Find the area of the shape below, the area of the shape shaded portion

$$\text{Area of sector} = \frac{\theta}{360^\circ} \pi r^2$$ $$\text{Area of sector} = \frac{90^\circ}{360^\circ} \pi (9 \text{ cm})^2$$ $$\text{Area of sector} = \frac{1}{4} \pi (81 \text{ cm}^2)$$ $$\text{Area of sector} = \frac{81\pi}{4} \text{ cm}^2$$

$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$$

Since it's a right-angled triangle with legs as radii:

$$\text{Area of triangle} = \frac{1}{2} \times 9 \text{ cm} \times 9 \text{ cm}$$ $$\text{Area of triangle} = \frac{81}{2} \text{ cm}^2$$ $$\text{Area of triangle} = 40.5 \text{ cm}^2$$

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$ $$\text{Area of segment} = \left(\frac{81\pi}{4} - 40.5\right) \text{ cm}^2$$

Using $\pi \approx 3.14159$:

$$\text{Area of segment} \approx \left(\frac{81 \times 3.14159}{4} - 40.5\right) \text{ cm}^2$$ $$\text{Area of segment} \approx (63.617 - 40.5) \text{ cm}^2$$ $$\text{Area of segment} \approx 23.117 \text{ cm}^2$$

The area of the circular segment is $\boxed{\left(\frac{81\pi}{4} - 40.5\right) \text{ cm}^2 \approx 23.12 \text{ cm}^2}$.

Height = 8 cm
Width = 4 cm

$$\text{Area}_1 = 8 \text{ cm} \times 4 \text{ cm} = 32 \text{ cm}^2$$

Height = 4 cm
Width = 4 cm

$$\text{Area}_2 = 4 \text{ cm} \times 4 \text{ cm} = 16 \text{ cm}^2$$

$$\text{Total Area} = \text{Area}_1 + \text{Area}_2$$ $$\text{Total Area} = 32 \text{ cm}^2 + 16 \text{ cm}^2$$ $$\text{Total Area} = 48 \text{ cm}^2$$

The area of the shaded L-shaped portion is $\boxed{48 \text{ cm}^2}$.

$$152 \text{ cm}^2 = \frac{1}{2} (6 \text{ cm} + 8 \text{ cm}) h$$ $$152 \text{ cm}^2 = \frac{1}{2} (14 \text{ cm}) h$$ $$152 \text{ cm}^2 = 7 \text{ cm} \cdot h$$

$$h = \frac{152 \text{ cm}^2}{7 \text{ cm}}$$ $$h = \frac{152}{7} \text{ cm}$$ $$h \approx 21.71 \text{ cm}$$

The height of the trapezium is $\boxed{\frac{152}{7} \text{ cm} \approx 21.71 \text{ cm}}$.

$$84 \text{ cm}^2 \times 2 = \text{base} \times 19 \text{ cm}$$ $$168 \text{ cm}^2 = \text{base} \times 19 \text{ cm}$$ $$\text{base} = \frac{168 \text{ cm}^2}{19 \text{ cm}}$$ $$\text{base} = \frac{168}{19} \text{ cm}$$ $$\text{base} \approx 8.84 \text{ cm}$$

The base of the triangle is $\boxed{\frac{168}{19} \text{ cm} \approx 8.84 \text{ cm}}$.