Question

1) The area ( ABC ) = 16cm2 and area ( DEF) =25cm2 and BC = 2.3cm find EF

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Answer 1

AnswEr :

$\underline{\bigstar\:\textsf{According \: to \: given\: in\: question:}}$

$\bullet\frak{Given}\begin{cases}\sf{ar(\triangle ABC) = 16cm^{2}}\\\sf{ar(\triangle DEF) = 25cm^{2}}\\\sf{BC = 2.3cm}\end{cases}$

$\bullet\normalsize\sf\ We \: have \: to \: find \: EF$

$\underline{\bigstar\:\textsf{ Let's \: head \: to \: the \: question \: now:}}$

$\bullet$ Here; we use Area of similar triangles theorem,which states that if area of two triangles are similar then they are equal to squares of corresponding sides$\normalsize\bf\ [\frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{AB^{2}}{DE^{2}} = \frac{BC^{2}}{EF^{2}} =\frac{AC^{2}}{DF^{2}}]$

$\large\ : \implies\sf\frac{ar(\triangle ABC) }{ar(\triangle DEF)} = \frac{BC^{2}}{EF^{2}}$

$\large\ : \implies\sf\frac{16}{25} = \frac{BC^2}{EF^2}$

$\large\ : \implies\sf\frac{(4)^2}{(5)^2} = \frac{BC^2}{EF^2}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ Cancel \: the \: squares \; both \; sides}) }$

$\large\ : \implies\sf\frac{4}{5} =\frac{2.3}{EF}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{--Cross multiplication--}) }$

$\normalsize\ : \implies\sf\ 4EF = 11.5 \\ \\ \normalsize\ : \implies\sf\frac{11.5}{4} =EF \\ \\ \normalsize\ : \implies\sf\ EF = 2.8cm$

$\underline{\therefore\:\textsf{Hence, \: EF = 2.8cm}}$