question_answer

Given that \[\text{PB}\bot \text{AB}\] and \[~\text{QA}\bot \text{AB}\] PO=4 cm and\[\text{QO}=\text{7cm}\], if area of\[\Delta \text{QAO}\] is\[\text{245 c}{{\text{m}}^{\text{2}}}\], what is the area of\[\Delta \text{PBO}\]?

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) \[AC+CD=\sqrt{10}+\sqrt{14}\,cm\]\[=\sqrt{2}(\sqrt{5}+\sqrt{7})\,cm\] \[\Delta \text{ABC}\tilde{\ }\Delta D\text{EF}\]\[\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{A{{P}^{2}}}{D{{Q}^{2}}}\]