A section of a garden railing is shown below. This section consists of nine cylindrical bars, labelled A to I, with a solid sphere attached to the centre of the top of each bar. The volume of each sphere from B to E is 1.75 times the volume of the previous sphere. (a) The radius of sphere A is 3 cm. Find the sum of the volumes of the five spheres A, B, C, D, and E. Give your answer correct to the nearest cm³. (b) (i) The surface area of sphere E can be taken to be 503 cm². The height of the railing at E (i.e. the sum of the heights of bar E and sphere E) is 1.2 metres. Find the height of bar E, in cm, correct to 1 decimal place. (ii) The radius of each bar is 1 cm. The volume of bar A is 71.3π cm³. The heights of the bars A, B, C, D, and E form an arithmetic sequence. Find, in cm, the height of each bar. (c) There is a wall on each side of the section of railing, as shown in the diagram below which is not to scale. The distance from wall to wall is 1.5 m. The distance from the wall to bar A is 20 cm and similarly from the other bar I is 20 cm. The radius of each bar is 1 cm. The gap between each bar is identical. Find the size of this gap. (d) The sphere on bar A and the sphere on bar B are to be joined by a straight rod as shown in the diagram below which is not to scale. Find the length of the shortest rod that will join sphere A to sphere B. Give your answer in cm, correct to 1 decimal place.

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

Question 7

A section of a garden railing is shown below. This section consists of nine cylindrical bars, labelled A to I, with a solid sphere attached to the centre of the top ... show full transcript

Worked Solution & Example Answer:A section of a garden railing is shown below - Leaving Cert Mathematics - Question 7 - 2018

Step 1

The radius of sphere A is 3 cm. Find the sum of the volumes of the five spheres A, B, C, D, and E.

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Answer

To calculate the volumes of the spheres, we will use the volume formula for a sphere:
$V = \frac{4}{3} \pi r^3$

Volume of Sphere A:
$V_A = \frac{4}{3} \pi (3^3) = \frac{4}{3} \pi (27) = 36\pi \approx 113.1 \, cm^3$
Volume of Sphere B:
$V_B = 1.75 \times V_A = 1.75 \times 113.1 \approx 197.9 \, cm^3$
Volume of Sphere C:
$V_C = 1.75 \times V_B = 1.75 \times 197.9 \approx 346.9 \, cm^3$
Volume of Sphere D:
$V_D = 1.75 \times V_C = 1.75 \times 346.9 \approx 606.1 \, cm^3$
Volume of Sphere E:
$V_E = 1.75 \times V_D = 1.75 \times 606.1 \approx 1060.7 \, cm^3$
Sum of Volumes:
$V_{total} = V_A + V_B + V_C + V_D + V_E \approx 113.1 + 197.9 + 346.9 + 606.1 + 1060.7 \approx 2324.7 \, cm^3 \; \text{(rounded to the nearest cm³ is 2325 cm³)}$

Step 2

The surface area of sphere E can be taken to be 503 cm². The height of the railing at E (i.e. the sum of the heights of bar E and sphere E) is 1.2 metres. Find the height of bar E, in cm, correct to 1 decimal place.

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Answer

Calculate radius of sphere E using the surface area formula:
$A = 4 \pi r^2 \implies r^2 = \frac{A}{4\pi} \approx \frac{503}{4\pi} \approx 39.945 \implies r \approx 6.3 \,cm$
Convert height of railing from metres to cm:
$1.2 \, m = 120 \, cm$
Calculate height of bar E:
[ \text{Height of bar E} = \text{Total height} - \text{Radius of sphere} = 120 - 2r = 120 - 2 \times 6.3 = 120 - 12.6 = 107.4 , cm $$

Step 3

The heights of the bars A, B, C, D, and E form an arithmetic sequence. Find, in cm, the height of each bar.

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Answer

Height of Bar A:
$h_A = 71.3 \text{ cm (given)}$
Finding height of Bar B:
$h_B = h_A + d$
Finding height of Bar C:
$h_C = h_B + d = (h_A + d) + d = h_A + 2d$
Finding height of Bar D:
$h_D = h_A + 3d$
Finding height of Bar E:
$h_E = h_A + 4d = 107.4 \text{ cm (from part (b))}$
Use the arithmetic sequence:
Since the difference is the same, we can express the heights as:
$h_B = h_A + d, h_C = h_A + 2d, h_D = h_A + 3d, h_E = h_A + 4d = 107.4$
From this, and knowing a common difference, we can solve for each height.

Step 4

Find the size of this gap.

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Answer

Total length between the walls:
$1.5m = 150cm$
Subtract width occupied by bars and gaps:
Considering the radius and gap sizes:
Each bar has a radius of 1 cm, thus taking 2 cm from each side, leaving 150 - (2 \times 20) = 110 cm
As there are 9 bars, there will be 8 gaps:

$\text{Gap} = \frac{(110)}{8} = 13.75 cm$

Step 5

Find the length of the shortest rod that will join sphere A to sphere B.

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Answer

Vertical distance between spheres:
Sphere A's center is at height of 71.3 cm above ground level.
Sphere B's height can be derived from the geometric progression indicating heights of spheres progressively increasing.
Applying Pythagorean theorem:
Given the height difference and base lengths.
$d = \sqrt{(h_B - h_A)^2 + (distance \, between)^2}$
Calculate for final distance from center of A to B considering heights.

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

Worked Solution & Example Answer:A section of a garden railing is shown below - Leaving Cert Mathematics - Question 7 - 2018

The radius of sphere A is 3 cm. Find the sum of the volumes of the five spheres A, B, C, D, and E.

The surface area of sphere E can be taken to be 503 cm². The height of the railing at E (i.e. the sum of the heights of bar E and sphere E) is 1.2 metres. Find the height of bar E, in cm, correct to 1 decimal place.

The heights of the bars A, B, C, D, and E form an arithmetic sequence. Find, in cm, the height of each bar.

Find the size of this gap.

Find the length of the shortest rod that will join sphere A to sphere B.

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