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

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(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$.By the geo

Vedclass · Accepted Answer

$4 \sqrt{13}$

Vedclass · Answer

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$.By the geometric mean theorem for right triangles,we have the relation $BM^2 = AM \cdot CM$.Given $BM = 12$ and $CM = 18$,we substitute these values into the equation:$12^2 = AM \cdot 18$$144 = AM \cdot 18$$AM = \frac{144}{18} = 8$.Now,consider $\Delta ABM$,which is a right-angled triangle at $M$ $(\angle AMB = 90^{\circ})$.Using the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.$AB^2 = 8^2 + 12^2$$AB^2 = 64 + 144$$AB^2 = 208$$AB = \sqrt{208} = \sqrt{16 \cdot 13} = 4 \sqrt{13}$.Thus,the correct option is $A$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 12$ and $CM = 18$,find $AB$.

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