Question

prove that ratio of area of 2similar triangle is equal to the square of the ratio of there corresponding sides
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Answer 1

Answer:

If there are two triangles say ΔABC and ΔPQR, then they are similar if,

i)    ∠A=∠P, ∠B=∠Q and ∠C=∠R

ii)    ABPQ = BCQR  = ACPR

If we have two similar triangles, then not only their angles and sides share a relationship but also the ratio of their perimeter, altitudes, angle bisectors, areas and other aspects are in ratio.

In the upcoming discussion, the relation between the areas of two similar triangles is discussed.

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 formula

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,

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 = 

Please mark as brilliant answer