question_answer

Divide the following polygons into parts (triangles and trapezium) to find out its area.

Answer:

(a) Given, a polygon EFGHI and FI is a diagonal of it. To find the area, we have to divide this polygon into triangle and trapezium. So, firstly draw perpendicular from opposite vertical on FI i.e. from points G, H and E to FI. Thus, we get perpendiculars GM, HN and EP respectively on FI and polygon is divided into 5 parts, out of which four are triangles and one is trapezium.

\[\therefore \]Area of polygon EFGHI = Area of \[\Delta GMF\]+ Area of trapezium GMNH

+ Area of \[\Delta HNI\]+ Area of \[\Delta EPI\]+ Area of \[\Delta EPF\]

\[\left( \frac{1}{2}\times FM\times GM \right)+\left[ \frac{1}{2}(GM+HN)\times MN \right]+\]

\[\frac{1}{2}\times NI\times HN+\left( \frac{1}{2}\times PI\times EP \right)+\left( \frac{1}{2}\times PF\times EP \right)\]

\[\left[ \begin{align}   & \because \,\,\,area\text{ }of\text{ }triangle\text{ }\frac{1}{2}\times \text{ b}ase\text{ }height\text{ }and\,\,\,\,\, \\  & area\text{ }of\text{ }trapezium\text{ }=\frac{1}{2}\left( sum\text{ }of\text{ }parallel\text{ }sides \right)\times height \\ \end{align} \right]\]

(b) Given, polygon is MNOPQR and NQ is a diagonal of it. Then, draw perpendiculars on diagonal QN from vertices O, M, P and R i.e. draw \[OA\bot NQ,\text{ }MB\bot NQ,\text{ }PC\bot NQ\text{ }and\text{ }RD\bot NQ.\]

Thus, polygon is divided into 6 parts, out of which four are triangles and two are trapeziums.

\[\therefore \] Area of polygon MNOPQR = Area of\[~\Delta OAN\] + Area of trapezium CPOA + Area of \[\Delta PCQ\]+ Area of \[\Delta RDQ\]+ Area of trapezium MBDR + Area of\[~\Delta MBN\]

\[=\left( \frac{1}{2}\times AN\times AO \right)+\left[ \frac{1}{2}\times (OA+PC)\times AC \right]\]                                           \[+\left( \frac{1}{2}\times QC\times PC \right)+\left( \frac{1}{2}\times QD\times RD \right)+\]

\[\left[ \frac{1}{2}\times (DR+MB)\times BD \right]+\left( \frac{1}{2}\times BN\times BM \right)\]