Questions: A square is cut as shown. What % is the middle section compared to the sum of the other two sections?

Solution Steps

The problem involves a square with a side length of 12 cm, which is divided into three sections by two lines originating from one corner and forming 30° angles with the sides of the square.

The area of the square is given by: \[ \text{Area of the square} = \text{side length}^2 = 12 \, \text{cm} \times 12 \, \text{cm} = 144 \, \text{cm}^2 \]

The square is divided into three sections by two lines forming 30° angles. These lines create three triangles within the square.

This is a 30°-60°-90° triangle. The side opposite the 30° angle is half the hypotenuse (which is the side of the square). \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \, \text{cm} \times 6\sqrt{3} \, \text{cm} = 18\sqrt{3} \, \text{cm}^2 \]

This is also a 30°-60°-90° triangle. The side opposite the 30° angle is half the hypotenuse (which is the side of the square). \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6\sqrt{3} \, \text{cm} \times 6 \, \text{cm} = 18\sqrt{3} \, \text{cm}^2 \]

This is a right triangle with the remaining area of the square. \[ \text{Area} = \text{Total area of the square} - \text{Area of Triangle 1} - \text{Area of Triangle 2} \] \[ \text{Area} = 144 \, \text{cm}^2 - 18\sqrt{3} \, \text{cm}^2 - 18\sqrt{3} \, \text{cm}^2 \] \[ \text{Area} = 144 \, \text{cm}^2 - 36\sqrt{3} \, \text{cm}^2 \]

The middle section is one of the triangles calculated above. \[ \text{Percentage} = \left( \frac{\text{Area of the middle section}}{\text{Sum of the areas of the other two sections}} \right) \times 100 \] \[ \text{Percentage} = \left( \frac{18\sqrt{3} \, \text{cm}^2}{144 \, \text{cm}^2 - 18\sqrt{3} \, \text{cm}^2} \right) \times 100 \]

Final Answer