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6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratioof their corresponding altitudes​

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Answer:

Hii friend,

Let ∆ABC and ∆DEF are two similar triangles.

Given :- ∆ABC similar to triangle DEF, AL Perpendicular to BC and DM Perpendicular to EF

Ar(∆ABC)/Ar(∆DEF) = AL²/DM²

Proof :- As we know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Therefore,

Ar(∆ABC)/Ar(∆DEF) = AB²/DE² ........(1)

In ∆ALB and ∆DME , we have

Angle ALB = Angle DME = 90°

and, Angle B = Angle E { ∆ABC similar∆DEF)

Therefore,

∆ALB similar to ∆DME { By AA similarity}

=> AB/DE = AL/DM

=> AB²/DE² = AL²/DM² ........(2)

From 1 and 2 we get,

Ar(∆ABC)/Ar(∆DEF) = AL²/DM²..... PROVED.....

HOPE IT WILL HELP YOU...... :-)

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