Question

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

• A. $12$
• B. $10$
• C. $5\sqrt{3}$
• D. $5\sqrt{5}$

Answer 1

Answer: (A)

$12$

$\angle BAC={90}^o$ [ Given ]

$AD\perp BC,DE\perp AC$ and $DF\perp AB$ [ Given ]

$\therefore$ $EDFA$ is a rectangle.

$AD=EF$ [ Diagonals of a rectangle are equal ]

$AB=15$ and $BC=25$ [ Given ]

$\Rightarrow$ $(BC{)}^2=(AC{)}^2+(AB{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(25{)}^2=(AC{)}^2+(15{)}^2$

$\Rightarrow$ $625=(AC{)}^2+225$

$\Rightarrow$ $(AC{)}^2=400$

$\therefore$ $AC=20$

Let $BD=(25-x)$

So, $CD=x$

In $△ADB$

$\Rightarrow$ $(AD{)}^2=(15{)}^2-(25-x{)}^2$ ----- ( 1 )

In $△ADC,$

$\Rightarrow$ $(AD{)}^2=(20{)}^2-x^2$ ----- ( 2 )

Equation ( 1 ) = Equation ( 2 )

$\Rightarrow$ $(15{)}^2-(25-x{)}^2=(20{)}^2-x^2$

$\Rightarrow$ $225-(625-50x+x^2)=400-x^2$

$\Rightarrow$ $225-625+50x-x^2=400-x^2$

$\Rightarrow$ $50x=800$

$\therefore$ $x=16$

$\therefore$ $CD=16$

In $△ADC,$

$(AC{)}^2=(AD{)}^2+(CD{)}^2$

$\Rightarrow$ $(AD{)}^2=(AC{)}^2-(CD{)}^2$

$\Rightarrow$ $(AD{)}^2=(20{)}^2-(16{)}^2$

$\Rightarrow$ $(AD{)}^2=400-256$

$\Rightarrow$ $(AD{)}^2=144$

$\therefore$ $AD=12$

$\therefore$ $AD=EF=12$