Question

37. Prove that if two triangles are similar then their area are proportional to the square
of the corresponding sides​

Answer 1

Ans:

Now, area of ΔABC =  1/2   × BC × AD

area of ΔPQR =  1/2   × QR × PE

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

area of ΔABC/ area of ΔPQR

= 1/2    ×BC×AD÷ 1/2   ×QR×PE

⇒  

area of ΔABC/ area of ΔPQR  =  

BC × AD/ QR × 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

⇒  

AD/ PE   =  AB/ PQ   …………….(2)

Since it is known that ΔABC~ ΔPQR,

AB /PQ   =  BC /QR  =  AC /PR   …………….(3)

Substituting this value in equation (1), we get

area of ΔABC /area of ΔPQR

=  AB /PQ  ×  AD /PE

Using equation (2), we can write

area of ΔABC /area of ΔPQR  

=  AB /PQ  × AB/PQ            

Also from equation (3),

area of ΔABC/ area of ΔPQR   =  ( AB/ PQ)²