1. **Problem statement:** We have two intersecting lines AB and CD at point E. Given: $|AE|=25$, $|CE|=24$, $|EB|=30$, and $\angle ACE=90^\circ$. We need to find: (i) $|\angle AEC|$ to 2 decimal places. (ii) Area of triangle ACE. (iii) Given area of triangle EBD equals area of ACE, find $|ED|$ to nearest whole number. 2. **Step (i): Find $|\angle AEC|$** Since $\angle ACE=90^\circ$, triangle ACE is right angled at C. Use the cosine rule in triangle AEC to find $\angle AEC$: $$AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos(\angle AEC)$$ But we don't know $AC$, so first find $AC$ using Pythagoras: $$AC = \sqrt{AE^2 - CE^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7$$ Now apply cosine rule: $$25^2 = 7^2 + 24^2 - 2 \cdot 7 \cdot 24 \cdot \cos(\angle AEC)$$ $$625 = 49 + 576 - 336 \cos(\angle AEC)$$ $$625 = 625 - 336 \cos(\angle AEC)$$ Subtract 625 both sides: $$0 = -336 \cos(\angle AEC)$$ Divide both sides by $\cancel{336}$: $$0 = -\cancel{336} \cos(\angle AEC) / \cancel{336}$$ $$0 = -\cos(\angle AEC)$$ So, $$\cos(\angle AEC) = 0$$ Therefore, $$\angle AEC = 90^\circ$$ 3. **Step (ii): Area of triangle ACE** Since $\angle ACE=90^\circ$, area is: $$\text{Area} = \frac{1}{2} \times AE \times CE = \frac{1}{2} \times 25 \times 24 = 300$$ 4. **Step (iii): Find $|ED|$ given area of triangle EBD equals area of ACE** Area of triangle EBD = 300. Base $EB = 30$ units. Area formula: $$\text{Area} = \frac{1}{2} \times EB \times ED$$ Substitute values: $$300 = \frac{1}{2} \times 30 \times ED$$ $$300 = 15 \times ED$$ Divide both sides by $\cancel{15}$: $$\frac{300}{\cancel{15}} = \cancel{15} ED / \cancel{15}$$ $$20 = ED$$ So, $$|ED| = 20$$ units (nearest whole number). --- **Final answers:** (i) $|\angle AEC| = 90.00^\circ$ (ii) Area of triangle ACE = 300 units$^2$ (iii) $|ED| = 20$ units