question_answer

ABCD is a rectangle of dimensions 8 units and 6 units. AEFC is a rectangle drawn in such a way that diagonal AC of the first rectangle is one side and side opposite to it is touching the first rectangle at D as shown in the given figure. What is the ratio of the area of rectangle ABCD to that of AEFC?   

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),\[A{{E}^{2}}=36-{{\left( \frac{18}{5} \right)}^{2}}\]

\[A{{E}^{2}}=36-\frac{324}{25}=\frac{900-324}{25}\]

\[\Rightarrow \]   \[A{{E}^{2}}=\frac{576}{25}\]\[\Rightarrow \]\[AE=\frac{24}{5}\]

\[\therefore \]\[\frac{\text{Area}\,\,\text{of}\,\,\text{rectangle}\,\,\text{ABCD}}{\text{Area}\,\,\text{of}\,\,\text{rectangle}\,\,\text{AEFC}}=\frac{8\times 6}{10\times \frac{24}{5}}=1\]