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ABCD is a square. With centres A, B, C and D, four quadrants (each touching two of the remaining three) are drawn inside the square ABCD as shown in the figure. If the area of the shaded region is $42$ cm$^2$, find the side of the square ABCD. Also, find the perimeter of the shaded region.
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Let the side of the square be $x$ cm
nRadius of each quadrant = $\frac{x}{2}$ (1/2 Mark)
nArea of shaded region = $x^2 - 4 \times \frac{90}{360} \times \frac{22}{7} \times (\frac{x}{2})^2$ (1 Mark)
n$42 = x^2 - \frac{22}{7} \times \frac{x^2}{4} = x^2 - \frac{11x^2}{14} = \frac{3x^2}{14}$ (1/2 Mark)
n$\implies x^2 = \frac{42 \times 14}{3} = 14 \times 14 = 196 \implies x = 14$ (1 Mark)
nPerimeter of the shaded region = $4 \times \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = 44$ cm (1/2 Mark)
nRadius of each quadrant = $\frac{x}{2}$ (1/2 Mark)
nArea of shaded region = $x^2 - 4 \times \frac{90}{360} \times \frac{22}{7} \times (\frac{x}{2})^2$ (1 Mark)
n$42 = x^2 - \frac{22}{7} \times \frac{x^2}{4} = x^2 - \frac{11x^2}{14} = \frac{3x^2}{14}$ (1/2 Mark)
n$\implies x^2 = \frac{42 \times 14}{3} = 14 \times 14 = 196 \implies x = 14$ (1 Mark)
nPerimeter of the shaded region = $4 \times \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = 44$ cm (1/2 Mark)