Question

Prove that the ratio of the areas of two similar triangles is equal to the ratio of their squares of their corresponding medians.

Answer 1

$<b>$
Hey there !!

Given :-)

$\bf{ → \triangle ABC \sim \triangle DEF }$

→ AP and DQ are the medians of ∆ABC and ∆DEF respectively.


To Prove :-)

$\bf{ → \frac{ ar( \triangle ABC ) }{ ar( \triangle DEF ) } = \frac{ {AP}^{2} }{ {DQ}^{2} } . }$


Proof :-)

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.


$\bf{ => \frac{ ar( \triangle ABC ) }{ ar( \triangle DEF ) } = \frac{ {AB}^{2} }{ {DE}^{2} } ........(1). }$



$\bf{ Now, \: \triangle ABC \sim \triangle DEF }$



$\bf{ => \frac{ AB }{ DE ) } = \frac{BC}{EF} = \frac{2BP}{2EQ} = \frac{BP}{EQ} .}$


$\bf{ => \frac{ AB }{ DE ) } = \frac{BP}{EQ} .}$


$\bf{ \angle B = \angle E ....[ => \triangle ABC \sim \triangle DEF }$


$\bf{ \triangle APB \sim \triangle DQE ....[ By \: SAS-similarity. }$



$\bf{ => \frac{ AB }{ DE ) } = \frac{AP}{DQ} .}$


[ Squaring both side. ]


$\bf{ \frac{ {AB}^{2} }{ {DE}^{2} } = \frac{ {AP}^{2} }{ {DQ}^{2} } ........(2). }$


▶ From equation (1) and (2), we get





$\large \boxed{ → \frac{ ar( \triangle ABC ) }{ ar( \triangle DEF ) } = \frac{ {AP}^{2} }{ {DQ}^{2} } . }$



✔✔Hence, it is proved ✅✅.

____________________________________




$\huge \boxed{ \mathbb{THANKS}}$



$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

→ triangle ABC similar triangle DEF

Prove that:-
→ ar( triangle ABC )/ ar( triangle DEF ) = AP^2/DQ^2.

Solution:-

ar( triangle ABC / ar(triangle DEF=AB^2/DE^2 ........(1)

triangle ABC similar triangle DEF

AB/ DE =BC/EF = BP/EQ

AB/DE = BP/EQ.

angle B = angle E ....[ => triangle ABC similar triangle DEF )

triangle APB similar triangle DQE ....[ By SAS. )

AB / DE = AP/DQ

AB^2/DE^2 = AP^2/DQ^2........(2).

From equation (1) and (2), we get

→ar(triangle ABC ) / ar( triangle DEF ) = AP^2/DQ^2