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

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(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM \perp AC$.By the geometric mean theorem in right triangles,$BM^2 = AM \cdot CM$.Given $BM = 10$ an

Vedclass · Accepted Answer

$25 + 15\sqrt{5}$

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(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM \perp AC$.By the geometric mean theorem in right triangles,$BM^2 = AM \cdot CM$.Given $BM = 10$ and $CM = 5$,we have $10^2 = AM \cdot 5$,which gives $100 = 5 \cdot AM$,so $AM = 20$.The hypotenuse $AC = AM + CM = 20 + 5 = 25$.Using the Pythagorean theorem in $\Delta BMC$,$BC^2 = BM^2 + CM^2 = 10^2 + 5^2 = 100 + 25 = 125$,so $BC = \sqrt{125} = 5\sqrt{5}$.Using the Pythagorean theorem in $\Delta AMB$,$AB^2 = BM^2 + AM^2 = 10^2 + 20^2 = 100 + 400 = 500$,so $AB = \sqrt{500} = 10\sqrt{5}$.The perimeter of $\Delta ABC = AB + BC + AC = 10\sqrt{5} + 5\sqrt{5} + 25 = 25 + 15\sqrt{5}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 10$ and $CM = 5$,find the perimeter of $\Delta ABC$.

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