In the given figure, (\mathrm{OD}=2 \times \mathrm{OB}), (\mathrm{OC}=2 \times \mathrm{OA}) and (\mathrm{CD}=2 \times \mathrm{AB}) then (\triangle \mathrm{AOB} \sim \triangle \mathrm{COD}) by ______.

In the given figure, OD = 2 × OB, OC = 2 × OA and CD = 2 × AB then △ AOB ~ △ COD by ______.

MCQ

रिकाम्या जागा भरा

SSS

Explanation:

Given,

⇒ OD = 2 × OB

⇒ `(OD)/(OB)` = 2 .......(1)

⇒ OC = 2 × OA

⇒ `(OC)/(OA)` = 2 .......(2)

⇒ CD = 2 × AB

⇒ `(CD)/(AB)` = 2 .......(3)

From equation (1), (2) and (3), we get:

⇒ `(OD)/(OB) = (OC)/(OA) = (CD)/(AB)`

∴ △ AOB ~ △ COD (By S.S.S. postulate)

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पाठ 15: Similarity (With Applications to Maps and Models) - EXERCISE 15 (A) [पृष्ठ २१५]

पाठ 15 Similarity (With Applications to Maps and Models)
EXERCISE 15 (A) | Q 1. (i) | पृष्ठ २१५