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

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(B) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other. According to the geometric mean theorem,$BD^2 = AD \cdot CD$. Given $BD = 2 \sqrt{30}$ and $CD = 6$. Substituting the values: $(2 \sqrt{30})^2 = AD \cdot 6$. $4 \cdot 30 = 6 \cdot AD$. $120 = 6 \cdot AD$. $AD = 20$. Since $AC = AD + CD$,we have $AC = 20 + 6 = 26$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. If $BD = 2 \sqrt{30}$ and $CD = 6$,then $AC = \ldots$

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