Area · Lesson 4 of 12
Triangles Between Parallel Lines
“Understand equal-area triangles on a common base between parallels and use reflection to solve minimum-perimeter and shortest-path problems.”
• Explain why triangles on the same base and between the same parallel lines have equal areas. • Distinguish what stays constant and what changes as a triangle's top vertex moves along a parallel line. • Analyse perimeter variation among equal-area triangles. • Use reflection to convert a broken-path problem into a straight-line shortest-path problem. • Apply the same reflection idea to a practical route that touches a boundary.
A triangle can change shape dramatically without changing area. One of the clearest examples appears when its base is fixed and the opposite vertex slides along a line parallel to that base. The triangle becomes left-leaning, centred, or right-leaning, yet the perpendicular distance between the parallel lines remains fixed.
This setting also creates a deeper question. If all such triangles have equal area, do they have equal perimeter? They do not. To locate the triangle with the least perimeter, we will use reflection and the fact that a straight segment is the shortest route between two points.
Same Base, Same Height
Let BC be a fixed base, and let point A move along a line l parallel to BC. Because parallel lines stay the same perpendicular distance apart, the height from A to line BC is constant. The base BC is also constant. Therefore ½ × base × height is unchanged.
Problem
Base BC is 14 cm. A parallel line is 6 cm away from BC. Point A can be anywhere on that line. What is the area of triangle ABC?
- 1.The base is always 14 cm.
- 2.The perpendicular height is always the distance between the parallel lines, 6 cm.
- 3.Area = ½ × 14 × 6 = 42 cm² for every position of A.
What Changes When the Vertex Moves?
Although area stays fixed, the side lengths AB and AC usually change as A moves. Therefore the perimeter changes. A very far-left or far-right vertex makes one or both sloping sides longer. The most balanced position gives a shorter total path from B to A to C.
This is an important contrast: equal area does not force equal perimeter. The area depends only on base and perpendicular height here, while perimeter depends on the actual sloping side lengths.
Problem
Two triangles share base BC = 10 cm and have their third vertices on a line 4 cm above BC. Explain what can be said without measuring the sloping sides.
- 1.Every triangle has area ½ × 10 × 4 = 20 cm².
- 2.The perimeters need not be equal because AB and AC vary with the position of A.
- 3.So area is fixed while perimeter can change.
Finding the Minimum Perimeter by Reflection
Because BC is common to all the triangles, minimising the perimeter is the same as minimising AB + AC. Reflection turns that two-segment broken path into an equivalent path to a reflected point.
Reflect C across line l to a point C′. Reflection preserves distance, so AC = AC′ for every point A on l. Therefore AB + AC has exactly the same length as AB + AC′. Among all paths from B to C′ that touch line l at A, the shortest is the straight segment BC′.
The minimum-perimeter triangle is therefore obtained where the straight line from B to C′ crosses l. Because B and C are equally far below l, the crossing point lies halfway horizontally between them. In this parallel-line setting, that point is also on the perpendicular bisector of BC.
A transformation that flips a figure across a line so that each point and its image are the same perpendicular distance from the mirror line.
Problem
B and C lie 12 cm apart on a line. A must lie on a parallel line above them. Which position of A minimises AB + AC?
- 1.Reflect C across the upper parallel line to C′.
- 2.Any broken path B → A → C has the same length as B → A → C′.
- 3.The shortest B-to-C′ route is a straight line.
- 4.Its intersection with the upper line is directly above the midpoint of BC, so the minimum occurs when A lies on the perpendicular bisector of BC.
Shortest Path to a Boundary and Then to a Destination
The same reflection idea solves routes such as house → river → water tank. The river bank acts like the mirror line. Reflect the destination across the bank, draw a straight segment from the starting point to the reflected destination, and use the intersection with the bank as the touching point.
Problem
A house and water tank lie on the same side of a straight river bank. Describe how to find the shortest route from the house to the river bank and then to the tank.
- 1.Reflect the water tank across the river-bank line to an image point.
- 2.Draw a straight segment from the house to that reflected point.
- 3.Where the segment meets the river bank is the optimal touching point.
- 4.Reflecting the second part back gives the shortest house → river → tank route.
All triangles with the common base and third vertex on the parallel line have the same area, but only one symmetric position gives the least perimeter. Do not infer a perimeter fact from an area fact.
Quiz
Why do triangles on the same base between the same parallel lines have equal areas?
As the top vertex moves along the parallel line, which quantity stays fixed?
What does reflection preserve in the shortest-path argument?
Why is a straight segment used after reflection?
Where is the minimum-perimeter vertex in the symmetric parallel-line setup?
Practice Problems
- Base BC is 18 cm and the parallel line is 7 cm away. Find the area of every triangle whose third vertex lies on that line.
- Explain why moving the third vertex 20 cm sideways does not change the area in the previous question.
- Draw two equal-area triangles on the same base that clearly have different perimeters.
- Use a reflection diagram to explain why the minimum-perimeter vertex lies above the midpoint of the base.
- A straight wall must be touched on a route from point P to point Q, both on the same side. Describe the construction for the shortest route.
- State one quantity that remains constant and two quantities that can change as the top vertex slides along the parallel line.
Key Takeaways
• Triangles on the same base between the same parallels have equal base and equal perpendicular height. • Their areas are equal even though their shapes and perimeters may differ. • Minimum perimeter reduces to minimising the sum of the two changing side lengths. • Reflection converts a broken path into an equivalent route to a reflected point. • A straight line then identifies the shortest possible route. • The same construction works for practical shortest-path problems involving a straight boundary.