Question

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Answer 1

Question:

If ΔDEF ∼ ΔABC, area of ΔDEF = 4/9 (Area of ΔABC) then BC/EF = ?

Answer:

BC/EF = 3/2

Concept used:

$\maltese \: \: \boxed{\tt{\dfrac{Area \: of \: \triangle ABC}{Area \: of \: \triangle DE F}=\bigg(\dfrac{AB}{DE}\bigg)^{2}=\bigg(\dfrac{BC}{EF}\bigg)^{2}=\bigg(\dfrac{AC}{DF}\bigg)^{2}}}$

Step-by-step explanation:

★ According to the question,

$\sf{Area \: of \: \triangle DE F=\dfrac{4}{9}\:(Area \: of \: \triangle ABC)}$

Dividing by (Area of ΔABC) on both sides,

$\implies \sf{\dfrac{Area \: of \: \triangle DE F}{Area \: of \: \triangle ABC}=\dfrac{4}{9}}$

The question is asked to find BC/EF. So let's take reciprocal on both the sides.

$\implies \sf{\dfrac{Area \: of \: \triangle ABC}{Area \: of \: \triangle DE F}=\dfrac{9}{4}}$

Now,

$\implies \sf{\dfrac{Area \: of \: \triangle ABC}{Area \: of \: \triangle DE F}=\bigg(\dfrac{BC}{EF}\bigg)^{2}}$

$\implies \sf{\dfrac{9}{4}=\bigg(\dfrac{BC}{EF}\bigg)^{2}}$

$\implies \sf{\bigg(\dfrac{BC}{EF}\bigg)^{2}=\dfrac{9}{4}}$

$\implies \sf{\dfrac{BC}{EF}=\sqrt{\dfrac{9}{4}}}$

$\implies \sf{\dfrac{BC}{EF}=\dfrac{3}{2}}$

$\therefore \: \underline{\boxed{\bf{\dfrac{BC}{EF}=\dfrac{3}{2}}}}$

Answer 2

$\sf\huge\bold{Answer:}$

Given :

• ∆DEF ∼∆ABC
• ar.∆DEF=4/9(ar.∆ABC)

To find :

• BC/EF

Solution :

We know that when two triangles are similar, then square of ratio of corresponding sides of two triangles is equal to the ratio of their areas.

$\boxed{ \tt \red{ \frac{ar. \triangle \: DEF }{ar. \triangle \: ABC} = \bigg( \frac{DE }{AB} \bigg) {}^{2} = \bigg( \frac{EF}{BC}\bigg) { }^{2} = \bigg( \frac{DF}{AC}\bigg) { }^{2} }}$

Given that

$:⟶ \tt\ ar. \triangle \: DEF = \frac{4}{9}(ar. \triangle \: ABC)$

$\tt {:⟶ \frac{ar. \triangle \: DEF }{ar. \triangle \: ABC} = \frac{4}{9} }$

Doing reciprocal, we get

$\tt {:⟶ \frac{ar. \triangle \:ABC }{ar. \triangle \: DEF } = \frac{9}{4} }...(i)$

Now, we know

$\tt {:⟶ \frac{ar. \triangle \: DEF }{ar. \triangle \: ABC} = \bigg( \frac{EF}{BC}\bigg) { }^{2} }$

Reciprocal gives :

$\tt {:⟶ \frac{ar. \triangle \:ABC }{ar. \triangle \: DEF } = \bigg( \frac{BC}{EF}\bigg) { }^{2} }...(ii)$

Comparing (i) & (ii)

$\tt {:⟶ \frac{ar. \triangle \:ABC }{ar. \triangle \: DEF } = \bigg( \frac{BC}{EF}\bigg) { }^{2} } = \frac{9}{4}$

We know that 9=3² & 4=2²

So,

$\tt { :⟶ \bigg( \frac{BC}{EF}\bigg) { }^{2} } = \frac{ {3}^{2} }{ {2}^{2} }$

Similarly,

$\tt { :⟶ \bigg( \frac{BC}{EF}\bigg) { }^{2} } = \bigg(\frac{ {3}}{ {2} } \bigg) {}^{2}$

Simplifying by cutting square on both sides :

$\tt {:⟶ \bigg( \frac{BC}{EF}\bigg) \cancel{{ }^{2}} } = \bigg(\frac{ {3}}{ {2} } \bigg) \cancel{{}^{2}}$

$\tt {:⟶ \bigg( \frac{BC}{EF}\bigg) = \bigg(\frac{ {3}}{ {2} } \bigg) }$

$\boxed{ \underline{\tt \bold\purple{\frac{BC}{EF}=\frac{3}{2}}}}$