WAEC 2017 · Paper 2 · Q8

  1. (a)

    The probabilities that Akafi and Iniola will pass an examination are 34\frac34 and 35\frac35 respectively. Find the probability that only Iniola will pass the examination.

  2. (b)

    In the diagram, circle XYZXYZ is inscribed in an equilateral triangle PQRPQR, touching PQPQ at XX, PRPR at YY and QRQR at ZZ. If OO is the centre of the circle and ∣XY∣=10 cm|XY| = 10\text{ cm}, calculate, correct to the nearest whole number: (i) ∠XOY\angle XOY; (ii) the area of the major sector XZYXZY. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

    10 cmPQROXYZ

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. "Only Iniola passes" means Iniola passes and Akafi fails.
  2. P(Akafi fails)=1−34P(\text{Akafi fails}) = 1 - \frac34
    =14= \frac14, so the probability is 14×35=320\frac14 \times \frac35 = \frac{3}{20}.

(b)(i)

  1. A radius meets a tangent at 90∘90^\circ, so ∠OXP=∠OYP=90∘\angle OXP = \angle OYP = 90^\circ.
  2. The triangle is equilateral, so ∠XPY=60∘\angle XPY = 60^\circ.
  3. The angles of quadrilateral PXOYPXOY add up to 360∘360^\circ: ∠XOY=360∘−90∘−90∘−60∘\angle XOY = 360^\circ - 90^\circ - 90^\circ - 60^\circ
    =120∘= 120^\circ.

(ii)

  1. Find the radius from the chord XY=10XY = 10 cm.
  2. The perpendicular from OO to XYXY halves both the chord and the angle, giving a right-angled triangle with half-chord 5 cm opposite an angle of 60∘60^\circ: r=5sin⁡60∘r = \frac{5}{\sin 60^\circ}
    ≈5.7735\approx 5.7735 cm.
  3. The major sector XZYXZY has angle 360∘−120∘=240∘360^\circ - 120^\circ = 240^\circ.
  4. Area =240360×227×5.77352= \frac{240}{360} \times \frac{22}{7} \times 5.7735^2
    ≈69.8\approx 69.8, which is 70 cm270\text{ cm}^2 to the nearest whole number.

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