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Tshego is renovating her home - NSC Mathematical Literacy - Question 3 - 2022 - Paper 2

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Tshego is renovating her home. She is removing the wooden-framed windows and is replacing them with aluminium-framed windows. The dimensions and the shape of two of... show full transcript

Worked Solution & Example Answer:Tshego is renovating her home - NSC Mathematical Literacy - Question 3 - 2022 - Paper 2

Step 1

Determine the perimeter of the rectangular window frame.

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Answer

To find the perimeter of the rectangular window frame, we will use the formula for the perimeter of a rectangle:

extPerimeter=2(extlength+extwidth) ext{Perimeter} = 2( ext{length} + ext{width})

Substituting the given dimensions:

  • Total length = A + B + C = 55 ext{ cm} + 99 ext{ cm} + 55 ext{ cm} = 209 ext{ cm}
  • Width = 149 ext{ cm}

Now calculate the perimeter:

extPerimeter=2(209extcm+149extcm)=2imes358extcm=716extcm ext{Perimeter} = 2(209 ext{ cm} + 149 ext{ cm}) = 2 imes 358 ext{ cm} = 716 ext{ cm}

Step 2

Calculate the inner area in cm² of the circular window frame.

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Answer

To calculate the area of the circular window frame, we first need to find the radius, which is half of the diameter:

  • Diameter = 605 ext{ mm} = 60.5 ext{ cm}
  • Radius = \frac{60.5 ext{ cm}}{2} = 30.25 ext{ cm}

Now we can use the area formula of a circle:

extArea=3.142×(radius)2 ext{Area} = 3.142 \times (radius)^2

Now substituting the radius:

extArea=3.142×(30.25)23.142×915.06252875.14extcm2 ext{Area} = 3.142 \times (30.25)^2 \approx 3.142 \times 915.0625 \approx 2875.14 ext{ cm}^2

Step 3

One of the windowpanes of the rectangular window frame broke. Write the probability, as a decimal, that it is NOT one of the windowpanes that can open.

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Answer

In the rectangular window frame, there are four windowpanes. Since only two of them can open (A and B), the probability that a randomly selected windowpane that broke is NOT one of the windowpanes that can open is calculated as follows:

  • Number of windowpanes that do NOT open = 2
  • Total number of windowpanes = 4

Thus, the probability is given by:

extProbability=extNumberofwindowpanesthatdoNOTopenextTotalnumberofwindowpanes=24=0.5 ext{Probability} = \frac{ ext{Number of windowpanes that do NOT open}}{ ext{Total number of windowpanes}} = \frac{2}{4} = 0.5

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