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    				| 1. | In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If `AD = sqrt25` cm then area of the rectangle is A. `32 cm^(2)`B. `40 cm^(2)`C. `44 cm^(2)`D. `48 cm^(2)` | 
| Answer» Correct Answer - B `AB^(2)=AC^(2)-BC^(2)=(AC^(2)-AD^(2))={(10^(2))-2(sqrt5)^(2)}=(100-20)=80` `rArr AB=sqrt80 = sqrt(16xx5) = 4sqrt5 cm`. `therefore ar("rect. ABCD")=(ABxxAD)=(4sqrt5xx2sqrt5)cm^(2) = 40 cm^(2)` | |