KCSE 2025 Maths P1 Q22 — Trapezium Construction and Area Outside an Arc

KCSE 2025 Form 4 Geometry

Published

The Question

“In the figure, AB = 10 cm and AD = 5 cm are part of a trapezium ABCD in which AB is parallel to DC, with angle DAB = 65 degrees. Using a ruler and a pair of compasses only, complete the trapezium given that angle ABC = 67.5 degrees and measure the length of DC. Then drop a perpendicular from D to meet AB at E and measure DE. Draw an arc of centre A and radius 5 cm so that the clockwise arc joins D to a point F on AB. Finally, calculate the area of the trapezium that lies outside this arc, taking pi as 3.142.”

1

Draw the base and fix D

Start with the given base line AB. At A, use the pair of compasses to set off the given angle of 65 degrees, then mark the given length AD along that new line to fix the vertex D. This locks in one slanting side of the trapezium exactly as the question describes.

AB=10 cm,AD=5 cm,∠DAB=65∘AB = 10\text{ cm}, \quad AD = 5\text{ cm}, \quad \angle DAB = 65^\circ
2

Complete the trapezium and measure DC

At B, construct the given angle of 67.5 degrees. Because DC must be parallel to AB, draw a line through D parallel to AB; where it meets the ray from B is the vertex C. Measuring the top side DC on the accurate drawing gives about 6.01 cm.

∠ABC=67.5∘,DC∥AB\angle ABC = 67.5^\circ, \quad DC \parallel AB
DC≈6.01 cmDC \approx 6.01\text{ cm}
3

Drop the perpendicular and measure the height

From D, drop a perpendicular to meet AB at E. This perpendicular is the height of the trapezium, since it is the distance between the two parallel sides. Measuring DE on the construction gives about 4.53 cm.

DE≈4.53 cmDE \approx 4.53\text{ cm}
4

Draw the arc of centre A

Keeping the compass radius at 5 cm and the point at A, draw the clockwise arc starting at D and swinging round to a point F on AB. Because AD is 5 cm, D already lies on this arc, and F is where the arc meets the base. The arc sweeps through the 65 degree angle at A, so it cuts off a circular sector from the trapezium.

arc: centre A, radius 5 cm, angle 65∘\text{arc: centre } A,\ \text{radius } 5\text{ cm},\ \text{angle } 65^\circ
5

Find the area of the whole trapezium

Use the trapezium area rule: half the sum of the two parallel sides multiplied by the height between them. Substitute AB = 10 cm, DC = 6.01 cm and height DE = 4.53 cm.

Atrap=12(AB+DC)×DEA_{\text{trap}} = \frac{1}{2}(AB + DC)\times DE
Atrap=12(10+6.01)×4.53≈36.28 cm2A_{\text{trap}} = \frac{1}{2}(10 + 6.01)\times 4.53 \approx 36.28\text{ cm}^{2}
6

Subtract the sector cut off by the arc

The arc encloses a sector of the circle with centre A, radius 5 cm and angle 65 degrees. Find its area as the fraction 65 over 360 of a full circle, then subtract it from the trapezium area to leave the region lying outside the arc.

Asector=65360×π×52≈14.18 cm2A_{\text{sector}} = \frac{65}{360}\times \pi \times 5^{2} \approx 14.18\text{ cm}^{2}
Aoutside=36.28−14.18≈22.1 cm2A_{\text{outside}} = 36.28 - 14.18 \approx 22.1\text{ cm}^{2}

Final Result

The top side DC measures about 6.01 cm, the height DE about 4.53 cm, and the area of the trapezium lying outside the arc is about 22.1 square centimetres.

Why this method works

The method works because a trapezium's area depends only on its two parallel sides and the perpendicular distance between them, which is exactly what DE measures. The arc traces every point 5 cm from A, so together with the two radii along AB and AD it bounds a sector — a slice of a circle whose size is simply the angle-fraction 65 over 360 of the full circle of radius 5. Since that sector sits entirely inside the trapezium, removing its area leaves precisely the part of the trapezium outside the arc.

✓

Sector fraction 65/360 ≈ 0.1806, times 3.142 × 25 ≈ 14.18 cm²; and 36.28 − 14.18 ≈ 22.1 cm², confirming the result.