In a triangle ABC,angle BAC = 90^{circ}; AD i | vedclass

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(A) Given that in $	riangle ABC$,$\angle BAC = 90^{\circ}$ and $AD$ is the altitude from $A$ onto $BC$.Since $AB = 15$ and $BC = 25$,by the Pythagore

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$12$

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(A) Given that in $	riangle ABC$,$\angle BAC = 90^{\circ}$ and $AD$ is the altitude from $A$ onto $BC$.Since $AB = 15$ and $BC = 25$,by the Pythagorean theorem:$AC = \sqrt{BC^2 - AB^2} = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20$.The area of $	riangle ABC$ can be calculated in two ways:Area $= \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes 15 	imes 20 = 150$.Also,Area $= \frac{1}{2} 	imes BC 	imes AD = \frac{1}{2} 	imes 25 	imes AD$.Equating the two areas: $\frac{1}{2} 	imes 25 	imes AD = 150 \implies AD = \frac{300}{25} = 12$.In quadrilateral $AEDF$,$\angle FAE = 90^{\circ}$,$\angle AFD = 90^{\circ}$,and $\angle AED = 90^{\circ}$. Thus,$AEDF$ is a rectangle.In a rectangle,the diagonals are equal. Therefore,$EF = AD$.Since $AD = 12$,the length of $EF$ is $12$.

In a triangle $ABC$,$\angle BAC = 90^{\circ}$; $AD$ is the altitude from $A$ onto $BC$. Draw $DE$ perpendicular to $AC$ and $DF$ perpendicular to $AB$. Suppose $AB = 15$ and $BC = 25$. Then the length of $EF$ is

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