Question

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding perimeter​

Answer 1

Answer:

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

According to the stated theorem,

area of ΔABC / area of ΔPQR = (AB/PQ)^2 = (BC/QR)^2 = (CA/RP)^2

As, Area of triangle = 1/2 × Base × Height

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

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

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

Also from equation (3),

area of ΔABC / area of ΔPQR = (AB/PQ)^2 = (BC/QR)^2 = (CA/RP)^2

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

Step-by-step explanation: