Question 1210155
Absolutely! Let's break down this problem step-by-step.

**1. Finding the area of Rhombus ADEF**

* **Properties of a Rhombus:**
    * All sides are equal.
    * Diagonals bisect each other at right angles.
* **Given:**
    * EF = 17 (side of the rhombus)
    * FO = 5 (half of diagonal FD)
* **Finding EO:**
    * Since triangle EOF is a right-angled triangle, we can use the Pythagorean theorem: $EF^2 = EO^2 + FO^2$
    * $17^2 = EO^2 + 5^2$
    * $289 = EO^2 + 25$
    * $EO^2 = 289 - 25 = 264$
    * $EO = \sqrt{264} = 2\sqrt{66}$
* **Finding Diagonals:**
    * FD = 2 * FO = 2 * 5 = 10
    * AE = 2 * EO = 2 * $2\sqrt{66}$ = $4\sqrt{66}$
* **Area of Rhombus ADEF:**
    * Area = (1/2) * (diagonal 1) * (diagonal 2)
    * Area = (1/2) * 10 * $4\sqrt{66}$
    * Area = $20\sqrt{66}$

**2. Finding the area of Parallelogram ABCE**

* **Properties of a Parallelogram:**
    * Opposite sides are equal and parallel.
* **Given:**
    * CD = 11
    * EF = 17
* **Finding AE:**
    * since ADEF is a rhombus, AE=EF. Therefore AE = 17.
* **Finding the Height of the Parallelogram:**
    * The height of the parallelogram is the same as the length EO.
    * Height = $2\sqrt{66}$
* **Finding the Area of the Parallelogram:**
    * Area = base * height
    * Area = AE * EO
    * Area = 17 * $2\sqrt{66}$
    * Area = $34\sqrt{66}$

**3. Finding the Area of the Composite Figure**

* **Total Area:**
    * Area of composite figure = Area of Rhombus ADEF + Area of Parallelogram ABCE
    * Total Area = $20\sqrt{66} + 34\sqrt{66}$
    * Total Area = $54\sqrt{66}$

**Therefore, the area of the composite figure is $54\sqrt{66}$.**