A) \[3+\sqrt{5}\]units
B) \[3-\sqrt{5}\]units
C) \[2-\sqrt{5}\]units
D) \[\sqrt{3}\]units
Correct Answer: B
Solution :
Given, \[A{{C}^{2}}=AB\times CB\] |
\[\Rightarrow \] \[{{x}^{2}}=2\times \,\,(2-x)\] |
\[\Rightarrow \] \[{{x}^{2}}=4-2x\] |
\[\Rightarrow \] \[{{x}^{2}}+2x-4=0\] |
\[\Rightarrow \] \[x=\frac{-\,\,2\pm \sqrt{4+16}}{2\times 1}\] |
\[\Rightarrow \] \[x=-1\pm \sqrt{5}\] |
Now, \[BC=2-(-1\pm \sqrt{5})\] |
\[=3-\sqrt{5}\] (neglect\[3+\sqrt{5}\because 3+\sqrt{5}>2\]) |
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