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Paradigm Specialising in O Level Mathematics 5 In the diagram, ABC is a triangle where M is the midpoint of AC. The point P lies on the line BM such that BP=(1)/(3)BM. The point Q lies on the line BC such that BQ:QC=1:4. Given that vec(AB)=b and vec(AC)=c. (a) Express, as simply as possible, in terms of b and or c, (i) vec(BC), (ii) vec(BM), (iii) vec(AP), (iv) vec(AQ). (b) Write down two facts about A,P and Q. (c) Find the numeral value of (" Area of "/_\ABP)/(" Area of quadrilateral "CMPQ),

Paradigm\newlineSpecialising in O Level Mathematics\newline55 In the diagram, ABC A B C is a triangle where M M is the midpoint of AC A C .\newlineThe point P P lies on the line BM B M such that BP=13BM B P=\frac{1}{3} B M .\newlineThe point Q Q lies on the line BC B C such that BQ:QC=1:4 B Q: Q C=1: 4 .\newlineGiven that ABundefined=b \overrightarrow{A B}=b and M M 00.\newline(a) Express, as simply as possible, in terms of M M 11 and or M M 22,\newline(i) M M 33,\newline(ii) M M 44,\newline(iii) M M 55,\newline(iv) M M 66.\newline(b) Write down two facts about M M 77 and Q Q .\newline(c) Find the numeral value of M M 99,

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Q. Paradigm\newlineSpecialising in O Level Mathematics\newline55 In the diagram, ABC A B C is a triangle where M M is the midpoint of AC A C .\newlineThe point P P lies on the line BM B M such that BP=13BM B P=\frac{1}{3} B M .\newlineThe point Q Q lies on the line BC B C such that BQ:QC=1:4 B Q: Q C=1: 4 .\newlineGiven that ABundefined=b \overrightarrow{A B}=b and M M 00.\newline(a) Express, as simply as possible, in terms of M M 11 and or M M 22,\newline(i) M M 33,\newline(ii) M M 44,\newline(iii) M M 55,\newline(iv) M M 66.\newline(b) Write down two facts about M M 77 and Q Q .\newline(c) Find the numeral value of M M 99,
  1. Calculate Vector BC: BC⃗=AC⃗−AB⃗=c−b\vec{BC} = \vec{AC} - \vec{AB} = c - b.
  2. Calculate Vector BM: BM⃗=12BC⃗=12(c−b).\vec{BM} = \frac{1}{2} \vec{BC} = \frac{1}{2} (c - b).
  3. Calculate Vector BP: BP⃗=13BM⃗=13(12(c−b))=16(c−b).\vec{BP} = \frac{1}{3} \vec{BM} = \frac{1}{3} \left(\frac{1}{2} (c - b)\right) = \frac{1}{6} (c - b).
  4. Calculate Vector AP: AP⃗=AB⃗+BP⃗=b+16(c−b)=16c+(56)b.\vec{AP} = \vec{AB} + \vec{BP} = \mathbf{b} + \frac{1}{6} (\mathbf{c} - \mathbf{b}) = \frac{1}{6} \mathbf{c} + \left(\frac{5}{6}\right) \mathbf{b}.
  5. Calculate Vector BQ: BQ⃗=15BC⃗=15(c−b).\vec{BQ} = \frac{1}{5} \vec{BC} = \frac{1}{5} (c - b).
  6. Calculate Vector AQ: AQ⃗=AB⃗+BQ⃗=b+15(c−b)=15c+(45)b.\vec{AQ} = \vec{AB} + \vec{BQ} = b + \frac{1}{5} (c - b) = \frac{1}{5} c + \left(\frac{4}{5}\right) b.
  7. Facts about AA, PP, QQ: Two facts about AA, PP, and QQ are: PP lies on the line segment BMBM and QQ lies on the line segment BCBC.
  8. Area of Triangle ABP: Area of triangle ABP is 12×\frac{1}{2} \times base ×\times height. Since BPBP is 13\frac{1}{3} of BMBM and MM is the midpoint of ACAC, the base of triangle ABP is 13\frac{1}{3} of ACAC. The height is the same for both triangles ABPABP and ×\times00.
  9. Area of Triangle ABC: Area of triangle ABP=12×(13AC)×height=16×base(AC)×height.ABP = \frac{1}{2} \times \left(\frac{1}{3} AC\right) \times \text{height} = \frac{1}{6} \times \text{base}(AC) \times \text{height}.
  10. Area of Quadrilateral CMPQ: Area of triangle ABC=12×base(AC)×height.ABC = \frac{1}{2} \times \text{base}(AC) \times \text{height}.
  11. Ratio of Areas: Area of quadrilateral CMPQCMPQ = Area of triangle ABCABC - Area of triangle ABPABP.
  12. Ratio of Areas: Area of quadrilateral CMPQ = Area of triangle ABC - Area of triangle ABP.AreaextofquadrilateralCMPQ=12×base(AC)×height−16×base(AC)×height=12×base(AC)×height−16×base(AC)×height=(12−16)×base(AC)×height.Area ext{ of quadrilateral }CMPQ = \frac{1}{2} \times base(AC) \times height - \frac{1}{6} \times base(AC) \times height = \frac{1}{2} \times base(AC) \times height - \frac{1}{6} \times base(AC) \times height = (\frac{1}{2} - \frac{1}{6}) \times base(AC) \times height.
  13. Ratio of Areas: Area of quadrilateral CMPQ = Area of triangle ABC - Area of triangle ABP. Area of quadrilateral CMPQ = $\frac{\(1\)}{\(2\)} \times \text{base}(AC) \times \text{height} - \frac{\(1\)}{\(6\)} \times \text{base}(AC) \times \text{height} = \frac{\(1\)}{\(2\)} \times \text{base}(AC) \times \text{height} - \frac{\(1\)}{\(6\)} \times \text{base}(AC) \times \text{height} = \left(\frac{\(1\)}{\(2\)} - \frac{\(1\)}{\(6\)}\right) \times \text{base}(AC) \times \text{height}. Area of quadrilateral CMPQ = \left(\frac{\(3\)}{\(6\)} - \frac{\(1\)}{\(6\)}\right) \times \text{base}(AC) \times \text{height} = \frac{\(2\)}{\(6\)} \times \text{base}(AC) \times \text{height} = \frac{\(1\)}{\(3\)} \times \text{base}(AC) \times \text{height}.
  14. Ratio of Areas: Area of quadrilateral CMPQ = Area of triangle ABC - Area of triangle ABP. Area of quadrilateral CMPQ = \(\frac{1}{2} \times \text{base}(AC) \times \text{height} - \frac{1}{6} \times \text{base}(AC) \times \text{height} = \frac{1}{2} \times \text{base}(AC) \times \text{height} - \frac{1}{6} \times \text{base}(AC) \times \text{height} = \left(\frac{1}{2} - \frac{1}{6}\right) \times \text{base}(AC) \times \text{height}\). Area of quadrilateral CMPQ = \(\left(\frac{3}{6} - \frac{1}{6}\right) \times \text{base}(AC) \times \text{height} = \frac{2}{6} \times \text{base}(AC) \times \text{height} = \frac{1}{3} \times \text{base}(AC) \times \text{height}\). Ratio of areas \(\left(\text{Area of triangle ABP}\right)/\left(\text{Area of quadrilateral CMPQ}\right) = \left(\frac{1}{6} \times \text{base}(AC) \times \text{height}\right) / \left(\frac{1}{3} \times \text{base}(AC) \times \text{height}\right)\).
  15. Ratio of Areas: Area of quadrilateral CMPQ = Area of triangle ABC - Area of triangle ABP.\(Area ext{ of quadrilateral }CMPQ = \frac{1}{2} \times \text{base}(AC) \times \text{height} - \frac{1}{6} \times \text{base}(AC) \times \text{height} = \frac{1}{2} \times \text{base}(AC) \times \text{height} - \frac{1}{6} \times \text{base}(AC) \times \text{height} = \left(\frac{1}{2} - \frac{1}{6}\right) \times \text{base}(AC) \times \text{height}.\)\(Area ext{ of quadrilateral }CMPQ = \left(\frac{3}{6} - \frac{1}{6}\right) \times \text{base}(AC) \times \text{height} = \frac{2}{6} \times \text{base}(AC) \times \text{height} = \frac{1}{3} \times \text{base}(AC) \times \text{height}.\)\(\text{Ratio of areas }\left(\text{Area of triangle ABP}\right)/\left(\text{Area of quadrilateral CMPQ}\right) = \left(\frac{1}{6} \times \text{base}(AC) \times \text{height}\right) / \left(\frac{1}{3} \times \text{base}(AC) \times \text{height}\right).\)\(\text{Ratio of areas} = \left(\frac{1}{6}\right) / \left(\frac{1}{3}\right) = \frac{1}{6} \times \frac{3}{1} = \frac{1}{2}.\)

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