Geometry path · G04
Before this lesson: Geometric language and ratios, Angle chasing
Your goal: Translate a length ratio into an area ratio using an identified common altitude.
Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.
Triangle area ratios: the key idea
Write [ABC] for the area of triangle ABC. Triangles with the same altitude have areas in the ratio of their bases; triangles sharing a base have areas in the ratio of their perpendicular heights. The height must be measured to the same line, including its extension when needed. Adding or subtracting a shared region can turn a difficult ratio into two simple ones. Length ratios along a side and area ratios should be connected with the explicit common-altitude reason.
A worked example
D lies on BC with BD:DC=2:3. Find [ABD]:[ADC].
Both triangles have vertex A and bases on line BC, so their perpendicular heights are equal. Their area ratio is BD:DC=2:3. Thus [ABD] is 2/5 of [ABC].
Your turn: change one thing
If [ABC]=40 and BD:DC=2:3, find [ADC].
Try this on paper before opening the explanation.
Compare your reasoning
The area has five equal ratio units; ADC receives three. Its area is (3/5)·40=24.
Pause and check
A trap to avoid: Using a base ratio when the heights are different.
Practise and adjust the level
Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.
Interactive practice loads here. You can also use the complete question set below.
Write a complete argument
Prove that a median divides a triangle into two equal-area triangles.
Planning hint
List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.
Read the full solution after your attempt
If M is the midpoint of BC, then BM=MC. Triangles ABM and ACM have these equal bases on the same line and share the perpendicular altitude from A. By the area formula one half times base times height, their areas are equal.
My proof notebook
Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.
All nine practice questions
Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.
1. With a shared altitude, triangle areas are proportional to what?
Foundation
- The squares of their bases
- Their perimeters
- Their largest angles
- Their bases
Hint
Use area=base·height/2.
Answer and reasoning
Their bases. The common factor height/2 cancels in the ratio.
2. The height of a triangle is measured how?
Foundation
- Along any side
- Along the median always
- Along the angle bisector always
- Perpendicularly to the base line
Hint
The area formula uses perpendicular distance.
Answer and reasoning
Perpendicularly to the base line. A sloping segment is not generally the altitude.
3. BD:DC=2:3. What is [ABD]:[ADC]?
Core
- 4:9
- 3:2
- 1:1
- 2:3
Hint
They share the altitude from A to BC.
Answer and reasoning
2:3. The area ratio equals the base ratio.
4. If [ABC]=40 and BD:DC=2:3, what is [ADC]?
Core
- 30
- 24
- 16
- 20
Hint
Use three of the five ratio parts.
Answer and reasoning
24. (3/5)·40=24.
5. A median divides the area in which ratio?
Stretch
- 1:2
- 2:3
- It depends on side lengths
- 1:1
Hint
Its two base segments are equal.
Answer and reasoning
1:1. Equal bases and a common altitude give equal areas.
6. Can triangles with equal bases have different areas?
Stretch
- Only for obtuse triangles
- Yes, if their heights differ
- No
- Only if one base is zero
Hint
Both base and perpendicular height determine area.
Answer and reasoning
Yes, if their heights differ. Equal base alone is insufficient without equal altitude.
7. Triangles sharing a base have areas proportional to what?
Foundation
- Their perpendicular heights
- Their perimeters
- Their squared heights
- Their angles alone
Hint
The common base factor cancels.
Answer and reasoning
Their perpendicular heights. Area is base times perpendicular height divided by two.
8. A median divides a triangle of area 30 into areas what?
Core
- 15 and 15
- 10 and 20
- 5 and 25
- 30 and 30
Hint
Its two base segments are equal.
Answer and reasoning
15 and 15. The equal-area triangles must each have half the total area.
9. Why compare triangles with bases on the same line?
Stretch
- All their sides become equal
- Their bases must be equal
- They must be right triangles
- A shared opposite vertex then gives the same altitude
Hint
Perpendicular distance from one point to one line is fixed.
Answer and reasoning
A shared opposite vertex then gives the same altitude. This supplies the common height needed for the base-ratio argument.
Choose your next step
Continue to The midpoint theorem. If this felt difficult, return to a prerequisite above. Every lesson stays open.
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