In the right triangle B is …
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Sia ? 6 years, 6 months ago
Here we are given that
{tex}\therefore \quad A B + A D = B C + C D{/tex}
or, {tex}AD = BC+CD-AB{/tex}
or, {tex}AD =h+d-x{/tex}
In the right angled {tex}\Delta A C D,{/tex}
{tex}A D ^ { 2 } = A C ^ { 2 } + D C ^ { 2 }{/tex}
or, {tex}( h + d - x ) ^ { 2 } = ( x + h ) ^ { 2 } + d ^ { 2 }{/tex}
or, {tex}( h + d - x ) ^ { 2 } - ( x + h ) ^ { 2 } = d ^ { 2 }{/tex}
{tex}( h + d - x - x - h ) ( h + d - x + x + h ) = d ^ { 2 }{/tex}
{tex} Because \ a^2-b^2=(a-b)(a+b){/tex}
or, {tex}( d - 2 x ) ( 2 h + d ) = d ^ { 2 }{/tex}
or, {tex}2 h d + d ^ { 2 } - 4 h x - 2 x d = d ^ { 2 }{/tex}
or, {tex}2hd = 4hx+2xd{/tex}
{tex}2hd= 2x(2h+d){/tex}
Hence {tex}x = \frac { h d } { 2 h + d }{/tex}
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