(a)
In triangle \( ADC \), drop perpendicular from \( D \) to \( M \) on \( CD \).
\( AM = \frac{110 – 70}{2} = 20 \, \text{m} \), \( \angle ADC = 60^\circ \), \( AD = 40 \, \text{m} \).
Compute:
\( \sin 60^\circ = \frac{h}{40} \)
\( h = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.6 \, \text{m} \).
Result: \( h = 20\sqrt{3} \approx 34.6 \, \text{m} \) [2].

(b)
Volume = area of trapezoid \( \times 200 \).
Area:
\( \frac{1}{2} \times (70 + 110) \times 20\sqrt{3} \approx 90 \times 34.6410 \approx 3117.69 \, \text{m}^2 \)
Volume:
\( 3117.69 \times 200 \approx 623,538 \approx 624,000 \, \text{m}^3 \).
Result: \( 624,000 \, \text{m}^3 \) (three significant figures) [3].

(c)
Common difference:
\( d = 43,200 – 45,000 = -1800 \, \text{m}^3 \).
Result: \( d = -1800 \, \text{m}^3 \) [1].

(d)
Sequence:
\( u_n = 45,000 + (n-1)(-1800) \)
For \( n = 13 \):
\( u_{13} = 45,000 + 12 \times (-1800) = 23,400 \, \text{m}^3 \).
Result: \( 23,400 \, \text{m}^3 \) [2].

(e)(i)
Set:
\( u_n = 45,000 + (n-1)(-1800) = 0 \)
Solve:
\( 1800(n-1) = 45,000 \)
\( n-1 = 25 \)
\( n = 26 \).
Result: \( n = 26 \) [2].

(e)(ii)
Pump stops after 25 hours (\( u_{26} = 0 \)).
Result: 25 hours [1].

(f)
Sum:
\( S_8 = \frac{8}{2} [2 \times 45,000 + (8-1)(-1800)] \)
\( = 4 \times 77,400 = 309,600 \, \text{m}^3 \).
Verify:
\( S_2 = 45,000 + 43,200 = 88,200 \, \text{m}^3 \).
Result: \( 309,600 \, \text{m}^3 \) [2].

(g)
Compute:
\( S_{25} = \frac{25}{2} [90,000 + 24(-1800)] \)
\( = 12.5 \times 46,800 = 585,000 \, \text{m}^3 \)
Since \( 585,000 < 624,000 \) and \( u_n \leq 0 \) for \( n > 25 \), tank never fills.
Result: Tank never fills, as maximum sum is 585,000 m³ [3].