If A C=13 {\cm}, B C=5 {\cm} and A E=4 {\cm} . Find D E and A D .

ICSE Class X Math 2015 Solved Paper - Question 48

$A B C$ is a right angled triangle with $∠ A B C=$ $90^{∘} . D$ is any point on $A B$ and $D E$ is perpendicular to AC. Prove that :

Question : 48 of 52

Marks: +1, -0

A C=13 {\cm}, B C=5 {\cm}

and

A E=4

{\cm}

. Find

D E

and

A D

Solution:

\;\text " In "\; △ {\ABC},

( {\AC})^2\;=( {\AB})^2+( {\BC})^2

169\;=( {\AB})^2+25

{\AB}\;=12 {\cm}

∵ \;△ {\ADE}\;∼ ∆ {\ACB}

∴ \;\; \;{ {\DE}}/{ {\BC}}=\;{ {\AD}}/{ {\AC}}\;=\;{ {\AE}}/{ {\AB}}

∴ \;\; \;{ {\DE}}/{ {\BC}}\;=\;{ {\AE}}/{ {\AB}}

\;{ {\DE}}/{5}\;=\;{4}/{12}

{\DE}\;=\;{20}/{12}=\;{5}/{3}=1.67 {\cm} .

Now,

\;{ {\AD}}/{ {\AC}}\;=\;{ {\AE}}/{ {\AB}}

\;{ {\AD}}/{13}\;=\;{4}/{12}

{\AD}\;=\;{13 × 4}/{12}=\;{13}/{3}

\;=4.33 {\cm} .

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