The Flower Bed Path That Was Not a Bisector
A triangular flower bed ABC in a school garden has base BC = 24 m and a perpendicular height from A to BC of 18 m. The gardener lays a straight path DE parallel to BC, with D on AB and E on AC, at a distance of 9 m from A measured along the height. He tells the principal that the path divides the bed into two parts of equal area, and asks for manure for each part on that basis. On the same afternoon, a 1.5 m measuring stick held vertically casts a shadow 2.4 m long, while the flagpole at the corner of the garden casts a shadow 9.6 m long on the same level ground. The principal is doubtful about the equal-area claim, wants the correct position of a parallel path if equal areas were really needed, and also wants the flagpole's height fixed from the shadow data.
1. Test the gardener's claim that DE divides the triangular bed into two equal areas.
Read the worked answer
DE is parallel to BC, so triangle ADE is similar to triangle ABC (AA similarity: angle A common, angle ADE = angle ABC as corresponding angles).
Linear ratio = AD/AB = height of ADE / height of ABC = 9/18 = 1/2.
For similar figures, area ratio = (linear ratio)² = (1/2)² = 1/4.
Area of ABC = (1/2) × 24 × 18 = 216 m². Area of ADE = 216/4 = 54 m². Area of trapezium DBCE = 216 − 54 = 162 m².
The two parts are in the ratio 54 : 162 = 1 : 3, not 1 : 1. The claim is false; the lower strip is three times the upper triangle, so manure for the two parts should be in the ratio 1 : 3, not split equally.
2. Where should a parallel path be drawn so that the two parts really do have equal areas?
Read the worked answer
Let the path be at height h from A along the 18 m altitude. For equal areas, area of the upper triangle = half of 216 = 108 m².
By similarity, (h/18)² = 108/216 = 1/2.
So h/18 = 1/√2, giving h = 18/√2 = 9√2 ≈ 12.73 m from A.
So the path must be about 12.73 m below A, not 9 m. Check: area at that height = 216 × (1/2) = 108 m², and the remaining trapezium is also 108 m².
Also note the general trap: doubling the height from A multiplies the area by four, so the 'half-height = half-area' intuition fails; equal areas require the linear ratio 1 : √2.
3. Find the length of DE and the height of the flagpole.
Read the worked answer
Length of DE: by similarity DE/BC = 1/2, so DE = 12 m.
Flagpole: the stick and the flagpole are both vertical on level ground, and the sun's rays are parallel, so the triangles formed by object, shadow and ray are similar.
height/shadow is the same for both: 1.5/2.4 = h/9.6.
h = 1.5 × 9.6/2.4 = 14.4/2.4 = 6 m.
Check the scale factor: 9.6/2.4 = 4, and 4 × 1.5 = 6 ✔. So the flagpole is 6 m tall. The key assumption to state is that both shadows are measured at the same time of day on level ground.