Question

Find the area of a triangle if its two sides measures 6 inches and 9 inches, and the bisector of the angle between the sides is 4√3

Answer 1

$l= \cfrac{1}{b+c}\sqrt{bc[(b+c)^2-a^2]}$

is the formula for bisector, where a, b, c are lengths of sides of the triangle .

Hence if b= 6, c=9 and l= $4\sqrt{3}$

$a=(b+c)\sqrt{\cfrac{bc-l^2}{bc}}$ ,

$a=(6+9)\sqrt{\cfrac{6\cdot 9- (4\sqrt{3})^2}{6\cdot 9}}=5$ .

The area of a triangle is

$S=\sqrt{p(p-a)(p-b)(p-c)}$ , where

$p=\cfrac{a+b+c}{2}=\cfrac{5+6+9}{2}=10$ .

$S=\sqrt{10(10-5)(10-6)(10-9)}=\sqrt{10\cdot 5\cdot 4\cdot 1},$

$S=10\sqrt{2}$

Answer: $10\sqrt{2}$