WAEC 2023 · Paper 2 · Q12

  1. (a)

    In the diagram, PP, QQ, RR and SS are points on the circle with centre OO; QR∥OSQR \parallel OS, ∠QOR=2m\angle QOR = 2m, ∠QPR=n\angle QPR = n and ∠SOR=54∘\angle SOR = 54^\circ. Find the values of mm and nn.

    2m54°nOSRQP

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

  2. (b)

    The length of a rectangle is 4 cm4\text{ cm} more than its width. If the perimeter is 40 cm40\text{ cm}, find its area.

Worked solution (try it first)

(a)

  1. QR∥OSQR \parallel OS, so ∠ORQ=∠SOR=54∘\angle ORQ = \angle SOR = 54^\circ (alternate angles).
  2. ∣OQ∣=∣OR∣|OQ| = |OR| (radii), so triangle OQROQR is isosceles and ∠OQR=∠ORQ=54∘\angle OQR = \angle ORQ = 54^\circ.
  3. Angles in triangle OQROQR: ∠QOR=180∘−54∘−54∘\angle QOR = 180^\circ - 54^\circ - 54^\circ
    =72∘= 72^\circ.
  4. So 2m=72∘2m = 72^\circ and m=36∘m = 36^\circ.
  5. The angle at the centre is twice the angle at the circumference on the same arc QRQR, so n=12×72∘=36∘n = \frac12 \times 72^\circ = 36^\circ.

(b)

  1. Let the width be ww cm.
  2. The length is then (w+4)(w + 4) cm.
  3. Perimeter: 2(w+w+4)=402(w + w + 4) = 40, so 4w+8=404w + 8 = 40 and w=8w = 8.
  4. The length is 8+4=128 + 4 = 12 cm, so the area is 12×8=9612 \times 8 = 96.
  5. The area is 96 cm296\text{ cm}^2.

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