हिंदी

In the given figure: (i) △ ABE ~ △ ADE (ii) △ ADE ~ △ ABC (iii) △ ADE ~ △ BAC (iv) △ ADE ~ △ CAB

Advertisements
Advertisements

प्रश्न

In the given figure:

विकल्प

  • △ ABE ~ △ ADE

  • △ ADE ~ △ ABC

  • △ ADE ~ △ BAC

  • △ ADE ~ △ CAB

MCQ
Advertisements

उत्तर

△ ADE ~ △ ABC

Explanation:

From figure,

⇒ ∠BAE = x (let)

⇒ ∠DAB = ∠EAC = y (let)

⇒ ∠AED = ∠ACB = z (let)

⇒ ∠DAE = ∠DAB + ∠BAE = y + x

⇒ ∠BAC = ∠BAE + ∠EAC = x + y

⇒ ∠DAE = ∠BAC = x + y

In △ ABC and △ ADE,

⇒ ∠DAE = ∠BAC (Proved above)

⇒ ∠AED = ∠ACB (Both equal to z)

∴ △ ADE ~ △ ABC (By A.A. axiom)

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Similarity (With Applications to Maps and Models) - EXERCISE 15 (A) [पृष्ठ २१५]

APPEARS IN

सेलिना Concise Mathematics [English] Class 10 ICSE
अध्याय 15 Similarity (With Applications to Maps and Models)
EXERCISE 15 (A) | Q 1. (g) | पृष्ठ २१५
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×