Question

Prove that “That ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”

Answer 1

Consider ∆ABC ~ ∆DEF

and height of ∆ABC be h₁ and ∆DEF be h₂

`"AB"/"DE" = "BC"/"EF" = "AC"/"DF"` ....corresponding sides of similar triangles (i)

∠ABC = ∠DEY … corresponding angles of similar triangles(ii)

Consider ∆ABX and ∆DEY

∠AXB = ∠DYE = 90°

From equation (ii)

∠ABC = ∠DEY

∴by AA test for similarity ∆ABX ~ ∆DEY

`"AB"/"DE" = "BX"/"EY" = "AX"/"DY"`....corresponding sides of similar triangles

But from figure AX = h₁ and DY = h₂

`"AB"/"DE" = "BX"/"EY" = "h"_1/"h"_2`...(iii)

A(∆ABC) = (1/2)×BC×h₁

A(∆DEF) = (1/2)×EF×h₂

`(A(∆ABC))/(A(∆DEF)) = (AB)/(DE) xx (AB)/(DE) = (AB^2)/(DE^2)` ...(iv)

Squaring equation (i) and using it in (iv)

`(A(∆ABC))/(A(∆DEF)) = (AB^2)/(DE^2) = (BC^2)/(EF^2) = (AC^2)/(DF^2)`

Hence proved

Therefore, the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.