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Punt B, C en E lê in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2021 - Paper 2
Question 7
Punt B, C en E lê in dieselfde horizontale vlak. ABCD is 'n reghoekige stuk plank. CDE is 'n driehoekige stuk plank met 'n regte hoek by C. Elk van die stukke plank ... show full transcript
Worked Solution & Example Answer:Punt B, C en E lê in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2021 - Paper 2
Step 1
7.1 Toon dat DC = \( \frac{BC}{4 \cos x} \)
Answer
In triangle ABC, we can use the sine rule:
[ \frac{CE}{BC} = \frac{\sin \angle BEC}{\sin \angle CEB} ]
Given that ( \angle BEC = 30° ) and ( \angle CEB = 2x ), we have:
[ CE = \frac{BC \cdot \sin(30°)}{\sin(2x)} ]
Using the double angle formula for sine, ( \sin(2x) = 2 \sin x \cdot \cos x ), we can express CE as:
[ CE = \frac{BC \cdot \frac{1}{2}}{2 \sin x \cos x} = \frac{BC}{4 \sin x \cos x} ]
Next, in triangle ACD:
[DC = \tan \angle DCE = \frac{CE}{BC} = \frac{BC \cdot \sin(30°)}{\sin(2x)} \Rightarrow DC = \frac{CE \cdot \cos x}{\sin x}\Rightarrow DC = \frac{4 \sin x \cos x}{4}\Rightarrow DC = \frac{BC}{4 \cos x} ] Therefore, we have shown that ( DC = \frac{BC}{4 \cos x} ).
Step 2
7.2 As x = 30°, toon dat die oppervlakte van ABCD = 3AB².
Answer
Substituting ( x = 30° ) into our previous equation results in:
[ DC = \frac{BC}{4 \cos(30°)} = \frac{BC}{4 \cdot \frac{\sqrt{3}}{2}} = \frac{BC}{2\sqrt{3}} ]
Given that AB is equal to DC and also to BC, we can equate:
[ AB = DC = BC ]
This gives us the area of rectangle ABCD as:
[ Area = AB \cdot DC = (AB)(BC) = (AB)(3AB) = 3AB² ]
Thus, we have demonstrated that the area of ABCD equals to ( 3AB² ).