Grade 12 Maths · Euclidean Geometry · Combine ratios through a shared part

Combine ratios through a shared part Grade 12

Link ratios from different parallel lines through a segment they share, rewriting them with the same number of parts, to find the ratio asked for (for example AD : DB = 2 : 3 and DE : EB = 1 : 2 give AD : DE : EB = 10 : 5 : 10).

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Grade 12 Ratio and Proportion | Part 5 of 5 Applying the Proportionality Theorem | NTENTE \ New Technological Education

Grade 12 Ratio and Proportion |Part 1 of 5 Applying the Proportionality Theorem| NTENTE \ New Technological Education

Grade 12 Proportionality Theorem: #6Kevinmathscience

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Question 10.2 · September 2026 · Paper 2 · Provincial · 3 marks

In the diagram, PQRS is a parallelogram with UT

∥

VW and

\frac{Q V}{V R} = \frac{2}{3}

.

Determine, with reasons, the value of: $Diagram for this question$

10.2

\frac{P U}{U V}

(3)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: combine ratios through a shared part

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3. Question 11.1.2 · November 2025 · Paper 2 · National (DBE) · 4 marks
Provide reasons for your statements.
11.1 In the diagram, $Δ A BC$ is drawn. BA is produced to E. F and D are points on BC such that AD $∥$ EF. AC and EF intersect at G. $\frac{CF}{FB} = \frac{2}{5}$ and $\frac{CG}{G A} = \frac{3}{2}$ .

Calculate, with reasons, the value of: $Diagram for question 11.1$
11.1.2 $\frac{B A}{E A}$ (4)
Show the memo (4 ticks)Memo
1. ✓
  FD = 4x/3 $F D = \frac{2}{3} CF$
  Let $CF = 2 x$ and $FB = 5 x$ :
  $F D = \frac{2}{3} (2 x) = \frac{4}{3} x$
  1 mark · Combine ratios through a shared part
2. ✓
  S $\frac{B A}{E A} = \frac{B D}{F D}$ [prop theorem; AD $∥$ EF]
  1 mark · Use the proportionality theorem
3. ✓
  Substitution $\frac{B A}{E A} = \frac{5 x - \frac{4}{3} x}{\frac{4}{3} x}$
  1 mark · Combine ratios through a shared part
4. ✓
  Answer $= \frac{11}{3} \times \frac{3}{4} = \frac{11}{4}$
  1 mark · Combine ratios through a shared part
4. Question 4.2.3 · March 2022 · Provincial · 3 marks
4.2 In the figure, $Δ A BC$ has $H G ∥ A C$ and $E D ∥ BC$ . $E D$ and $H G$ intersect at $K$ . $\frac{A D}{D C} = \frac{3}{2}$ and $BG = 2 GC$ . $A B = 15$ units.
4.2.3 Determine with reasons the value of $\frac{G K}{KH}$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Calculate HE $H E = A E - A H = 9 - 5 = 4$ units
  $B H = A B - A H = 15 - 5 = 10$ units
  $BE = A B - A E = 15 - 9 = 6$ units
  1 mark · Combine ratios through a shared part
2. ✓
  Setup ratio $\frac{G K}{KH} = \frac{BE}{E H}$ (Prop. Thm; $E K ∥ BG$ )
  $\frac{G K}{KH} = \frac{6}{4} = \frac{3}{2}$
  2 marks · Use the proportionality theorem
5. Question 5.2 · March 2021 · Provincial · 4 marks
In the figure $A F = 2 CG$ and $FE ∥ GB$ . $\frac{A E}{A B} = \frac{2}{5}$ .
5.2 Determine $\frac{C H}{H E}$ with reasons.(4)
Show the memo (4 ticks)Memo
1. ✓
  Express lengths in terms of variablesLet $A F = 2 b$ and $FG = 3 b$ (from 5.1). Then $A G = 5 b$ .
  Given $A F = 2 CG ⟹ 2 b = 2 CG ⟹ CG = b$ .
  1 mark · Combine ratios through a shared part
2. ✓
  Use proportionality theorem $\frac{C H}{H E} = \frac{CG}{GF}$ (prop theorem; $G H ∥ FE$ )
  $= \frac{b}{3 b} = \frac{1}{3}$
  3 marks · Use the proportionality theorem
6. Question 9.2 · March 2021 · Provincial · 6 marks
The diagram is the top view design of a new railway system. AF || BE and AC || GD. $\frac{A B}{BC} = \frac{4}{7}$ and $\frac{A G}{A F} = \frac{9}{17}$ .
9.2 If the straight line distance of the track from F to C is 374 km and it takes 50 hours to build one km of the track, determine the number of hours it will take to build the section from E to D.(6)
Show the memo (6 ticks)Memo
1. ✓
  $FE = 4 p$ and $EC = 7 p$ Let $FE = 4 p$ and $EC = 7 p$ (since $\frac{FE}{FC} = \frac{4}{11}$ )
  1 mark · Use a ratio, rate or proportion
2. ✓
  $p = 34$ $11 p = 374 ⟹ p = 34$
  $FE = 4 (34) = 136$ km
  1 mark · Solve a linear equation
3. ✓
  $F D = 8 m$ and $D C = 9 m$ Let $F D = 8 m$ and $D C = 9 m$ (since $\frac{C D}{D F} = \frac{9}{8}$ )
  1 mark · Use a ratio, rate or proportion
4. ✓
  $m = 22$ $17 m = 374 ⟹ m = 22$
  $C D = 9 (22) = 198$ km
  1 mark · Solve a linear equation
5. ✓
  $E D = 40$ km $E D = 374 - 4 p - 9 m = 374 - 136 - 198 = 40$ km
  1 mark · Combine ratios through a shared part
6. ✓
  Answer: 2 000 hoursTime $= 40 \times 50 = 2000$ hours
  1 mark · Use a ratio, rate or proportion

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