The problem described in the image is related to an isosceles trapezoid $ABCD$, with the following properties:

• $AB \parallel CD$ (AB and CD are parallel)
• $AD = BC$ (the non-parallel sides are equal)
• $AB = 24 \, \text{cm}$ (the larger base)
• $AC \perp BC$ (diagonal $AC$ is perpendicular to base $BC$)

You are asked to calculate the following:

a) The area of the trapezoid $ABCD$.

b) The distance from the vertex $D$ to the side $BC$.

c) The area of triangle $MAB$, where $M$ is the point where $AD$ intersects $BC$.

I will help you solve these one by one.

Part a) Calculating the area of trapezoid $ABCD$:

The area of an isosceles trapezoid is given by the formula:

$A = \frac{1}{2} \times (AB + CD) \times h$

Where:

• $AB$ is the larger base.
• $CD$ is the smaller base.
• $h$ is the height, which can be calculated or derived from the problem.

To proceed further, I'll need information about either the height or the smaller base $CD$, as they are not provided directly in the problem.

Do you have the value for $CD$ or the height? If not, we can calculate them based on the perpendicular diagonal condition.