Question

areas of two similar triangles are proportional to squares of corresponding sides​

Answer 1

Answer:

Theorems on the area of similar triangles

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 sides.

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

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 × ADPE

⇒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 areas of two similar triangles is proportional to the squares of the corresponding sides of both the triangles

thanks

Answer 2

Answer:

Hope this helps

Step-by-step explanation: