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

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Tshepo 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:Tshepo 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 use the formula:

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

Given that the total length (A + B + C) is:

55extcm+99extcm+55extcm=209extcm55 ext{ cm} + 99 ext{ cm} + 55 ext{ cm} = 209 ext{ cm}

And the width is 149 cm. Thus, we calculate:

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

Therefore, the perimeter of the rectangular window frame is 716 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 from the inner diameter:

ext{Radius} = rac{ ext{Diameter}}{2} = rac{605 ext{ mm}}{2} = 302.5 ext{ mm} = 30.25 ext{ cm}

Now, using the area formula for a circle:

extArea=3,142imes(extRadius)2 ext{Area} = 3,142 imes ( ext{Radius})^2

Calculating the area gives:

extArea=3.142imes(30.25extcm)2 =3,142imes915.0625extcm2 extAreaext(rounded)extisapproximately2,872.07extcm2 ext{Area} = 3.142 imes (30.25 ext{ cm})^2 \ = 3,142 imes 915.0625 ext{ cm}^2 \ ext{Area} ext{ (rounded)} ext{ is approximately } 2,872.07 ext{ cm}^2

Thus, the inner area of the circular window frame is approximately 2,872.07 cm².

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, out of which four can open. Hence, there are no windowpanes that do not open.

The probability that it is NOT one of the windowpanes that can open is calculated as follows:

P( ext{NOT opening}) = rac{ ext{Number of non-opening panes}}{ ext{Total number of panes}} = rac{0}{4} = 0

Therefore, the probability that it is NOT one of the windowpanes that can open is 0.0.

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