What is the unshaded area of the figure in \[\text{c}{{\text{m}}^{\text{2}}}\]? (Take\[\pi =\text{3}.\text{14}\].)
A) 40
B) 43
C) 25
D) 33
Correct Answer: B
Solution :
Radius of each circle \[f(p)={{\text{p}}^{\text{3}}}+\text{6}{{\text{p}}^{\text{2}}}+\text{lip}+\text{6}\] Area of shaded region\[f(p)\] \[p(x)={{x}^{2}}+3x-2\] Breadth of rectangle = Diameter of circle \[\Rightarrow \] Area of rectangle = Diameter of circle \[p(-1)={{(-1)}^{2}}+3(-1)-2=(-4)\] \[p(-1)=-4\]Area of the unshaded region \[({{m}^{2}}+9){{x}^{2}}+13x+6m\] Hence, the unshaded area of the figure is\[({{m}^{2}}+9){{x}^{2}}+13x+6m\].
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