Question

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding altitude​

Answer 1

GIVEN:

• Two triangles ABC and PQR such that triangle ABC ∽ triangle PQR and AM, PN are their medians.

TO PROVE:

• Ar(Triangle ABC)/Ar(Triangle PQR) = AM²/PN²

PROOF:

Since,the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\rm{\therefore \dfrac{Area \: ( \triangle \: ABC ) }{Area ( \triangle PQR)} = \dfrac{AB^2}{PQ^2}....1)}$

Now,

$\rm{\triangle ABC \sim \triangle PQR }$

$\rm{\longrightarrow \dfrac{AB}{PQ} = \dfrac{BC}{QR}}$

$\rm{\longrightarrow \dfrac{AB}{PQ} = \dfrac{2BM}{2QN} = \dfrac{BM}{QN} ....2)}$

Thus, in triangles AMB and PNQ

$\rm{\longrightarrow \dfrac{AB}{PQ} = \dfrac{BM}{QN} \: and \: \angle B = \angle Q }$ ( triangle ABC ∽ triangle PQR)

So, by SAS criterian of similarity, we have

$\rm{ \triangle AMB \sim \triangle PQN}$

$\rm{\longrightarrow \dfrac{BM}{QN} = \dfrac{AM}{PN} }$ ....❸

From Equation 2) and 3), we get

$\rm{ \dfrac{AB}{PQ} = \dfrac{AM}{PN}}$

$\rm{\longrightarrow \dfrac{AB^2}{PQ^2} = \dfrac{AM^2}{PN^2}}$ .....❹ 

From Equation 1) and 4), we get

$\bf{ \dfrac{Area ( \triangle ABC)}{ Area ( \triangle PQR)} = \dfrac{ AP^2 }{PN^2}}$

$\large\underline\mathbf{HENCE \: PROVED....}$

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