Question

Prove that : When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the

squares of their corresponding sides.​

Answer 1

To prove this theorem, consider two similar triangles ΔABC and ΔPQR;

According to the stated theorem,

area of ΔABCarea of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2

As, Area of triangle = 12 × Base × Height

To find the area of ΔABC and ΔPQR, draw the altitudes AD and PE from the vertex A and P of ΔABC andΔPQR, respectively, as shown in figure.

Now, area of ΔABC = 12 × BC × AD

area of ΔPQR = 12 × QR × PE

The ratio of the areas of both the triangles can now be given as:

area of ΔABCarea of ΔPQR = 12×BC×AD12×QR×PE

⇒ area of ΔABCarea of ΔPQR = BC × ADQR × PE ……………. (1)

Now in ∆ABD and ∆PQE, it can be seen that:

∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)

∠ADB = ∠PEQ (Since both the angles are 90°)

From AA criterion of similarity ∆ADB ~ ∆PEQ

⇒ ADPE = ABPQ …………….(2)

Since it is known that ΔABC~ ΔPQR,

ABPQ = BCQR = ACPR …………….(3)

Substituting this value in equation (1), we get

area of ΔABCarea of ΔPQR = ABPQ × ADPE

Using equation (2), we can write

area of ΔABCarea of ΔPQR = ABPQ × ABPQ

⇒area of ΔABCarea of ΔPQR =(ABPQ)2

Also from equation (3),

area of ΔABCarea of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2

This proves that the ratio of the area of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.