A) \[\frac{\sqrt{3}}{\sqrt{2}}\]
B) \[\frac{\sqrt{3}}{2}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
In \[\Delta ABC\] \[AB=y+1\], \[AC=1-y\], \[BC=1\] |
\[A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}\] |
\[{{(y+1)}^{2}}={{(1-y)}^{2}}+{{(1)}^{2}}\] |
\[{{(1+y)}^{2}}-{{(1-y)}^{2}}=1\] |
\[4y=1\Rightarrow y=\frac{1}{4}\]Hence, radius of circle \[T=\frac{1}{4}\] |
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