4.Prove that the ratios of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.
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4.Prove that the ratios of the areas of two similar triangles is equal to the ratio of the squares of the
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
[tex] \large\color{brown}{ \mathtt{Proof :- }}[/tex]
Given :-
To prove :-
[tex] \\ \mathtt{ \implies \frac{ar(ABC)}{ar(PQR)} = { (\frac{ AB}{PQ }) }^{2} = { (\frac{ BC}{ QR }) }^{2} = ({ \frac{ CA}{R P}) }^{2}}[/tex]
Construction:-
Now ,
[tex] \\ \mathtt{ \implies \: ar (ABC)= \frac{1}{2} BC × AM}[/tex]
[tex] \\ \mathtt{\implies ar ( PQR) = \frac{1}{2} QR × PN}
[tex] \\ \mathtt{\implies ar ( PQR) = \frac{1}{2} QR × PN}[/tex]
So,
[tex] \\ \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)[/tex]
[tex] \\ \mathtt{ \implies \frac{ BC \times AM}{ QR \times PN} }[/tex]
Now , In ΔABM & ΔPQN,
∠B = ∠Q. ......( As ΔABC∼ΔPQR)
∠M = ∠N. ....( Each is of 90°)
So,
ΔABM∼ΔPQN. .....................(AA similarity criterion)
Therefore,
[tex] \\ \mathtt{ \implies \frac{AM }{ PN } = \frac{ AB }{ PQ} }............(2)[/tex]
Also,
ΔABC∼ΔPQR. .......................(Given)
So,
[tex] \\ \mathtt{ \implies \frac{AB}{ PQ } = \frac{ BC }{QR} = \frac{ CA }{RP} } \: ..............(3)[/tex]
Therefore,
[tex] \\ \mathtt{ \implies \frac{ar(ABC) }{ar( PQR ) } = \frac{ AB }{ PQ} \times \frac{AM }{ PN }.......[ From (1) and (3)]}
[/tex]
[tex] \\ \mathtt{ \implies \frac{AB } { PQ } \times \frac{AB } { PQ }..........[From(2)]}[/tex]
[tex] \\ \mathtt{\implies { (\frac{AB}{PQ} )}^{2} }[/tex]
Now, using( 3), we get
[tex] \\ \mathtt{ \implies \frac{ar(ABC)}{ar(PQR)} = { (\frac{ AB}{PQ }) }^{2} = { (\frac{ BC}{ QR }) }^{2} = ({ \frac{ CA}{R P}) }^{2}}[/tex]
[tex] \color{orange}{\mathtt{Hence , \: \: Proved }}[/tex]