Question

bc and bde are two equilateral triangle such that d is the mid point on bc .if bd :dc =2:1 ,then the ratio of the area to that of the ∆ bde​

Answer 1

Given:-

$\bf \triangle ABC \: is \: a \: equilateral \\ \bf \triangle BDE \: is \:a \: equilateral$

$\sf BD = \dfrac{1}{2} \: BC \: as \: D \: is \: midpoint \: of \: BC$

To Find:-

$\sf : \implies \blue{\dfrac{ar \triangle \: ABC}{ar \triangle \: BDE} }$

Solution:-

Since,

ABC and BDE are Equilateral,

Their sides would be in the same ratio.

$\sf : \implies \green{\dfrac{ab}{be} = \dfrac{AC}{ED} = \dfrac{BC}{BD}}$

Hence,

By SSS similarity

△ABC ≈ △BDE

And,

We know that ratio of area of triangle is equal To the ratio of square of corresponding sides.

So,

$\sf : \orange{\implies \dfrac{area \: of \: \triangle \: ABC}{area \: of \: \triangle \: BDE} = \dfrac{(BC {)}^{2} }{ (BD {)}^{2} } } \\ \\ \\ \sf : \implies \red{\dfrac{(BC {)}^{2} }{ (\dfrac{BD}{2} {)}^{2} }} \: \: \: \bf since \: BD \: = \dfrac{1}{2} BC \\ \\ \\ \sf : \implies \green{ \dfrac{B {C}^{2} }{ \dfrac{B {C}^{2} }{4} } } \\ \\ \\ \sf : \implies \pink{\dfrac{4B {C}^{2} }{B {C}^{2} } } \\ \\ \\ \sf : \implies \blue{ \dfrac{4}{1} }$

Hence,

$\sf \green{\dfrac{area \: of \: \triangle ABC}{area \: of \: \triangle BDE}} \\ \\ \sf : \implies \purple{ \dfrac{4}{1}} \\ \\ \bf \red{i.e. \: 4 \ratio 1}$