Question

prove that areas of similar triangle are proportional to the squares on the corresponding sides ​

Answer 1

it is a theorem in similar triangles chapter...THEOREM:The ratio of areas of two similar triangles is equal to the squares of their corresponding sides

Answer 2

Step-by-step explanation:

Given : AB/PQ =AC/PR = BC/QR      .........................i

InΔABC & ΔPQR

ar.ΔABC/ar.ΔPQR = (1/2× base × height )/(1/2×base ×height)

⇒  1/2 × BC× AD/ 1/2× QR× PM

⇒ BC×AD/QR×PM          ............................ii

InΔABD &ΔPQM

∠ADB  =  ∠PNQ                  =  90°

∠ABD  =  ∠PQM                  =   ΔABC~ ΔPQR

SO, ΔABD ~ ΔPQM  by AA

AB/PQ = BD/QM = AD/PM

AB/PQ = AD/PM

So, BC/QR = AD/PM                             From  (i)

Put value of AD/PM in eq   ii

ar.ΔABC/ar.ΔPQR = BC/QR × AD/PM

⇒       BC/QR× BC/QR

⇒       BC^2/QR^2

Hence proved