Grade 10 Maths · Euclidean Geometry

Use similar triangles' sides and angles

Use triangles of the same shape to find a missing side from matching sides in the same ratio, for example AB/DE = BC/EF.

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Question 10.1 · May/June 2024 · Paper 1 · National (DBE) · 4 marks

EHGF is a rectangle. HE is produced

x^{2}

cm to N and EH is produced

x^{2}

cm to P. NF produced intersects PG produced at M to form an isosceles triangle MNP with NM = MP. D lies on NP where MD

⊥

NP. NP = 4 cm and MD = 3 cm. Diagram for this question

10.1 Show that the area of

EFG H

is given by

A (x) = 6 x^{2} - 3 x^{4}

.(4)

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

More past paper questions: use similar triangles' sides and angles

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3. Question 10.2.2 · May/June 2025 · Paper 2 · National (DBE) · 2 marks
10.2 In the diagram, O is the centre of the circle. Points D and B lie on the circle. Points A and C lie outside the circle such that side AC of $Δ A D C$ passes through B. F is a point on BC such that $FO ∥ B D$ . $D R = RB$ and RO is drawn.
10.2.2 If it is given that $R \hat{D} O = F \hat{C} O$ , show, with reasons, that $FO \cdot C D = CO \cdot BC$ .(2)
Show the memo (1 tick)Memo
1. ✓
  Use the similarity ratio and given angle equalityFrom 10.2.1: $\frac{FO}{B D} = \frac{CO}{C D}$ [ $∣∣∣$ $Δ$ s]
  $∴ FO \cdot C D = CO \cdot B D$
  
  But $R \hat{D} O = F \hat{C} O$ [given]
  $∴ B D = BC$ [sides opp equal $∠$ s]
  
  Substitute BC for BD:
  $∴ FO \cdot C D = CO \cdot BC$
  2 marks · Use similar triangles' sides and angles
4. Question 5.5.2 · March 2022 · Provincial · 3 marks
In the figure: $PQ$ is the diameter of the circle. $SP$ is the tangent to the circle at $P$ . $QT$ intersects the circle at $W$ and with $T$ on the line $SP$ . $QS$ intersects the circle at $R$ . $\hat{P}_{1} = x$ .
5.5 If $Q W = 5$ cm, $T W = 1$ cm, $QR = 4$ cm and $W R = 2$ cm, calculate the lengths of:
5.5.2 SR(3)
Show the memo (2 ticks)Memo
1. ✓
  Use similarity ratio to find QS $\frac{Q W}{QS} = \frac{4}{6} = \frac{2}{3}$
  $\frac{5}{QS} = \frac{2}{3}$
  $QS = 7, 5$ cm
  2 marks · Use similar triangles' sides and angles
2. ✓
  Calculate SR $SR = QS - QR = 7, 5 - 4 = 3, 5$ cm
  1 mark · Use similar triangles' sides and angles
5. Question 8.2.2 · March 2022 · Provincial · 4 marks
8.2 In the diagram, $A B$ is the diameter of circle with centre $P$ . $EP ⊥ A B$ .
8.2.2 Hence, prove that $BE = \frac{P E ^{2} \cdot B A \cdot B D}{BP}$ .(4)
Show the memo (3 ticks)Memo
1. ✓
  Use similarity ratios $\frac{BP}{B D} = \frac{PE}{D A} = \frac{BE}{B A}$
  1 mark · Use similar triangles' sides and angles
2. ✓
  Express DAFrom $\frac{BP}{B D} = \frac{PE}{D A}$ , we get $D A = \frac{PE \cdot B D}{BP}$ ... (1)
  1 mark · Use similar triangles' sides and angles
3. ✓
  Express DA from other ratioFrom $\frac{PE}{D A} = \frac{BE}{B A}$ , we get $D A = \frac{PE \cdot B A}{BE}$ ... (2)
  Wait, the prompt requires proving $BE = \dots$ . Let's check the target expression units. LHS is length. RHS is $L^{2} \cdot L \cdot L / L = L^{3}$ . Dimensional mismatch in the user prompt question text?
  Let's check the PDF source text.
  Source [920]: $BE = \frac{P E ^{2} \cdot B A \cdot B D}{BP}$ ?
  Wait, let's re-read the memo logic.
  Memo Step 1: $D A = \frac{BP}{B D \cdot PE}$ (This seems wrong dimensionally).
  Let's assume the question text is exactly as per image/PDF.
  Actually, checking similarity: $BP / B D = PE / D A$ . Thus $D A = PE \cdot B D / BP$ .
  Also $PE / D A = BE / B A ⟹ D A = PE \cdot B A / BE$ .
  Equate: $PE \cdot B D / BP = PE \cdot B A / BE$ .
  $B D / BP = B A / BE ⟹ BE = B A \cdot BP / B D$ .
  This doesn't match the required proof.
  Let's look at the Memo [1224].
  Memo says: $D A = \frac{BP}{B D . PE}$ (Equation 1 - this is algebra error in memo or transcription? $BP / B D = PE / D A ⟹ BP \cdot D A = B D \cdot PE ⟹ D A = B D \cdot PE / BP$ ).
  Memo Equation 2: $D A = \frac{PE . B A}{BE}$ .
  Memo equating: $\frac{BP}{B D . PE} = \frac{PE . B A}{BE}$ .
  Then $BP \cdot BE = P E^{2} \cdot B A \cdot B D$ ?
  If the memo has an error, I must still output the solution steps as per the memo's logic to satisfy the 'extract' requirement, or fix it if it's obvious.
  The target proof in the prompt text is likely $BE = \frac{P E ^{2} \cdot B A \cdot B D}{BP}$ .
  Let's work backwards.
  If $BE = \frac{P E ^{2} \cdot B A \cdot B D}{BP}$ , then $BE \cdot BP = P E^{2} \cdot B A \cdot B D$ .
  From similarity: $BE \cdot BP$ is not a standard product.
  $BP \cdot B A = BE \cdot B D$ ? No ( $BP / B D = BE / B A ⟹ BP \cdot B A = BE \cdot B D$ ).
  So $BE = BP \cdot B A / B D$ .
  The question asks to prove $BE = \frac{P E ^{2} \cdot B A \cdot B D}{BP}$ ... This dimensionally is $L = L^{2} \cdot L \cdot L / L = L^{3}$ . This is physically impossible for lengths.
  However, I must transcribe what is there.
  Memo [1224] shows the derivation leading to $BE = \frac{P E ^{2} \cdot B A \cdot B D}{BP}$ by equating the erroneous $D A$ expressions.
  I will transcribe the steps as they appear in the memo/logic provided.
  2 marks · Use similar triangles' sides and angles
6. Question 2.2.4 · March 2021 · Provincial · 2 marks
2.2 In the diagram, a circle with a tangent CD is drawn. A, B, D, and E are points on the circumference of the circle. $A E = A B$ and $A B ∥ E D$ . $\hat{A}_{1} = x$ .
2.2.4 Hence, prove that $A E^{2} = BC \cdot E D$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Substitute DC from (a) into (b) $E A = (\frac{BC \times A D}{D B}) (\frac{D E}{A D})$
  1 mark · Use similar triangles' sides and angles
2. ✓
  Simplify $E A = \frac{BC \times D E}{D B}$
  Since $D B = E A$ (from 2.2.3b logic)
  $E A = \frac{BC \times D E}{E A}$
  $E A^{2} = BC \times D E$
  1 mark · Use similar triangles' sides and angles

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