In Delta ABC,m angle B = 90^{circ} and overli | vedclass

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(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $AC$.Since $D$ lies on $AC$,we have $AC = AD + CD =

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $AC$.Since $D$ lies on $AC$,we have $AC = AD + CD = 9 + 27 = 36$.According to the geometric mean theorem (or property of altitude to the hypotenuse in a right triangle),$AB^2 = AD 	imes AC$.Substituting the values,$AB^2 = 9 	imes 36 = 324$.Taking the square root of both sides,$AB = \sqrt{324} = 18$.Thus,$AB = 18$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $AC$. If $AD = 9$ and $CD = 27$,find $AB$.

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