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WAEC 2013 · Paper 2 · Q12

  1. (a)

    An open rectangular tank is made from a steel plate of area 1440 m21440\text{ m}^2. Its length is twice its width. If the depth of the tank is 4 m4\text{ m} less than the width, find the length of the tank.

  2. (b)

    In the diagram, ∣XU∣=10 cm|XU| = 10\text{ cm}, ∣WU∣=8 cm|WU| = 8\text{ cm}, ∣UV∣=3.5 cm|UV| = 3.5\text{ cm} and XVXV is perpendicular to WZWZ. If the radius of the circle is 7 cm7\text{ cm}, calculate, correct to 2 decimal places, the length of the arc WVWV. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

    10 cm3.5 cm8 cmXVWZU
Worked solution (try it first)

(a)

  1. Let the width be ww m.
  2. Then the length is 2w2w m and the depth is (w−4)(w - 4) m.
  3. The tank is open, so the plate makes the base and four sides but no top.
  4. Base: 2w×w=2w22w \times w = 2w^2.
  5. Two long sides: 2×2w(w−4)=4w2−16w2 \times 2w(w - 4) = 4w^2 - 16w.
  6. Two short sides: 2×w(w−4)=2w2−8w2 \times w(w - 4) = 2w^2 - 8w.
  7. Add them: the total area is 8w2−24w8w^2 - 24w.
  8. Set it equal to the plate: 8w2−24w=14408w^2 - 24w = 1440.
  9. Divide by 8: w2−3w−180=0w^2 - 3w - 180 = 0.
  10. Factorise: (w−15)(w+12)=0(w - 15)(w + 12) = 0.
  11. A width can't be negative, so w=15w = 15.
  12. The length is 2×15=302 \times 15 = 30.
  13. The tank is 30 m30\text{ m} long.

(b)

  1. In the right-angled triangle WUXWUX, tan⁡∠WXU=∣WU∣∣XU∣\tan\angle WXU = \frac{|WU|}{|XU|}
    =810= \frac{8}{10}, so ∠WXV=38.66∘\angle WXV = 38.66^\circ.
  2. The arc WVWV subtends ∠WXV\angle WXV at the circumference, so it subtends twice as much at the centre: 2×38.66∘=77.32∘2 \times 38.66^\circ = 77.32^\circ.
  3. Arc length =77.32360×2×227×7= \frac{77.32}{360} \times 2 \times \frac{22}{7} \times 7
    =9.450= 9.450.
  4. The arc WVWV is 9.45 cm9.45\text{ cm} long.

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