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ICSE Class X Math 2013 Paper

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In the given figure, ABA B and DED E are perpendicular to BCB C.
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Question : 13 of 46
Marks: +1, -0
Prove that △ABC∼△DEC\triangle ABC \sim \triangle DEC
Solution:  
From △ABC\triangle ABC and △DEC\triangle DEC ,
∠ABC  =  ∠DEC\angle ABC\;=\;\angle DEC
  =90∘      (Given)  \;=90^{\circ}\;\;\;\text{(Given)}\;
  and      ∠ACB=  ∠DCE  (Common)  \;\text{and}\;\;\;\angle ACB=\;\angle DCE \;\text{(Common)}\;
∴    △ABC  ∼  △DEC\therefore\;\;\triangle ABC\;\sim\;\triangle DEC
  (  By AA similarity)  \;(\;\text{By AA similarity})\;
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