The answer
(a) \(\triangle APS \equiv \triangle BQP\) by AAS
(b) \(AP = 24\text{ cm}\), \(PB = 7\text{ cm}\)
O-Level E-Math 2025 Paper 1 Question 20 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 20 of the O-Level E-Math 2025 Paper 1. It tests proving congruent triangles (aas), in the Congruence & similarity / Pythagoras area. It is worth 6 marks: (a) 3 + (b) 3. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(a) In triangles \(APS\) and \(BQP\): - \(\angle PAS = \angle QBP = 90^{\circ}\) (interior angles of the larger square \(ABCD\)); - \(PS = QP\) (sides of the same smaller square \(PQRS\)); - \(\angle SPA = \angle PQB\). Reason: at \(P\), the angles on the straight line \(AB\) give \(\angle SPA + \angle SPQ + \angle QPB = 180^{\circ}\), and \(\angle SPQ = 90^{\circ}\) (angle of the smaller square), so \(\angle SPA + \angle QPB = 90^{\circ}\). In right-angled triangle \(BQP\), \(\angle PQB + \angle QPB = 90^{\circ}\) too. Hence \(\angle SPA = \angle PQB\).
With one equal side (\(PS = QP\)) and two equal angles, \(\triangle APS \equiv \triangle BQP\) by AAS.
(b) Larger square side \(= \dfrac{124}{4} = 31\text{ cm}\), so \(AP + PB = 31\). Smaller square side \(PS = \dfrac{100}{4} = 25\text{ cm}\).
By congruence \(AS = PB\), so let \(AP = x\) and \(PB = 31 - x = AS\). Triangle \(APS\) is right-angled at \(A\), so by Pythagoras \(AP^2 + AS^2 = PS^2\): \[x^2 + (31 - x)^2 = 25^2 \Rightarrow x^2 + 961 - 62x + x^2 = 625 \Rightarrow 2x^2 - 62x + 336 = 0 \Rightarrow x^2 - 31x + 168 = 0.\] Factorise: \((x - 7)(x - 24) = 0\), so \(x = 7\) or \(x = 24\). From the diagram \(AP\) is the longer part, so \(AP = 24\text{ cm}\) and \(PB = 7\text{ cm}\).
Answer: (a) \(\triangle APS \equiv \triangle BQP\) by AAS
(b) \(AP = 24\text{ cm}\), \(PB = 7\text{ cm}\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a proving congruent triangles (aas) question from Congruence & similarity / Pythagoras, worth 6 marks: (a) 3 + (b) 3.
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