### (a)

**Step 1 — Assume the variable.**
Let the original duration of the tour be $x$ days.

**Step 2 — Write the daily expenses in both cases.**
$$\text{Original daily expense} = \frac{4200}{x}$$
$$\text{New daily expense (for } x+3 \text{ days)} = \frac{4200}{x+3}$$

**Step 3 — Form the equation from the given condition.**
$$\frac{4200}{x} - \frac{4200}{x+3} = 70$$

**Step 4 — Simplify.**
$$4200\left[\frac{(x+3) - x}{x(x+3)}\right] = 70 \ \Rightarrow\ \frac{4200 \times 3}{x(x+3)} = 70$$
$$12600 = 70x(x+3) \ \Rightarrow\ 180 = x^2 + 3x$$

**Step 5 — Form the quadratic equation.**
$$x^2 + 3x - 180 = 0$$

**Step 6 — Factorise.**
$$x^2 + 15x - 12x - 180 = 0$$
$$x(x+15) - 12(x+15) = 0 \ \Rightarrow\ (x+15)(x-12) = 0$$

**Step 7 — Solve and reject the invalid root.**
$$x = -15 \ (\text{rejected, duration cannot be negative}) \quad\text{or}\quad x = 12$$

$$\boxed{\text{Original duration of the tour} = 12 \text{ days}}$$

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### (b)

**Step 1 — Assume the variables.**
Let the altitude $= x$ cm, then the base $= (x + 10)$ cm.

**Step 2 — Use the area formula.**
$$\frac12 \times \text{base} \times \text{altitude} = 600$$
$$\frac12 \times (x+10) \times x = 600$$

**Step 3 — Form the quadratic equation.**
$$x^2 + 10x = 1200 \ \Rightarrow\ x^2 + 10x - 1200 = 0$$

**Step 4 — Factorise.**
$$x^2 + 40x - 30x - 1200 = 0$$
$$x(x+40) - 30(x+40) = 0 \ \Rightarrow\ (x+40)(x-30) = 0$$

**Step 5 — Solve.**
$$x = -40 \ (\text{rejected}) \quad\text{or}\quad x = 30$$

So altitude $= 30$ cm and base $= 40$ cm.

**Step 6 — Find the hypotenuse by Pythagoras theorem.**
$$h = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ cm}$$

$$\boxed{\text{Altitude} = 30 \text{ cm},\quad \text{Base} = 40 \text{ cm},\quad \text{Hypotenuse} = 50 \text{ cm}}$$