Leon designed a logo as shown. The logo is made up of two identical small semi-circles, a circle and a big quadrant. YZ is the diameter of the circle and O is the centre of the circle. X is the centre of the big quadrant and XY = XZ. The radius of the small semi-circle is 7 cm. (Take $\pi = \tfrac{22}{7}$)

1. Determine the diameter of the small semi-circle

Diameter of small semi-circle = $2 \times 7 = 14$ cm

2. Determine the length of YZ

The two small semi-circles span the diameter of the large circle. Thus, $YZ = 2 \times 14 = 28$ cm.

1. Determine the length of YZ from part (a)

From part (a), the diameter of the circle $YZ$ is equal to $4 \times 7 = 28\text{ cm}$.

2. Identify the properties of triangle XYZ

Since $X$ is the centre of the big quadrant and $XY, XZ$ are its radii, $\angle YXZ = 90^\circ$. Thus, triangle $XYZ$ is a right-angled isosceles triangle with hypotenuse $YZ = 28\text{ cm}$.

3. Calculate the area of triangle XYZ

Let the height of triangle $XYZ$ from $X$ to $YZ$ be $h$. Since it is a right-angled isosceles triangle, the height to the hypotenuse is half the hypotenuse. So $h = \frac{28}{2} = 14\text{ cm}$.
Area of $\triangle XYZ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 28 \times 14 = 196\text{ cm}^2$.
Alternatively, $XY^2 + XZ^2 = YZ^2 \implies 2r^2 = 28^2 = 784 \implies r^2 = 392$.
Area of $\triangle XYZ = \frac{1}{2} \times XY \times XZ = \frac{1}{2} r^2 = \frac{1}{2} \times 392 = 196\text{ cm}^2$.

1. Find the area of the two small semi-circles.

Area of two small semi-circles = $2 \times \frac{1}{2} \times \pi \times r^2 = \pi \times 7^2$
$= \frac{22}{7} \times 49 = 154\text{ cm}^2$

2. Find the square of the radius of the big quadrant ($XY^2$).

From the figure, the diameter of the circle $YZ = 7 + 7 + 7 + 7 = 28\text{ cm}$.
Since $X$ is the centre of the quadrant, $\angle YXZ = 90^\circ$ and $XY = XZ$.
Using Pythagoras' theorem in $\triangle XYZ$:
$XY^2 + XZ^2 = YZ^2$
$2XY^2 = 28^2 = 784$
$XY^2 = 392$

3. Find the area of the big quadrant and the total shaded area.

Area of the big quadrant = $\frac{1}{4} \times \pi \times XY^2$
$= \frac{1}{4} \times \frac{22}{7} \times 392 = 308\text{ cm}^2$
Total shaded area = $154 + 308 = 462\text{ cm}^2$