1. **Problem statement:** We have a right triangle ABC with vertical side AB = 2 m, horizontal side BC = 3 m, and hypotenuse AC. We need to find: (a) The length of AC in surd form using the Pythagorean theorem. (b) The angle of elevation $\theta$ at vertex C between AC and BC, in degrees to two decimal places. (c) Using $|BD| = 2|DC|$, find the area of quadrilateral BEFD in the form $\frac{a}{b}$ m². 2. **Formula and rules:** - Pythagorean theorem for right triangle: $$AC^2 = AB^2 + BC^2$$ - To find angle $\theta$, use trigonometric ratios, here cosine since $\theta$ is between AC and BC: $$\cos(\theta) = \frac{BC}{AC}$$ - Area of a polygon can be found by subtracting or adding areas of triangles and rectangles. 3. **Step (a): Find $|AC|$ using Pythagoras** $$AC^2 = AB^2 + BC^2 = 2^2 + 3^2 = 4 + 9 = 13$$ $$AC = \sqrt{13}$$ So, $|AC| = \sqrt{13}$ meters in surd form. 4. **Step (b): Find angle $\theta$** Using cosine: $$\cos(\theta) = \frac{BC}{AC} = \frac{3}{\sqrt{13}}$$ Calculate $\theta$: $$\theta = \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$$ Numerical value: $$\theta \approx \cos^{-1}(0.83205) \approx 33.69^\circ$$ 5. **Step (c): Find area of shape BEFD** Given $|BD| = 2|DC|$, and BC = 3 m, so: Let $|DC| = x$, then $|BD| = 2x$, and $BD + DC = BC = 3$, so: $$2x + x = 3 \Rightarrow 3x = 3 \Rightarrow x = 1$$ Thus, $|DC| = 1$ m and $|BD| = 2$ m. Points B, D, C lie on horizontal line BC. E and F lie vertically below B and D respectively, at height 2 m (since AB = 2 m). Area of rectangle BEFD is: $$\text{length} = |BD| = 2$$ $$\text{height} = 2$$ $$\text{Area}_{BEFD} = 2 \times 2 = 4$$ But the problem states BEFD is under the stairs, so we must confirm shape and area carefully. Since E and F are vertically below B and D at height 2 m, BEFD is a rectangle with base BD = 2 m and height 2 m. Therefore, area of BEFD: $$= 2 \times 2 = 4 \text{ m}^2$$ 6. **Final answers:** - (a) $|AC| = \sqrt{13}$ m - (b) $\theta \approx 33.69^\circ$ - (c) Area of BEFD = $4$ m², which can be written as $\frac{4}{1}$ m².