in figure ABCD is a rectangle …
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Sia ? 5 years ago
We have, AC = 10 cm and AD = 2{tex}\sqrt{5}{/tex} cm
{tex}\therefore{/tex} AC2 = AD2 + CD2
{tex}\Rightarrow{/tex} CD = {tex}\sqrt{AC^2 - AD^2}{/tex} = {tex}\sqrt{(10)^{2}-(2 \sqrt{5})^{2}}{/tex} = 4{tex}\sqrt{5}{/tex} cm
{tex}\therefore{/tex} ar(rect ABCE) = AD {tex}\times{/tex} CD = 2{tex}\sqrt{5}{/tex} {tex}\times{/tex} 4{tex}\sqrt{5}{/tex} cm2
= 8 {tex}\times{/tex} 5 cm2 = 40 cm2
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