Question

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

Answer 1

Given, ΔABC ∼ ΔPQR

To Prove: `(ar (ΔABC))/(ar (ΔPQR)) = ((AB)/(PQ))^2`

= `((BC)/(QR))^2`

= `((AC)/(PR))^2`

Construction: Draw AD ⊥ BC and PE ⊥ QR.

Proof: ΔABC ∼ ΔPQR

⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R and 

`"AB"/"PQ" = "BC"/"QR" = "AC"/"PR"`   ...(i) [∵ Similar triangles are equiangular and their corresponding sides are proportional]

In ΔADB and ΔPEQ,

∠B = ∠Q   ...[From (i)]

∠ADB = ∠PEQ   ...[Each 90°]

∴ ΔADB ∼ ΔPEQ   ...[AA similarity]

⇒ `"AD"/"PE" = "AB"/"PQ"`   ...(ii) [Corresponding sides of similar triangles]

From equations (i) and (ii),

`"AB"/"PQ" = "BC"/"QR" = "AC"/"PR" = "AD"/"PE"`   ...(iii)

Now, `(ar (ΔABC))/(ar (ΔPQR)) = (1/2 xx BC xx AD)/(1/2 xx QR xx PE)`

= `((BC)/(QR)) xx ((AD)/(PE))`

= `"BC"/"QR" xx "BC"/"QR"`

⇒ `(ar (ΔABC))/(ar (ΔPQR)) = "BC"^2/"QR"^2`   ...(iv) [From equations (iii)]

From equations (iii) and (iv),

`(ar (ΔABC))/(ar (ΔPQR)) = ((AB)/(PQ))^2`

= `((BC)/(QR))^2`

= `((AC)/(PR))^2`