Question

prove that areas of similar triangles is equal to the squareof the ratio of corresponding circumstances radii

Answer 1

Step-by-step explanation:

For proportionality,

Angle(A) = Angle(P),

Angle(B) = Angle(Q) and

Angle(C) = Angle(R)

As both ∆s are proportional, so

AB/PQ= BC/QR = AC/PR = k

Area(∆ABC) = (1/2) × AO × BC

Area(∆PQR) = (1/2) × PO × QR

Divide both,

[Area(∆ABC)/Area(∆PQR)] = [(1/2) × AO × BC] / [(1/2) × PO × QR]

= [AO × BC] / [PO × QR]

= (AO/PO) × k [As BC/QR = k]

Now in ∆ABO and ∆PQO, it can be seen that:

∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)

∠AOB = ∠POQ (Since both the angles are 90°)

From AA criterion of similarity ∆AOB ~∆POQ

So, AO/PO = AB/PQ = k

Therefore,

[Area(∆ABC)/Area(∆PQR)] = (AO/PO) × k = (k)^2 = (AB/PQ)^2

Hence Prooved.