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FIGUUR I toon 'n oprit wat na die ingang van 'n gebou lei - NSC Mathematics - Question 8 - 2022 - Paper 2

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FIGUUR I toon 'n oprit wat na die ingang van 'n gebou lei. B, C en D lê op dieselfde horisontale vlak. Die loodregte hoogte (AC) van die oprit is 0,5 m en die hoogte... show full transcript

Worked Solution & Example Answer:FIGUUR I toon 'n oprit wat na die ingang van 'n gebou lei - NSC Mathematics - Question 8 - 2022 - Paper 2

Step 1

8.1 Bereken die lengte van AB.

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Answer

To find the length of AB, we use the sine ratio:

0.5AB=sin⁡(15°)\frac{0.5}{AB} = \sin(15°)

Rearranging gives us:

AB=0.5sin⁡(15°)AB = \frac{0.5}{\sin(15°)}

Calculating this yields:

AB≈1.93 mAB \approx 1.93 \, \text{m}

Step 2

8.2 Bereken die lengte van BE.

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Answer

Using the cosine rule to find BE:

BE2=AB2+AE2−2(AB)(AE)cos⁡(120°)BE^2 = AB^2 + AE^2 - 2(AB)(AE) \cos(120°)

Substituting the values:

BE2=(1.93)2+(0.915)2−2(1.93)(0.915)cos⁡(120°)BE^2 = (1.93)^2 + (0.915)^2 - 2(1.93)(0.915)\cos(120°)

Calculating this gives:

BE≈2.52 mBE \approx 2.52 \, \text{m}

Step 3

8.3 Bereken die oppervlakte van ABFD.

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Answer

Given that BF = FD = \frac{5}{7}BE, we find:

BF=FD=57(2.52)≈1.80 mBF = FD = \frac{5}{7}(2.52) \approx 1.80 \, \text{m}

Area of triangle ABFD:

AreaABFD=12(BF)(FD)sin⁡(75°)\text{Area}_{ABFD} = \frac{1}{2}(BF)(FD) \sin(75°)

Substituting:

AreaABFD=12(1.80)(1.80)sin⁡(75°)≈1.56 m2\text{Area}_{ABFD} = \frac{1}{2}(1.80)(1.80) \sin(75°) \approx 1.56 \, \text{m}^2

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