4.Prove that the ratios of the areas of two similar triangles is equal to the ra...

Question

Prove that the ratios of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Answer

$\large\color{brown}{ \mathtt{Proof :- }}$

Given :-

We are given two triangles ABC and PQR such that ΔABC∼ΔPQR.

To prove :-

$\\ \mathtt{ \implies \frac{ar(ABC)}{ar(PQR)} = { (\frac{ AB}{PQ }) }^{2} = { (\frac{ BC}{ QR }) }^{2} = ({ \frac{ CA}{R P}) }^{2}}$

Construction:-

Draw altitudes AM & PM of the triangles.

Now ,

$\\ \mathtt{ \implies \: ar (ABC)= \frac{1}{2} BC × AM}$

$\\ \mathtt{\implies ar ( PQR) = \frac{1}{2} QR × PN}[tex] \\ \mathtt{\implies ar ( PQR) = \frac{1}{2} QR × PN}$

So,

$\\ \mathtt{ \implies \frac{ar(ABC) }{ar( PQR )} = \frac{ \cancel\frac{1}{2} \times \: BC \times AM}{ \cancel{\frac{1}{2}} \times \: QR \times PN} }.........(1)$

$\\ \mathtt{ \implies \frac{ BC \times AM}{ QR \times PN} }$

Now , In ΔABM & ΔPQN,

∠B = ∠Q. ......( As ΔABC∼ΔPQR)

∠M = ∠N. ....( Each is of 90°)

So,

ΔABM∼ΔPQN. .....................(AA similarity criterion)

Therefore,

$\\ \mathtt{ \implies \frac{AM }{ PN } = \frac{ AB }{ PQ} }............(2)$

Also,

ΔABC∼ΔPQR. .......................(Given)

So,

$\\ \mathtt{ \implies \frac{AB}{ PQ } = \frac{ BC }{QR} = \frac{ CA }{RP} } \: ..............(3)$

Therefore,

$\\ \mathtt{ \implies \frac{ar(ABC) }{ar( PQR ) } = \frac{ AB }{ PQ} \times \frac{AM }{ PN }.......[ From (1) and (3)]}$

$\\ \mathtt{ \implies \frac{AB } { PQ } \times \frac{AB } { PQ }..........[From(2)]}$

$\\ \mathtt{\implies { (\frac{AB}{PQ} )}^{2} }$

Now, using( 3), we get

$\\ \mathtt{ \implies \frac{ar(ABC)}{ar(PQR)} = { (\frac{ AB}{PQ }) }^{2} = { (\frac{ BC}{ QR }) }^{2} = ({ \frac{ CA}{R P}) }^{2}}$

$\color{orange}{\mathtt{Hence , \: \: Proved }}$

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