1. **Problem statement:** We have an equilateral triangle ABC with side length 10 cm. A circle centered at point A intersects the midpoints of sides AB and AC. We need to find the shaded area of the triangle formed by these intersections. 2. **Key information:** - Side length of equilateral triangle ABC: $10$ cm - Circle centered at A intersects midpoints of AB and AC 3. **Step 1: Calculate the height of the equilateral triangle ABC.** The height $h$ of an equilateral triangle with side length $s$ is given by: $$h = \frac{\sqrt{3}}{2} s$$ Substitute $s=10$: $$h = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \approx 8.660$$ cm 4. **Step 2: Coordinates setup for clarity.** Place point A at the origin $(0,0)$. - Point B at $(10,0)$ - Point C at $(5, 5\sqrt{3})$ 5. **Step 3: Find midpoints of AB and AC.** - Midpoint of AB, $M_{AB}$: $$\left( \frac{0+10}{2}, \frac{0+0}{2} \right) = (5,0)$$ - Midpoint of AC, $M_{AC}$: $$\left( \frac{0+5}{2}, \frac{0+5\sqrt{3}}{2} \right) = \left(2.5, \frac{5\sqrt{3}}{2} \right) \approx (2.5, 4.33)$$ 6. **Step 4: Radius of the circle centered at A.** The circle passes through these midpoints, so radius $r$ is the distance from A to either midpoint. Calculate distance from A to $M_{AB}$: $$r = \sqrt{(5-0)^2 + (0-0)^2} = 5$$ Similarly, distance from A to $M_{AC}$: $$r = \sqrt{(2.5-0)^2 + \left(\frac{5\sqrt{3}}{2} - 0\right)^2} = \sqrt{2.5^2 + \left(4.33\right)^2} = \sqrt{6.25 + 18.75} = \sqrt{25} = 5$$ 7. **Step 5: Identify the shaded triangle.** The shaded area is the triangle formed by points A, $M_{AB}$, and $M_{AC}$. 8. **Step 6: Calculate the area of triangle $AM_{AB}M_{AC}$.** Use the formula for area of triangle given coordinates: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Let: - $A = (0,0)$ - $M_{AB} = (5,0)$ - $M_{AC} = \left(2.5, \frac{5\sqrt{3}}{2}\right)$ Calculate: $$\text{Area} = \frac{1}{2} |0(0 - \frac{5\sqrt{3}}{2}) + 5\left(\frac{5\sqrt{3}}{2} - 0\right) + 2.5(0 - 0)|$$ $$= \frac{1}{2} |0 + 5 \times \frac{5\sqrt{3}}{2} + 0| = \frac{1}{2} \times \frac{25\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}$$ 9. **Step 7: Calculate numerical value.** $$\frac{25\sqrt{3}}{4} \approx \frac{25 \times 1.732}{4} = \frac{43.3}{4} = 10.825$$ 10. **Step 8: Final answer rounded to 3 significant figures:** $$\boxed{10.8 \text{ cm}^2}$$ This is the shaded area of the triangle formed by A and the midpoints of AB and AC.