Question

17. The ratio of the areas of two similar triangles is equal to the square of the
ratio of their corresponding sides

Answer 1

Given:-

• ∆ ABC ~ ∆ DEF

To prove:-

ar(∆ABC)/ar(DEF) = AB²/DE² = AC²/DF² = BC²/EF²

$\bf\:CONSTRUCTION:-$

Draw AL ⊥ BC and DM ⊥EF

Proof:-

∆ABC ~ ∆ DEF,

It follows that they are equiangular and their sides are proportional.

∴ ∠A = ∠D , ∠B = ∠E , ∠C = ∠F and

AB/DE = BC/EF = AC/DF.....................(i)

Now,

ar(∆ABC) = (1/2 × BC × AL)

ar(∆DEF) = (1/2 × EF × DM)

∴ ar(∆ABC)/ar(∆DEF) = 1/2 × BC × AL / 1/2 × EF × DM = BC/EF × AL/DM ........(ii)

In ∆ ALB and ∆ DME ,we have:

∠ALB = ∠DME = 90°

and ∠B = ∠E (From eq (i) )

∴ ∆ALB ~ ∆DME

consequently,

AL/DM = AB/DE

then,

AB/DE = BC/EF (from eq (i) )

∴ AL/DM = BC/EF .......................(iii)

Using eq (iii) in eq(ii)

ar(∆ABC)/ar(∆DEF) = (BC/EF × BC/EF)

=BC²/EF²

Similarly,

ar(∆ABC)/ar(∆DEF) = AB²/DE² and,

ar(∆ABC)/ar(∆DEF) = AC²/DF²

hence,

ar(∆ABC)/ar(DEF) = AB²/DE² = AC²/DF² = BC²/EF²