| Column-l | Column-II |
(P)  | (i) \[262.50\text{ }c{{m}^{2}}\] |
(Q)  | (ii) \[271.04\text{ }c{{m}^{2}}\] |
(R)  | (iii) \[427.14\,\,c{{m}^{2}}\] |
A) (P) \[\to \](iii); (Q) \[\to \](ii); (R) \[\to \] (i)
B) (P) \[\to \] (ii); (Q) \[\to \] (i); (R)\[\to \] (iii)
C) (P) \[\to \] (ii); (Q) \[\to \] (iii); (R) \[\to \] (i)
D) (P) \[\to \] (iii); (Q) \[\to \](i); (R) \[\to \] (ii)
Correct Answer: C
Solution :
(P) Draw OD perpendicular to AD. Now, in \[\Delta \,{\mathrm O}D\Beta ,\]
\[\cos \,{{60}^{o}}=\frac{OD}{OB}\Rightarrow \frac{1}{2}=\frac{OD}{21}\Rightarrow OD=\frac{21}{2}cm\] And \[\sin \,{{60}^{o}}=\frac{DB}{OB}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}=\frac{DB}{21}\,\,\Rightarrow \,DB=\frac{21\sqrt{3}}{2}cm\] \[\therefore \] \[AB=2\times DB=21\sqrt{3}\,cm\] Now, area of \[\Delta OAB=\frac{1}{2}\times OD\times AB\] \[=\frac{1}{2}\times \frac{21}{2}\times 21\sqrt{3}=\frac{441\sqrt{3}}{4}\] Area of shaded region = Area of minor sector - Area of \[\Delta \,OAB\] \[=\frac{120}{360}\times \frac{22}{7}\times {{21}^{2}}-\frac{441\sqrt{3}}{4}\] \[=462-190.96=271.04\,c{{m}^{2}}\] (Q)Since, \[\angle AOB=\frac{{{360}^{o}}}{6}={{60}^{o}}\] And also \[\angle B=\angle A\] So, \[\angle A=\angle B=\angle AOB={{60}^{o}}\] \[\therefore \] \[\angle AOB\]is an equilateral triangle Area of shaded region \[=6\text{ }\times \][Area of minor sector - Area of \[\Delta \,OAB\]] \[=6\times \left[ \frac{{{60}^{o}}}{360}\times \frac{22}{7}\times {{28}^{2}}-\frac{\sqrt{3}}{4}\times {{28}^{2}} \right]\] \[=6\times [410.67-339.48]=6\times 71.19\] \[=427.14c{{m}^{2}}\] (R) Area of square \[=352\text{ }c{{m}^{2}}~~\] Radius of semicircle \[=\frac{35}{2}cm\] Area of semicircle \[=\frac{1}{2}\times \pi \times {{\left( \frac{35}{2} \right)}^{2}}=481.25c{{m}^{2}}\] Area of shaded region = Area of square \[-2\times \] (Radius of semicircle) \[=1225-2\times 481.25=262.50\,c{{m}^{2}}\]
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