The figure below is made up of a rectangle ABCD and a semicircle. AEC is a straight line. The arc of the semicircle touches AB at point F. DC = 16 cm and AF = FB. (Take $\pi = 3.14$)

1. Identify the radius of the semicircle

The diameter of the semicircle is the base of the rectangle, $DC = 16\text{ cm}$.
Radius $r = 16 \div 2 = 8\text{ cm}$.

2. Calculate the area of the semicircle

Area of semicircle $= \frac{1}{2} \times \pi \times r^2$
$= \frac{1}{2} \times 3.14 \times 8^2$
$= \frac{1}{2} \times 3.14 \times 64$
$= 3.14 \times 32$
$= 100.48\text{ cm}^2$

1. Identify the relationship between the given areas and the standard geometric shapes

Let the unshaded region inside the semicircle and below the diagonal $AEC$ be $U$.
Area of semicircle = $\text{Area } X + U$
Area of triangle $ADC$ (the region below the diagonal $AEC$) = $\text{Area } Y + U$
Therefore, $\text{Area } X - \text{Area } Y = \text{Area of semicircle} - \text{Area of triangle } ADC$.

2. Calculate the area of the semicircle

Radius of semicircle $r = 16 \div 2 = 8\text{ cm}$.
$\text{Area of semicircle} = \frac{1}{2} \times \pi \times r^2$
$= \frac{1}{2} \times 3.14 \times 8^2 = 100.48\text{ cm}^2$

3. Calculate the area of triangle ADC and find the difference

$\text{Area of triangle } ADC = \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times 16 \times 8 = 64\text{ cm}^2$
$\text{Difference} = 100.48 - 64 = 36.48\text{ cm}^2$