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WAEC 2012 · Paper 2 · Q11

A sector of a circle with radius 20 cm20\text{ cm} has an area of 396 cm2396\text{ cm}^2. Calculate, correct to 1 decimal place, the: [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

  1. (a)

    sectoral angle;

  2. (b)

    perimeter of the sector;

  3. (c)

    volume of the cone formed when the sector is bent such that its straight edges coincide.

Worked solution (try it first)

(a)

  1. θ360×227×202=396\frac{\theta}{360} \times \frac{22}{7} \times 20^2 = 396, so θ=396×360×722×400\theta = \frac{396 \times 360 \times 7}{22 \times 400}
    =113.4∘= 113.4^\circ.

(b)

  1. Arc =2×arear= \frac{2 \times \text{area}}{r}
    =2×39620= \frac{2 \times 396}{20}
    =39.6= 39.6 cm.
  2. Perimeter =39.6+20+20=79.6= 39.6 + 20 + 20 = 79.6 cm.

(c)

  1. Bent into a cone, the arc becomes the base circumference: 2×227×r=39.62 \times \frac{22}{7} \times r = 39.6, so r=6.3r = 6.3 cm.
  2. The sector's radius, 20 cm, becomes the slant height.
  3. Height: h=202−6.32h = \sqrt{20^2 - 6.3^2}
    =400−39.69= \sqrt{400 - 39.69}
    =360.31= \sqrt{360.31}
    ≈18.98\approx 18.98 cm.
  4. Volume =13×227×6.32×18.98= \frac13 \times \frac{22}{7} \times 6.3^2 \times 18.98
    ≈789.3 cm3\approx 789.3\text{ cm}^3.

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