| Column-l | Column-II |
(P)  | (1) \[54\text{ }c{{m}^{2}}\] |
(Q)  | (2) \[\text{90 }c{{m}^{2}}\] |
(R)  | (3) \[171\text{ }c{{m}^{2}}\] |
(S)  | (4) \[710\text{ }c{{m}^{2}}\] |
A) (P)\[\to \](3), (Q)\[\to \](1), (R)\[\to \](2), (S)\[\to \](4)
B) (P)\[\to \](1), (Q)\[\to \](2), (R)\[\to \](3), (S)\[\to \](4)
C) (P)\[\to \](2), (Q)\[\to \](3), (R)\[\to \](1), (S)\[\to \](4)
D) (P)\[\to \](3), (Q)\[\to \](1), (R)\[\to \](4), (S)\[\to \](2)
Correct Answer: D
Solution :
(P) Area of \[\Delta ABC\] \[=\frac{1}{2}\times 25\times 18=225c{{m}^{2}}\] = Area of \[\Delta BDC=\frac{1}{2}\times 18\times 6=54\,c{{m}^{2}}\] \[\therefore \] Area of shaded region \[=225-54=171\text{ }c{{m}^{2}}\] (Q) Area of shaded region \[=\frac{1}{2}\times AB\times AE+\frac{1}{2}\times AB\times DE\] \[=\frac{1}{2}\times AB\times (AE+DE)=\frac{1}{2}\times 6\times 18=54c{{m}^{2}}\] (R) Area of shaded region = Area of \[\Delta \,AED\]+ Area of square ABCD - Area of circle \[=\frac{1}{2}\times AD\times ED+DC\times DC-\pi \times {{(7)}^{2}}\] \[=\frac{1}{2}\times 24\times 24+{{(24)}^{2}}-\frac{22}{7}\times {{(7)}^{2}}\] \[288+576-154=710\text{ }c{{m}^{2}}\] (S) Shaded area = Area of rectangle ABCD - Area of rectangle EFGH \[=AB\times SC-EF\times FG\] \[=(18+3)\times (9+3)-18\times 9\] \[=252-162=90c{{m}^{2}}\]
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