According to the given question,

\[AP\text{ }=\text{ }\left( 1/3 \right)\text{ }PB\]

So, \[AP/PB\text{ }=\text{ }1/3\]

In \[\vartriangle \text{ }APQ\text{ }and\text{ }\vartriangle ABC\]

As\[PQ\text{ }||\text{ }BC\], corresponding angles are equal

\[\angle APQ\text{ }=\angle ABC\]and \[\angle AQP\text{ }=\angle ACB\]

Because, \[\vartriangle APQ\text{ }\sim\text{ }\vartriangle ABC\]by AA criterion for similarity

Thus,

(i) \[Area\text{ }of\text{ }\vartriangle ABC/\text{ }Area\text{ }of\text{ }\vartriangle APQ\]

\[=\text{ }A{{B}^{2}}/\text{ }A{{P}^{2}}\]

So,

\[=\text{ }{{4}^{2}}/{{1}^{2}}~=\text{ }16:\text{ }1\]

\[\left[ AP/PB\text{ }=\text{ }1/3\text{ }so,\text{ }AB/AP\text{ }=\text{ }4/1 \right]\]

(ii) \[Area\text{ }of\text{ }\Delta \text{ }APQ/Area\text{ }of\text{ }Trapezium\text{ }PBCQ\]

\[=\text{ }Area\text{ }of\text{ }\Delta \text{ }APQ/\]

\[\left( Area\text{ }of\text{ }\Delta \text{ }ABC\text{ }\text{ }Area\text{ }of\text{ }\Delta \text{ }APQ \right)\]

That is,

\[=\text{ }1/\text{ }\left( 16/\text{ }1 \right)\text{ }=\text{ }1:\text{ }16\]