To solve this, we analyze the given geometric figure and conditions.

Step 1: Understand the geometry

The figure is a rectangle $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $E$. For a rectangle:

• The diagonals are equal in length.
• The diagonals bisect each other, meaning: $DE = EC \quad \text{and} \quad BE = EA.$

Step 2: Use the given conditions

We are provided:

1. $DE = 4x + 1$,
2. $EB = 12x - 31$,
3. $CD = 28$ (this is also the length of $AB$, as opposite sides of a rectangle are equal).

Step 3: Solve for $x$

From the bisection property of the diagonals: $DE = EC \quad \text{and} \quad EB = EA.$ So, we equate $DE = EC$ and $EB = EA$.

Calculate $EC$:

Since $DE = EC$, and the total length of diagonal $BD$ is divided equally: $DE = EC = \frac{BD}{2}.$

Similarly, $EB = EA = \frac{AC}{2}$.

Now, $BD = AC$ (because the diagonals are equal in a rectangle).

From $CD = 28$, the length of the diagonal can be found using the Pythagorean theorem: $BD = \sqrt{AB^2 + AD^2}.$

Let's calculate this further:

1. Substituting $AB = CD = 28$, $BD = AC$, and using the equations $DE = 4x + 1$, we solve for (Let us reframe and analyze the problem step-by-step.

Step 1: Analyze the Geometry

• The figure is a rectangle $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $E$.
• Diagonals of a rectangle bisect each other, so: $DE = EC \quad \text{and} \quad EB = EA.$

Step 2: Solve for $x$

We are given:

1. $DE = 4x + 1$,
2. $EB = 12x - 31$,
3. $CD = 28$ (equal to $AB$).

Since $DE = EC$, the total diagonal length $BD$ is: $BD = DE + EC = 2 \cdot DE = 2 \cdot (4x + 1) = 8x + 2.$

Similarly, $EB = EA$, and the full diagonal $AC$ is: $AC = EB + EA = 2 \cdot EB = 2 \cdot (12x - 31) = 24x - 62.$

In a rectangle, the two diagonals are equal, so: $BD = AC.$ Substitute the expressions for $BD$ and $AC$: $8x + 2 = 24x - 62.$

Step 3: Solve for $x$

Simplify the equation: $8x + 2 = 24x - 62.$ $2 + 62 = 24x - 8x.$ $64 = 16x.$ $x = 4.$

Step 4: Find $AD$

Using the Pythagorean theorem for the rectangle: $BD = \sqrt{AB^2 + AD^2}.$ We already know $BD = 8x + 2$. Substitute $x = 4$: $BD = 8(4) + 2 = 32 + 2 = 34.$

Also, $AB = CD = 28$. Substitute these values into the Pythagorean theorem: $34 = \sqrt{28^2 + AD^2}.$ Square both sides: $34^2 = 28^2 + AD^2.$ $1156 = 784 + AD^2.$ $AD^2 = 1156 - 784 = 372.$ $AD = \sqrt{372} \approx 19.3.$

Final Answer:

$x = 4, \quad AD \approx 19.3.$

Let me know if you want a detailed explanation of any step or have additional questions! Here are some related questions you can explore:

1. What are the properties of diagonals in a rectangle?
2. How do you calculate the diagonal of a rectangle using the Pythagorean theorem?
3. How do diagonals bisect each other in different quadrilaterals?
4. What are the differences between rectangles and other parallelograms in terms of diagonal properties?
5. How would this problem change if the figure were a square?

Tip: Always check if the diagonals or sides are given explicitly in geometric problems—this helps simplify your calculations.