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O-Level E-Math · 2018 · P2 Q7 Circle properties & congruence / mensuration · SAS congruence via radii and vertically opposite angles 9 marks: (a) 3, (b) 3 + 3 · geometry & measurement (congruence, tangents, area of an annulus) difficulty 4 of 5

O-Level E-Math 2018 Paper 2, Question 7: SAS congruence via radii and vertically opposite angles

The answer

(a) congruent by SAS
(b)(i) \(\approx 10.6 \text{ cm}^2\)
(ii) \(\approx 115 \text{ cm}^2\)

O-Level E-Math 2018 Paper 2 Question 7 · Verified worked solution by the Genius Plus Academy teaching team

What this question tests

This is Question 7 of the O-Level E-Math 2018 Paper 2. It tests sas congruence via radii and vertically opposite angles, in the Circle properties & congruence / mensuration area. It is worth 9 marks: (a) 3, (b) 3 + 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.

Step-by-step solution

(a) In triangles \(OAC\) and \(OBD\): - \(OA = OB\) (radii of the large circle); - \(OC = OD\) (radii of the small circle); - \(\angle AOC = \angle BOD\) (vertically opposite angles, since \(AB\) and \(CD\) are straight lines crossing at \(O\)).

So \(\triangle OAC \equiv \triangle OBD\) by SAS (two sides and the included angle).

(b) \(C\) lies on the small circle and \(AC\) is a tangent to it, so the radius \(OC\) is perpendicular to \(AC\), i.e. \(\angle OCA = 90^{\circ}\). Triangle \(OAC\) is right-angled at \(C\) with hypotenuse \(OA = 7\) cm and \(\angle OAC = 30^{\circ}\).

(i) \(OC = OA \sin 30^{\circ} = 7 \times 0.5 = 3.5\) cm (this is the radius of the small circle), and \(AC = OA \cos 30^{\circ} = 7 \times \dfrac{\sqrt3}{2} = 6.062\) cm. Area \(= \dfrac12 \times OC \times AC = \dfrac12 \times 3.5 \times 6.062 = 10.6 \text{ cm}^2\) (3 s.f.).

(ii) The shaded region is the ring between the two circles: large radius \(7\) cm, small radius \(OC = 3.5\) cm. Shaded area \(= \pi(7^2 - 3.5^2) = \pi(49 - 12.25) = 36.75\pi = 115 \text{ cm}^2\) (3 s.f., using \(\pi = 3.142\) gives \(115.5\), which rounds to \(115\)).

Answer: (a) congruent by SAS
(b)(i) \(\approx 10.6 \text{ cm}^2\)
(ii) \(\approx 115 \text{ cm}^2\)

Same structure, different numbers

A question is hard because of its structure, not its surface.

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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Questions students ask

What does O-Level E-Math 2018 Paper 2 Question 7 test?

It is a sas congruence via radii and vertically opposite angles question from Circle properties & congruence / mensuration, worth 9 marks: (a) 3, (b) 3 + 3.

Is this the same as IP Math?

Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.

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