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

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(A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.Using the Pythagorean theorem,the hypotenuse $AC = \sqrt{A

Vedclass · Accepted Answer

$6.72$

Vedclass · Answer

(A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.Using the Pythagorean theorem,the hypotenuse $AC = \sqrt{AB^2 + BC^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.The area of $\Delta ABC$ can be calculated in two ways:Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes 7 	imes 24 = 84$.Also,using $AC$ as the base and $BM$ as the altitude,Area $= \frac{1}{2} 	imes AC 	imes BM = \frac{1}{2} 	imes 25 	imes BM$.Equating the two areas: $\frac{1}{2} 	imes 25 	imes BM = 84$.$25 	imes BM = 168$.$BM = \frac{168}{25} = 6.72$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. If $AB = 7$ and $BC = 24$,find the length of $BM$.

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