Exemplary Solutions
Problem 1: Find the area of a triangle whose sides are $13\text{ cm}, 14\text{ cm}, 15\text{ cm}$. Hence, find the altitude corresponding to the side of length $14\text{ cm}$.
Solution:
1. Calculate semi-perimeter $s$:
$$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\text{ cm}$$
2. Calculate $(s - a), (s - b), (s - c)$:
$$s - a = 21 - 13 = 8\text{ cm}, \quad s - b = 21 - 14 = 7\text{ cm}, \quad s - c = 21 - 15 = 6\text{ cm}$$
3. Apply Heron's formula:
$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21 \times 8 \times 7 \times 6}$$
Factorise under the radical for easy root extraction:
$$\text{Area} = \sqrt{(7 \times 3) \times (2^3) \times 7 \times (2 \times 3)} = \sqrt{7^2 \times 3^2 \times 2^4} = 7 \times 3 \times 4 = \mathbf{84\text{ cm}^2}$$
4. Find altitude $h$ to the side of $14\text{ cm}$:
$$\text{Area} = \frac{1}{2} \times \text{base} \times h \implies 84 = \frac{1}{2} \times 14 \times h$$
$$84 = 7h \implies h = \frac{84}{7} = \mathbf{12\text{ cm}}$$
Problem 2: The cross-section of a canal is a trapezium. If the canal is $10\text{ m}$ wide at the top, $6\text{ m}$ wide at the bottom, and the area of cross-section is $640\text{ m}^2$, find its depth.
Solution:
Parallel sides are $a = 10\text{ m}$ and $b = 6\text{ m}$. Area $= 640\text{ m}^2$. Let depth be $h$.
$$\text{Area} = \frac{1}{2}(a + b) \times h \implies 640 = \frac{1}{2}(10 + 6) \times h$$
$$640 = \frac{1}{2}(16) \times h = 8h \implies h = \frac{640}{8} = \mathbf{80\text{ meters}}$$
