Question

Prove that if the area of two similar triangles are equal then they are congruent

Answer 1

Answer:

all sides of the first triangle are equal to the all sides of the second sides then they are congruent

Answer 2

$\huge\colorbox{seagreen}{\tt{Answer ♥︎}}$

${\underline{\underline{\sf{\color{blue}{Given}}}}}$

$\sf{∆ ABC ∼ PQR}$

$\sf{ar ( ∆ ABC ) = ar ( ∆ PQR )}$

${\underline{\underline{\sf{\color{blue}{To \: prove}}}}}$

$\sf{∆ ABC ≅ ∆ PQR}$

${\underline{\underline{\sf{\color{blue}{Proof}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\sf{∴ \frac{ar ( ∆ ABC )}{ar(∆PQR)} = \frac{ {AB}^{2} }{ {PQ}^{2} } = \frac{ {BC}^{2} }{ {QR}^{2} } = \frac{ {AC}^{2} }{ {PR}^{2} } }}}$

( Ratio of area of similar triangles is equal to the square of corresponding sides )

$\: \: \: \: \: \: \: \: \: \: \sf{But, \: \frac{ar(∆ ABC)}{ar(∆PQR } = 1 \: \:──━━⧼ \: Given}$

$\: \: \: \: \: \: \sf{∴ \frac{ {AB}^{2} }{ {PQ}^{2} } = \frac{ {BC}^{2} }{ {QR}^{2} } = \frac{ {AC }^{2} }{ {PR}^{2} } = 1 }$

$\sf{So, \: {AB }^{2} = {PQ}^{2} \: or \: AB = PQ}$

$\: \: \: \: \: \: \: \: \sf{ {BC }^{2} = {QR}^{2} \: or \: BC= QR}$

$\: \: \: \: \: \: \: \: \sf{ {AC }^{2} = {PR}^{2} \: or \: AC= PR}$

$\sf{∴ By \: SSS \: congruency \: axiom,}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: {\underline{\underline{\sf{\bold{\color{red}{∴ ∆ ABC ≅ ∆ PQR}}}}}}$

$\sf{Hence, proved}$

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