A section of a garden railing is shown below - Leaving Cert Mathematics - Question 7 - 2018

Given the surface area of sphere E:
$4 \pi r_E^2 = 503 \Rightarrow r_E^2 = \frac{503}{4 \pi} \Rightarrow r_E \approx 6.3 \, cm$.
The height at E is 120 cm, where:\nHeight of bar E + radius of sphere E = 120 cm.
Thus, $\text{Height of bar E} = 120 - 2 \times 6.3 \approx 120 - 12.6 \approx 107.4 \, cm.$
Therefore, height of bar E is approximately 107.4 cm.

Let the height of bar A be h. The sequence of heights for the bars is given as:

• Height of bar A: $h = 71.3$.
• Height of bar B: $h + d$,
• Height of bar C: $h + 2d$, etc., where d is the common difference. Using the arithmetic sequence where:
  $h_B = h_A + d$,
  $h_C = h_A + 2d$,
  $h_D = h_A + 3d$,
  $h_E = h_A + 4d$.
  With the known heights: $71.3, \, 80.3, \, 89.3, \, 98.3, \, 107.3: \\ d = 9 \Rightarrow h = 71.3, \, h_B = 80.3, \, h_C = 89.3, \, h_D = 98.3, \, h_E = 107.3.$

The total width including bars and gaps is calculated as follows:

• Total distance from wall to wall: 150 cm.
• Distance taken by bars A to I (9 bars): 9 * 1 cm = 9 cm.
• Total distance of gaps: $150 - 9 \Rightarrow 141 cm.$
  The gaps are equal, so we divide by the number of gaps (8).
  Therefore, size of each gap is:
  $\frac{141}{8} = 17.625 cm \approx 17.6 cm.$

To find the length of the rod:

• The base coordinates are (1, h_A + 3) and (3, h_B + 3), with heights 104.3 cm for A and 110.3 cm for B.
• Thus, the height difference and horizontal difference are:
  Length of the rod: $|WY| = \sqrt{(3 - 1)^2 + (110.3 - 104.3)^2} \approx \sqrt{4 + 36} = \sqrt{40} \approx 6.3 cm.$ Therefore, the length of the rod is approximately 6.3 cm.