A) \[\text{6}0\text{ c}{{\text{m}}^{\text{2}}}\]
B) \[\text{4}0\text{ c}{{\text{m}}^{\text{2}}}\]
C) \[\text{125}\,\text{c}{{\text{m}}^{\text{2}}}\]
D) \[\text{8}0\,\text{c}{{\text{m}}^{\text{2}}}\]
Correct Answer: D
Solution :
Given\[-\left( \frac{1}{n}+\frac{2}{n}+........'n'\,terms \right)\]and \[=n-\left( \frac{1+2+3+.....+'n'terms}{n} \right)\], PO = 4 cm, QO = 7 cm and area of \[=n-\frac{n(n+1)}{2n}\]. In\[=n-\frac{n+1}{2}=\frac{n-1}{2}\]and\[\Delta ABC,AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\sqrt{10}\,cm\], \[\Delta ACD,CD=\sqrt{A{{D}^{2}}+A{{C}^{2}}}\] \[=\sqrt{14}cm\] \[\therefore \] (A.A. corollary) \[AC+CD=\sqrt{10}+\sqrt{14}\,cm\]\[=\sqrt{2}(\sqrt{5}+\sqrt{7})\,cm\] \[\Delta \text{ABC}\tilde{\ }\Delta D\text{EF}\]\[\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{A{{P}^{2}}}{D{{Q}^{2}}}\]You need to login to perform this action.
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