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Theorem 6.6 - Ratio of areas of two similar triangles.jpg

2 Theorem 6.7 if perpendicular is drawn from the vertex of the right angle of right triangle to the hypotenuse.jpg
3 Theorem 6.6 - Class 10 -r(ABC PQR) = 12 BC x AM = 12 QR  x PN.jpg4 Theorem 6.6 - Class 10 - Now again using AB  PQ  = BC  QR = AC PR.jpg

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Transcript

Theorem 6.6 :- The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides. Given :- ∆ABC ~ ∆PQR To Prove :- ar (ABC)﷮ar (PQR)﷯ = AB﷮PQ﷯﷯﷮2﷯ = BC﷮QR﷯﷯﷮2﷯= AC﷮PR﷯﷯﷮2﷯ Construction :- Draw AM ⊥ BC and PN ⊥ QR. Proof: Divide (1) & (2) ar (ABC)﷮ar (PQR)﷯ = 1﷮2﷯ × BC × AM﷮ 1﷮2﷯ × QR × PN ﷯ ar (ABC)﷮ar (PQR)﷯ = BC × AM﷮QR × PN﷯ Now, ∆ABM and ∆PQN ∠B = ∠Q ∠M = ∠N ∆ABM ∼ ∆PQN ∴ AB﷮PQ﷯ = AM﷮PN﷯ From (A) ar (ABC)﷮ar (PQR)﷯ = BC × AM﷮QR × PN﷯ ar (ABC)﷮ar (PQR)﷯ = BC﷮QR﷯ × AB﷮PQ﷯ Now, Given ∆ABC ~ ∆PQR ⇒ AB﷮PQ﷯ = BC﷮QR﷯ = AC﷮PR﷯ Putting in (C) ⇒ ar (ABC)﷮ar (PQR)﷯ = AB﷮PQ﷯ × AB﷮PQ﷯ = AB﷮PQ﷯﷯﷮2﷯ Now, again using AB﷮PQ﷯ = BC﷮QR﷯ = AC﷮PR﷯ ⇒ ar (ABC)﷮ar (PQR)﷯ = AB﷮PQ﷯﷯﷮2﷯= BC﷮QR﷯﷯﷮2﷯ = AC﷮PR﷯﷯﷮2﷯ Hence Proved.

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CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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