Delta ABC has integral side lengths. The inte | vedclass

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(D) By the Angle Bisector Theorem,the ratio of the sides $AB$ and $BC$ is equal to the ratio of the segments $AD$ and $DC$ on the side $AC$.$\frac{AB}

Vedclass · Accepted Answer

$33$

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(D) By the Angle Bisector Theorem,the ratio of the sides $AB$ and $BC$ is equal to the ratio of the segments $AD$ and $DC$ on the side $AC$.$\frac{AB}{BC} = \frac{AD}{DC} = \frac{3}{8}$.Let $AB = 3k$ and $BC = 8k$ for some positive constant $k$.The side $AC = AD + DC = 3 + 8 = 11$.Since the side lengths must be integers,$3k$ and $8k$ must be integers. This implies $k$ must be of the form $\frac{n}{m}$ where $m$ divides both $3$ and $8$,so $k$ must be an integer.By the triangle inequality in $\Delta ABC$:$1) AB + BC > AC \implies 3k + 8k > 11 \implies 11k > 11 \implies k > 1$.$2) AB + AC > BC \implies 3k + 11 > 8k \implies 11 > 5k \implies k Since $k$ is an integer and $1 Thus,$AB = 3(2) = 6$ and $BC = 8(2) = 16$.The perimeter of $\Delta ABC = AB + BC + AC = 6 + 16 + 11 = 33$.

$\Delta ABC$ has integral side lengths. The internal angle bisector of angle $B$ meets side $AC$ at $D$. If $AD = 3$ and $DC = 8$,then the perimeter of $\Delta ABC$ is:

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