Quadrilateral Area 7Fdb6F

1. **Problem Statement:** Calculate the area of the given quadrilateral-like shape with sides 4 cm, 4 cm, and an angle of 75° between them, where a segment from the top point to the bottom center is perpendicular to the diagonal. 2. **Understanding the shape:** The shape can be divided into two triangles by the diagonal from the left bottom point to the right point. 3. **Formula for area of a triangle:** The area of a triangle with two sides $a$ and $b$ and included angle $\theta$ is given by: $$\text{Area} = \frac{1}{2}ab\sin(\theta)$$ 4. **Applying the formula:** Given $a = 4$ cm, $b = 4$ cm, and $\theta = 75^\circ$, $$\text{Area} = \frac{1}{2} \times 4 \times 4 \times \sin(75^\circ)$$ 5. **Calculate $\sin(75^\circ)$:** $$\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$ 6. **Substitute back:** $$\text{Area} = \frac{1}{2} \times 4 \times 4 \times \frac{\sqrt{6} + \sqrt{2}}{4} = \cancel{\frac{1}{2}} \times \cancel{4} \times \cancel{4} \times \frac{\sqrt{6} + \sqrt{2}}{\cancel{4}} = 2 \times (\sqrt{6} + \sqrt{2})$$ 7. **Final area:** $$\text{Area} = 2(\sqrt{6} + \sqrt{2}) \approx 2(2.449 + 1.414) = 2 \times 3.863 = 7.726 \text{ cm}^2$$ **Answer:** The area of the quadrilateral-like shape is approximately $7.726$ cm$^2$.