A) 3 : 2
B) 4 : 3
C) 5 : 4
D) 3 : 1
Correct Answer: D
Solution :
[d] In right angled \[\Delta ABC,\] \[{{(a+b)}^{2}}={{(a-b)}^{2}}+{{a}^{2}}\] \[\Rightarrow \] \[4b=a\] Now, \[\frac{BD}{DC}=\frac{a-b}{b}=\frac{4b-b}{b}\] \[=\frac{3b}{b}=\frac{3}{1}=3:1\] |
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