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