Find, area of A D E : area of quadrilateral ...

ICSE Class X Math 2015 Paper

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

Question : 49 of 52

Marks: +1, -0

Find, area of

\triangle A D E

: area of quadrilateral BCED.

Solution:

\;\;\frac{\text{ Ar. of }\;(\triangle {ABC})}{\text{ Ar. of }\;(\triangle {ADE})}=\;\frac{AB^2}{AE^2}

=\;\frac{144}{16}=\;\frac{9}{1}

\;\;\frac{\text{ Ar. of }\;(\triangle {ADE}) + {Ar}\;\text{ . of }\;({BCED})}{{Ar}\;\text{ . of }\;(\triangle {ADE})}=9

\;1+\;\frac{\text{ Ar. of }\;(B C E D)}{\text{ Ar. of }\;(\triangle A D E)}=9

\;\;\frac{\text{ Ar. of }\;(\triangle {ADE})}{\text{ Ar. of }\;({BCED})}=\;\frac{1}{8}

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