Let ∆ABC~∆DEF and their areas be, …
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Sia ? 6 years, 6 months ago
{tex}\vartriangle {/tex} ABC {tex} \sim {/tex} {tex}\vartriangle {/tex} DEF ...............Given
{tex}\therefore {/tex} {tex}{{ar(\Delta ABC)} \over {ar(\Delta DEF)}} = {\left( {{{BC} \over {EF}}} \right)^2}{/tex} ..........[The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides]
{tex}\Rightarrow {/tex} {tex}\frac{{64}}{{121}} = {\left( {\frac{{BC}}{{15.4}}} \right)^2} \Rightarrow {\left( {\frac{8}{{11}}} \right)^2} = {\left( {\frac{{BC}}{{15.4}}} \right)^2}{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{8}{{11}} = \frac{{BC}}{{15.4}}{/tex} ...........Taking square root on both sides
{tex}\Rightarrow {/tex} BC={tex}\frac{{8 \times 15.4}}{{11}} \Rightarrow BC = 11.2\,cm{/tex}
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