A) \[425\,\,\text{c}{{\text{m}}^{2}}\]
B) \[425.75\,\,\text{c}{{\text{m}}^{2}}\]
C) \[428\,\,\text{c}{{\text{m}}^{2}}\]
D) \[\text{428}\text{.75}\,\,\text{c}{{\text{m}}^{\text{2}}}\]
Correct Answer: D
Solution :
In \[\Delta ABC,\] |
\[A{{C}^{2}}=\sqrt{{{28}^{2}}+{{21}^{2}}}=\sqrt{784+441}=\sqrt{1225}\] |
\[\Rightarrow \] \[AC=35\,\,\text{cm}\] |
Area of shaded portion = Area of semi-circle ACE |
+ Area of A ABC -Area of quadrant circle BCD |
\[=\frac{\pi {{r}^{2}}}{2}+\frac{1}{2}\times BC\times BA-\frac{\pi }{4}\times r_{1}^{2}\] |
\[=\frac{22}{7}\times \frac{1}{2}\times \frac{35}{2}\times \frac{35}{2}+\frac{1}{2}\times 21\times 28-\frac{22}{7\times 4}\times 21\times 21\]\[=\frac{5\times 11\times 35}{4}+\frac{1}{2}[21\times 28-33\times 21]\]\[=\frac{1925}{4}+\frac{1}{2}(-105)=48.125-5250=428.75\,\,\text{c}{{\text{m}}^{2}}\] |
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