. From here I'm lost

geometry

Answer 2

Find the area of a triangle if its two sides measure 6in.6in. and 9in.9in., and the bisector of the angle between the sides is 43√43 in. I'm thinking of using the formula AA=12bh12bh I can't find the base or height, I used the Angle bisector formula which is l=l=ab[(a+b)2−c2]√a+bab[(a+b)2−c2]a+b So i found out C which i think is the base should i multiply it by 22? because I think it's the half. From here I'm lost

geometry