A) 2
B) 3/2
C) 1
D) 8/9
Correct Answer: C
Solution :
Let ED = x |
Now, \[AC=\sqrt{{{8}^{2}}+{{6}^{2}}}=10\] |
In \[\Delta \,\,AED,\] |
\[A{{E}^{2}}=A{{D}^{2}}-{{x}^{2}}=36-{{x}^{2}}\] ...(i) |
And in \[\Delta CFD,\] |
\[C{{F}^{2}}={{(8)}^{2}}-{{(10-x)}^{2}}\] ...(ii) |
From Eqs. (i) and (ii), we get |
\[36-{{x}^{2}}=64-{{(10-x)}^{2}}\] \[(\because AE=FC)\] |
\[\Rightarrow \] \[36-{{x}^{2}}=64-(100+{{x}^{2}}-20x)\] |
\[\Rightarrow \] \[20x=72\Rightarrow x=\frac{18}{5}\] |
\[\therefore \]From Eq. (i),\[A{{E}^{2}}=36-{{\left( \frac{18}{5} \right)}^{2}}\] |
\[A{{E}^{2}}=36-\frac{324}{25}=\frac{900-324}{25}\] |
\[\Rightarrow \] \[A{{E}^{2}}=\frac{576}{25}\]\[\Rightarrow \]\[AE=\frac{24}{5}\] |
\[\therefore \]\[\frac{\text{Area}\,\,\text{of}\,\,\text{rectangle}\,\,\text{ABCD}}{\text{Area}\,\,\text{of}\,\,\text{rectangle}\,\,\text{AEFC}}=\frac{8\times 6}{10\times \frac{24}{5}}=1\] |
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