KCSE 2025 Maths P1 Q14 — Net of a Trapezoidal Prism

KCSE 2025 Form 4 Geometry

Published

The Question

“The figure shows a prism ABCDEFGH whose cross section is a trapezium. The prism is 5 cm long, so BC = CF = 5 cm. On the trapezium cross section the parallel sides measure 5 cm and 2 cm (AD = HE = 2 cm), while the two slanting sides measure 2.5 cm each (AB = DC = 2.5 cm). Draw a labelled net of the prism.”

1

Understand what a net is

A net is the flat shape you get by unfolding a solid so that every face lies in one plane. This prism has two identical trapezium end faces and rectangles wrapping around the sides that join them. To draw the net you must reproduce all six faces at their true sizes, joined along their shared edges.

2

Find the length of the side rectangles

Because the prism is 5 cm long, every rectangle that wraps around the sides is 5 cm in the direction of the length. Their other dimension equals the length of the trapezium edge they are attached to. So each side face is a rectangle 5 cm long by the width of its trapezium edge.

prism length=5 cm\text{prism length} = 5 \text{ cm}
3

List the four wrap-around rectangles

Unfolding the four sides of the trapezium gives a strip of four rectangles laid side by side. The bottom is the base rectangle, then a slanting face on each side, and the top face. Each takes the 5 cm length as one side and its trapezium edge as the other.

base=5×5\text{base} = 5 \times 5
two slanting faces=2.5×5\text{two slanting faces} = 2.5 \times 5
top=2×5\text{top} = 2 \times 5
4

Find the height of the trapezium ends

To draw the two trapezium ends accurately you need their perpendicular height. Drop a perpendicular from the short parallel side to the long one. The base overhangs the top by three centimetres in total, so each side overhangs by half of that, which is 1.5 cm. This forms a right-angled triangle with the 2.5 cm slant as the hypotenuse and 1.5 cm as the horizontal, so use Pythagoras to get the height.

overhang=5−22=1.5 cm\text{overhang} = \frac{5 - 2}{2} = 1.5 \text{ cm}
h=2.52−1.52=6.25−2.25=4=2 cmh = \sqrt{2.5^{2} - 1.5^{2}} = \sqrt{6.25 - 2.25} = \sqrt{4} = 2 \text{ cm}

The height 2 cm is needed to construct each trapezium end at its true shape.

5

Assemble and label the net

Draw the strip of four rectangles in a row, then attach one trapezium end to the base rectangle folding up on one side and the other trapezium end folding up on the opposite side. Label every vertex and mark the measurements so the net is clearly identified as belonging to this prism.

The net is made of four rectangles and two trapezium ends, giving six faces in total.

Final Result

The labelled net consists of four rectangles in a strip — a 5 cm by 5 cm base, two 2.5 cm by 5 cm slanting faces and a 2 cm by 5 cm top — together with two identical trapezium ends whose parallel sides are 5 cm and 2 cm, slanting sides 2.5 cm and perpendicular height 2 cm.

Why this method works

A net works because unfolding a solid does not change the size or shape of any face; it only rotates each face about a shared edge until they all lie flat. Every rectangle keeps the 5 cm length of the prism, and each trapezium end keeps its exact cross-section dimensions. Working out the 2 cm height by Pythagoras matters because the slanting sides alone do not tell you how tall to draw the trapezium — the height is what lets you construct the end faces at their true shape rather than by guesswork.

✓

The two parallel sides differ by 3 cm, split as 1.5 cm each side; since 1.5, 2 and 2.5 form a 3-4-5 triangle scaled by 0.5, the 2 cm height is confirmed.