4.Prove that the ratios of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.
ans???
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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 corresponding sides.
ans???
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Step-by-step explanation:
for proving theorem we use concept of area of triangle and property of Similarities of triangle
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]