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In the diagram below, two of the sides have a length of $s$, where $0 < s < 6$ and $s \in \mathbb{R}$ - Junior Cycle Mathematics - Question 11 - 2019

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Question 11

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In the diagram below, two of the sides have a length of $s$, where $0 < s < 6$ and $s \in \mathbb{R}$. All angles are $90^{\circ}$ or $270^{\circ}$. (a) Find a fo... show full transcript

Worked Solution & Example Answer:In the diagram below, two of the sides have a length of $s$, where $0 < s < 6$ and $s \in \mathbb{R}$ - Junior Cycle Mathematics - Question 11 - 2019

Step 1

Find a formula (in algebra) for the area of this shape, in terms of $s$.

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Answer

To find the area of the shape, we can split it into rectangles. The overall dimensions of the shape are 10 units in width and (10s)(10 - s) in height due to the part where the length is ss.

The area can be calculated as follows:

  1. The height is contributed by the full height minus the height where the length ss is present:
    • Area = Width × Height
    • Area = 10×(6)s×s10 \times (6) - s \times s
    • Area = 60s260 - s^2

Thus, the formula for the area is: extArea=60s2 ext{Area} = 60 - s^2

Step 2

Show that (or explain why) the perimeter of this shape is always 32, no matter what the value of $s$.

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Answer

To find the perimeter, we need to sum the lengths of all sides. The perimeter can be calculated as follows:

The lengths are:

  • Left Side: 66 units
  • Bottom Side: 1010 units
  • Right Side: (6s)(6 - s) units
  • Top Side: ss units (upward)

Thus, the perimeter PP is given by: P=10+6+(6s)+sP = 10 + 6 + (6 - s) + s This simplifies to: P=10+6+6=32P = 10 + 6 + 6 = 32

As we can see, ss cancels out, meaning the perimeter remains constant at 32 units regardless of the value of ss.

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