Question

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides prove the theorem​

Answer 1

Step-by-step explanation:

this the answer of this question

Answer 2

Answer:If two triangles are similar, then the ratio of the area of both triangles is proportional to square of the ratio of their corresponding sides.

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

According to the stated theorem

ar△ABC/△PQR  =(AB/PQ)^2=(BC)^2=(CA/RP)^2

Since the area of triangle =  1/2×base×altitude

To find the area of ΔABC and ΔPQR draw the altitudes AD and PE from the vertex A and P of ΔABC and ΔPQR  

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:

ar△ABC  /ar△PQR = 1/2×BC×AD\1/2×QR×PE

ar△ABC/ ar△PQR =BC×AD/QR×PE

Now in △ABD and △PQE it can be seen  

∠ABC=∠PQR (Since ΔABC∼ΔPQR )

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

From AA criterion of similarity ΔADB∼ΔPEQ

AD/ PE  = AB/PQ

Since it is known that ΔABC∼ΔPQR

AB /PQ = BC/ QR =CA/ RP

Substituting this value in equation , we get

ar△ABC/ ar△PQR=AB/ PQ × AD /PE

​we can write

ar△ABC /ar△PQR  =( AB /PQ)^2

Similarly we can prove  

ar△ABC /ar△PQR  =(AB/ PQ )^2=(BC/ QR ) ^2 =(CA/ RP)  ^2