In triangle ABC,angle A = 90^{circ} and AD pe | vedclass

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(C) In $	riangle ABC$,$\angle BAC = 90^{\circ}$ and $AD \perp BC$.Since $	riangle ABC \sim 	riangle ACD$ (by $AA$ similarity criterion,as $\angle C

Vedclass · Accepted Answer

$16:9$

Vedclass · Answer

(C) In $	riangle ABC$,$\angle BAC = 90^{\circ}$ and $AD \perp BC$.Since $	riangle ABC \sim 	riangle ACD$ (by $AA$ similarity criterion,as $\angle C = \angle C$ and $\angle ADC = \angle BAC = 90^{\circ}$),the ratio of their areas is equal to the square of the ratio of their corresponding sides.Specifically,$\frac{	ext{Area}(	riangle ABC)}{	ext{Area}(	riangle ACD)} = \left( \frac{BC}{AC} ight)^{2}$.Given $BC = 8 \ cm$ and $AC = 6 \ cm$.Therefore,$\frac{	ext{Area}(	riangle ABC)}{	ext{Area}(	riangle ACD)} = \left( \frac{8}{6} ight)^{2} = \left( \frac{4}{3} ight)^{2} = \frac{16}{9}$.Thus,the ratio is $16:9$.

In $\triangle ABC$,$\angle A = 90^{\circ}$ and $AD \perp BC$,where $D$ lies on $BC$. If $BC = 8 \ cm$ and $AC = 6 \ cm$,then the ratio of the area of $\triangle ABC$ to the area of $\triangle ACD$ is:

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