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Points B, C and E lie in the same horizontal plane - NSC Mathematics - Question 7 - 2021 - Paper 2
Question 7
Points B, C and E lie in the same horizontal plane. ABCD is a rectangular piece of board. CDE is a triangular piece of board having a right angle at C. Each piece of... show full transcript
Worked Solution & Example Answer:Points B, C and E lie in the same horizontal plane - NSC Mathematics - Question 7 - 2021 - Paper 2
Step 1
7.1 Show that DC = \( \frac{BC}{4 \cos^2 x} \)
Answer
To approach this problem, we use triangle ABC. We apply the sine rule:
In triangle ACD, we use the tangent function:
[DC = \tan \angle DEC]
Here, substituting for CE, we have:
[DC = \frac{BC \cdot \sin 30°}{\sin 2x} = \frac{BC \cdot \frac{1}{2}}{\sin 2x} = \frac{BC}{2 \sin 2x}]
Knowing that (\sin 2x = 2 \sin x \cos x), we can rewrite:
[DC = \frac{BC}{2(2 \sin x \cos x)} = \frac{BC}{4 \cos^2 x}].
Step 2
7.2 If x = 30°, show that the area of ABCD = 3AB².
Answer
Substituting (x = 30°) into our formula for DC, we find:
[DC = \frac{BC}{4 \cos^2(30°)} = \frac{BC}{4 \cdot \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{BC}{4 \cdot \frac{3}{4}} = \frac{BC}{3}]
Since AB = DC, it follows that (AB = \frac{BC}{3}). Thus, the area of rectangle ABCD can be calculated as:
[\text{Area} = AB \cdot BC = \left(\frac{BC}{3}\right) \cdot BC = \frac{BC^2}{3}]
Now, if we relate this back to the dimensions:
[BC = 3AB]
Therefore, the area is:
[\text{Area} = 3AB^2].