A) 8cm
B) 7cm
C) \[8\sqrt{2}\,cm\]
D) \[7\sqrt{3}\,cm\]
Correct Answer: C
Solution :
Join 0 to P and Q. Join P to R. Draw \[\pi \]. Now SP = QR, as they are opposite sides of rectangle PRQS. \[\text{4}0={{\text{2}}^{\text{3}}}\times \text{5}\] \[\text{6}0={{\text{2}}^{\text{2}}}\times \text{3}\times \text{5}\] \[\therefore \] \[={{2}^{2}}\times 3\times 5\times 2=120\] \[f(p)={{\text{p}}^{\text{3}}}+\text{6}{{\text{p}}^{\text{2}}}+\text{lip}+\text{6}\] \[f(p)\] \[p(x)={{x}^{2}}+3x-2\]
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