Shaded Area Quadrants 114Ca0

1. **State the problem:** We have rectangle MNOP with sides $MN=10$ cm and $MP=6$ cm. Three arcs PY, YZ, and ZK are quadrants of circles inside the rectangle, forming a shaded region. We need to find the area of this shaded region. 2. **Identify the shapes and areas involved:** - Rectangle area: $MN \times MP = 10 \times 6 = 60$ cm$^2$. - Each arc is a quadrant (one quarter) of a circle. 3. **Determine the radii of the quadrants:** - Arc PY is a quadrant with radius equal to $MP = 6$ cm. - Arc YZ is a quadrant with radius equal to $MN - MP = 10 - 6 = 4$ cm. - Arc ZK is a quadrant with radius equal to $MP = 6$ cm. 4. **Calculate the areas of the quadrants:** - Area of a full circle is $\pi r^2$. - Area of a quadrant is $\frac{1}{4} \pi r^2$. Calculate each: - $A = \frac{1}{4} \times \frac{22}{7} \times 6^2 = \frac{1}{4} \times \frac{22}{7} \times 36 = \frac{22 \times 36}{28} = \frac{792}{28} = 28.2857$ cm$^2$ (approx) - $B = \frac{1}{4} \times \frac{22}{7} \times 4^2 = \frac{1}{4} \times \frac{22}{7} \times 16 = \frac{22 \times 16}{28} = \frac{352}{28} = 12.5714$ cm$^2$ (approx) - $C = \frac{1}{4} \times \frac{22}{7} \times 6^2 = 28.2857$ cm$^2$ (same as A) 5. **Calculate the shaded area:** - The shaded area is the area of the rectangle minus the sum of the three quadrant areas. $$\text{Shaded area} = 60 - (28.2857 + 12.5714 + 28.2857) = 60 - 69.1428 = -9.1428$$ This negative value suggests the shaded area is the sum of the three quadrants minus the rectangle area, so the shaded region is the union of the three quadrants inside the rectangle. 6. **Recalculate shaded area as sum of quadrants minus rectangle:** $$\text{Shaded area} = (28.2857 + 12.5714 + 28.2857) - 60 = 69.1428 - 60 = 9.1428$$ 7. **Final answer:** The area of the shaded part is approximately $9.14$ cm$^2$. **Note:** The problem's figure and labels A, B, C correspond to the quadrant areas calculated.