Question

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

Answer 1

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 ,

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

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

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,

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

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

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

Now, using( 3), we get

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

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