ABC is a triangle. AB = 5 text{ cm},AC = sqrt | vedclass

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(A) Let $BD = x$. Then $DC = 8 - x$.In $	riangle ADB$,by Pythagoras theorem: $AD^2 = AB^2 - BD^2 = 5^2 - x^2 = 25 - x^2$.In $	riangle ADC$,by Pythag

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$6$

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(A) Let $BD = x$. Then $DC = 8 - x$.In $	riangle ADB$,by Pythagoras theorem: $AD^2 = AB^2 - BD^2 = 5^2 - x^2 = 25 - x^2$.In $	riangle ADC$,by Pythagoras theorem: $AD^2 = AC^2 - DC^2 = (\sqrt{41})^2 - (8 - x)^2 = 41 - (64 - 16x + x^2) = 41 - 64 + 16x - x^2 = 16x - x^2 - 23$.Equating the two expressions for $AD^2$:$25 - x^2 = 16x - x^2 - 23$$25 = 16x - 23$$16x = 48$$x = 3 	ext{ cm}$.So,$BD = 3 	ext{ cm}$.Now,$AD = \sqrt{25 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 	ext{ cm}$.Area of $	riangle ABD = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BD 	imes AD = \frac{1}{2} 	imes 3 	imes 4 = 6 	ext{ cm}^2$.

$ABC$ is a triangle. $AB = 5 \text{ cm}$,$AC = \sqrt{41} \text{ cm}$,and $BC = 8 \text{ cm}$. $AD$ is perpendicular to $BC$. What is the area (in $\text{cm}^2$) of triangle $ABD$?

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