ABC is an iscosceles triangle right …
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Sia ? 6 years, 5 months ago
ABC is an isosceles triangle right angled at B,
Let AB=BC=x cm
By Pythagoras theorem In {tex}\triangle {/tex} ABC
AC2 = AB2 + BC2
AC2 = x2 + x2
AC2 = 2x2
AC={tex}\sqrt 2 {/tex} x
{tex}\triangle{/tex}ACD{tex}\sim{/tex}{tex}\triangle{/tex}ABE (Given)
Thus, using Area Theorem
{tex}\therefore{/tex}{tex}\frac { \operatorname { ar } \Delta \mathrm { ABE } } { \operatorname { ar } \Delta A \mathrm { CD } } = \frac { \mathrm { AB } ^ { 2 } } { \mathrm { AC } ^ { 2 } } = \frac { x ^ { 2 } } { ( \sqrt { 2 } x ) ^ { 2 } }{/tex}
={tex}\frac { x ^ { 2 } } { 2 x ^ { 2 } }{/tex}={tex}\frac{1}{2}{/tex}=1:2.
Hence, Areas of the two given triangles are in the ratio 1:2
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