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WAEC 2021 · Paper 2 · Q6

The diagram shows the net of a rectangular pyramid: a 10 cm×4 cm10\text{ cm} \times 4\text{ cm} rectangle with a triangle on each side; the triangles on the 4 cm4\text{ cm} sides have height 6 cm6\text{ cm}. Calculate, correct to two decimal places, the:

6 cm10 cm4 cm
  1. (a)

    slant height (the length of a sloping edge);

  2. (b)

    perpendicular height;

  3. (c)

    total surface area of the pyramid.

Worked solution (try it first)

(a)

  1. A triangle on a 4 cm side has height 6 cm, so its sloping edge runs from a corner of the base to the apex: half the base (2 cm) and the height (6 cm) make a right-angled triangle.
  2. Slant edge =62+22= \sqrt{6^2 + 2^2}
    =40= \sqrt{40}
    ≈6.32\approx 6.32 cm.

(b)

  1. When the pyramid is folded up, the 6 cm height of that face runs from the middle of a short side to the apex.
  2. The middle of a short side is 5 cm (half of 10) from the centre of the base.
  3. Perpendicular height =62−52= \sqrt{6^2 - 5^2}
    =11= \sqrt{11}
    ≈3.32\approx 3.32 cm.

(c)

  1. The height of the triangles on the 10 cm sides: from the middle of a long side (2 cm from the centre), 11+22=15≈3.873\sqrt{11 + 2^2} = \sqrt{15} \approx 3.873 cm.
  2. Total surface area =10×4⏟base+2×12×4×6+2×12×10×15= \underbrace{10 \times 4}_{\text{base}} + 2 \times \frac12 \times 4 \times 6 + 2 \times \frac12 \times 10 \times \sqrt{15}
    =40+24+38.73= 40 + 24 + 38.73
    =102.73 cm2= 102.73\text{ cm}^2.

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