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Points B, C and E lie in the same horizontal plane - NSC Mathematics - Question 7 - 2021 - Paper 2

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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

Show that DC = \( \frac{BC}{4 \cos x} \)

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Answer

In triangle ABC:

  1. By the sine rule, we can write: CE=BCsin⁡30°sin⁡2xCE = \frac{BC \sin 30°}{\sin 2x}

  2. Since ( \sin 30° = \frac{1}{2} ), this becomes: CE=BC⋅12sin⁡2xCE = \frac{BC \cdot \frac{1}{2}}{\sin 2x} CE=BC2sin⁡2xCE = \frac{BC}{2 \sin 2x}

In triangle ACD:

  1. We know that ( DC = \tan \angle DCE ) So, ( tan DCE = \frac{CE}{BC} ) Therefore, substituting for CE: DC=BC⋅sin⁡30°2xDC = \frac{BC \cdot \sin 30°}{2x} Thus, substituting into the formula, we have: DC=BC⋅12tan⁡xDC = \frac{BC \cdot \frac{1}{2}}{\tan x}

Step 2

If x = 30°, show that the area of ABCD = 3AB²

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Answer

Substituting x = 30° into the expression for DC:

  1. We get: DC=BC4cos⁡30°DC = \frac{BC}{4 \cos 30°}

  2. Given that ( \cos 30° = \frac{\sqrt{3}}{2} ), it becomes: DC=BC4⋅32=BC23DC = \frac{BC}{4 \cdot \frac{\sqrt{3}}{2}} = \frac{BC}{2\sqrt{3}}

  3. Finding the area of rectangle ABCD: Area=AB⋅DC\text{Area} = AB \cdot DC

  4. With the dimensions substituting in the area formula, we find: =(AB)(BC)  ⟹  3AB2= (AB)(BC) \implies 3AB²

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