NA, OC, QD and RB are perpendiculars to diagonal MP.
Answer:
It is clear from figure that polygon MNOPQR is divided into six parts, out of which four are triangles and two are trapeziums. Area of polygon MNOPQR = Area of \[\Delta MAN\]+ Area of trapezium ACON + Area of \[\Delta OCP\]+ Area of \[\Delta PDQ\]+ Area of trapezium DBR.Q + Area of ARBM ?...(i) Now, area of \[\Delta MAN\] \[=\frac{1}{2}\times MA\times AN=\frac{1}{2}\times 2\times 2.5=2.5\,c{{m}^{2}}\] Area of trapezium ACON = \[\frac{1}{2}\left( AN+OC \right)\times AC\] \[=\frac{1}{2}\text{ }\left( AN+OC \right)\times \left( MC-MA \right)~~~~~~~\] \[\left[ \text{ }\because \text{ }AC=MC-MA \right]\] \[=\frac{1}{2}\text{ }\left( 2.5+3 \right)\times \left( 6-2 \right)=\frac{1}{2}\times \text{ }5.5\times 4\] \[=\text{ }5.5\times 2=11\text{ }c{{m}^{2}}\] 
Area of \[\Delta OCP\]\[=\text{ }\frac{1}{2}\times \text{ }CP\times OC\] \[=\frac{1}{2}(MP-MC)\times OC\] \[=\frac{1}{2}(9-6)\times 3=\frac{1}{2}\times 3\times 3\] \[[\because \,\,CP\,=MP-MC]\] \[=\frac{9}{2}=4.5\,c{{m}^{2}}\] Area of \[\Delta PDQ=\frac{1}{2}\times PD\times DQ\] \[=\frac{1}{2}\times (MP-MD)\times DQ\] \[[\because \,PD=MP-MD]\] \[=\frac{1}{2}\times (9-7)\times 2=2c{{m}^{2}}\] Area of trapezium \[DBRQ=\frac{1}{2}\times \left( DQ+BR \right)\times BD\] \[=\frac{1}{2}\times (DQ+BR)\times (MD-MB)\]\[[\because \,BD=MD-MB]\] \[=\frac{1}{2}\times (2+2.5)\times (7-4)\] \[=\frac{1}{2}\times 4.5\times 3=\frac{13.5}{2}=6.75\,c{{m}^{2}}\] Area of \[\Delta RBM=\frac{1}{2}\times MB\times BR\] \[=\frac{1}{2}\times 4\times 2.5=5c{{m}^{2}}\] On putting all these values in Eq. (i), we get Area of polygon MNOPQR \[=\left( 2.5+11+4.5+2+6.75+5 \right)\text{ }c{{m}^{2}}\] \[=31.75\text{ }c{{m}^{2}}\] Hence, area of polygon MNOPQR is 31.75 \[c{{m}^{2}}.\]
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