=
Same base and same height
The two area formulas contain exactly the same factors.
A₁ = A₂Lesson 5.1 — Equal Areas from Shared Bases and Heights
第5讲 · 同底等高与等积变形
A difficult shaded region often becomes easy when you replace it with a triangle of equal area. Learn to prove equality through a shared base, equal perpendicular heights, parallel lines, midpoints, and common-region subtraction.
Prove the base. Prove the height. Then transform the area.
Mission 1
Not completed
Equal-area transformations begin with one formula
For any triangle, area depends only on a chosen base and its perpendicular height:
Triangle area = ½ × base × perpendicular height
∥
Vertices on a parallel line
If two triangles share a base and their third vertices lie on a line parallel to it, their heights are equal.
same base + equal heights−
Equal wholes minus the same part
Removing one common region from equal-area wholes leaves equal-area remainders.
X + C = Y + C ⇒ X = YCongruent means exactly the same shape and size, even after moving or turning. Equal area alone does not imply congruence. In the equations above, the letters stand for areas.
Mission 2
Not completed
Slide a vertex without changing the area
The base remains 10 units long. The top vertex moves along a line parallel to the base, always 6 units away.
30square units — unchanged
Invariant: the shape changes, but the base and perpendicular height do not. Therefore the area stays ½ × 10 × 6 = 30.
square units
Mission 3
Not completed
One shared base, two vertices on a parallel line
Triangles AED and CED share base ED. Points A and C move along a line parallel to ED, so their perpendicular heights to ED are equal.
samebase ED
sameperpendicular height
=triangle areas
Mission 4
Not completed
Equal wholes minus one common triangle
In trapezoid ABCD, AD is parallel to BC. Diagonals AC and BD meet at O. Area(COD) = 3 square units and Area(BOC) = 6 square units. For this mission, find Area(AOB); not every given number is needed.
Diagram not necessarily to scale. Use the parallel-line and shared-base facts, not visual measurement.
1. Equal wholes: triangles ADB and ADC share base AD. Their third vertices B and C lie on BC, a line parallel to AD. Therefore their heights—and areas—are equal.
2. Common part: both large triangles contain triangle AOD.
3. Equal remainders: removing AOD from both leaves triangle AOB equal in area to triangle COD.
AOB + AOD=COD + AOD
↓ remove AOD from both
AOB=COD
square units
Mission 5
Not completed
Transform through a parallelogram and a shared base
Triangle ABC has area 56 cm². Parallelogram DEFC has half that area, and DE is parallel to AC. The shaded target is triangle AED.
Area(DEFC) = 56 ÷ 2 = 28
Diagonal EC halves the parallelogram: Area(DEC) = 28 ÷ 2 = 14
Triangles AED and CED share base ED, and DE ∥ AC, so their heights are equal
Area(AED) = Area(CED) = 14
cm²
cm²
cm²
Mission 6
Not completed
Two midpoints create a quarter-area triangle
D is the midpoint of BC, and E is the midpoint of AB. The shaded triangle is AED.
D is midpoint of BC: triangles ABD and ACD share the height from A and have equal bases BD and DC, so each is half of triangle ABC.
E is midpoint of AB: inside triangle ABD, triangles AED and BED share the height from D and have equal bases AE and EB, so each is half of triangle ABD.
Two halvings: Area(AED) = ½ × ½ × Area(ABC) = ¼ Area(ABC).
square units
square units
Mission 7
Not completed
A shaded area that ignores the unknown top base
A trapezoid has bottom base 10 cm and height 6 cm. A point P lies between the endpoints of the bottom base. Joining P to the two top vertices creates a central white triangle and two shaded side triangles.
Trapezoid area = ½ × (top + 10) × 6
Central triangle area = ½ × top × 6
Shaded total = trapezoid − central triangle
Shaded total = ½ × 10 × 6 = 30
30cm² — unchanged while you move the sliders
cm²
Mission 8
Not completed
Equal-area transformation detective
Decide which conclusion is justified. Do not trust appearance alone.
Case A
Two triangles share a base; their third vertices lie on a line parallel to that base.
Case B
Two triangles share a base, but their third vertices are at unknown distances from the base.
Case C
Two equal-area figures both contain the same region R.
Case D
A median divides a triangle by joining a vertex to the midpoint of the opposite side.
Case E
Two triangles have equal area.
Case F
A picture makes two heights look equal, but no right-angle or parallel-line fact is given.
Extra practice — put the relationships together
This additional practice combines the relationships used above. Work through the steps before the workshop; these extra checks do not add to the ten mission count.
Add equal-area triangles across two strips
original Exercise 2
The outer figure is a parallelogram. Its horizontal base is 8 cm, and its perpendicular height is 6 cm. The dividing line is parallel to the bases; it need not be halfway between them. In each strip, the shaded triangles do not overlap, their bases partition a full 8 cm edge, and their opposite vertices lie on the dividing line. Find the total shaded area.
Use the stated relationships; do not measure the drawing.
Try each step first. Open a hint if you need a starting point.
Hint — choose the first relationship
In each strip, the shaded triangle bases together cover one complete 8 cm edge. Their heights within that strip are equal.
Worked solution — compare after trying
- Call the upper and lower strip heights h₁ and h₂. Together they make the full height: h₁ + h₂ = 6.
- The upper shaded triangles share height h₁. Add their bases: their total area is ½ × 8 × h₁.
- The lower shaded triangles share height h₂ and their bases also total 8. Their total area is ½ × 8 × h₂.
- Combine before calculating: ½ × 8 × (h₁ + h₂) = ½ × 8 × 6 = 24 cm². The middle line need not halve the height.
Mission 9
Not completed
Equal-area transformation workshop
Solve at least 6 of the 8 questions correctly. All lengths below are in cm and areas in cm². Use the figure from the named mission where specified.
1Two triangles share the same base and perpendicular height. One area is 18. Find the other.
2Two triangles share a base of 12 and both have height 5. Compare their areas.
3In Mission 4’s trapezoid, Area(COD) = 3. Find Area(AOB).
4Use Mission 5’s figure, where DE is parallel to AC. Triangle ABC has area 56 cm², and parallelogram DEFC has half that area. Find shaded Area(AED).
5In Mission 6’s midpoint figure, shaded Area(AED) = 2. Find Area(ABC).
6A trapezoid’s bottom base is 14 and height is 8. The two side regions are formed as in Mission 7. Find their total area.
7Two equal-area wholes each have area 40 and share the same overlap of area 17. Find each remainder.
8Two triangles share a height. Their bases are 4 and 7. The first area is 12. Find the second area. With the same height, each base unit contributes the same area. Four units make 12, so one makes 12 ÷ 4. Use seven units for the other triangle.Hint — use equal base units
Transformation checklist
- Name the two triangles or complete figures.
- Identify the shared base or prove equal perpendicular heights.
- Use parallel lines, midpoints, or a common region as justification.
- Replace the difficult region with an equal-area one only after the equality is proved.
- Never measure an unstated length from the picture.
Workshop hint — choose your first step
Name the shared base and perpendicular height. If there is a midpoint, compare the two base parts before calculating.
Worked workshop solutions — open after trying
Compare your first different step, then close this panel and try again.
- Same base and equal perpendicular height give equal areas: 18.
- Each area is 12 × 5 ÷ 2 = 30. The areas are equal.
- ADB and ADC have equal areas. Remove their common AOD: AOB = COD = 3.
- DEFC = 56 ÷ 2 = 28. Its diagonal half is 14. The shared-base, parallel-line transfer gives AED = 14.
- AED is one quarter of ABC, so ABC = 2 × 4 = 8.
- Subtract the central triangle from the trapezoid. The top-base terms cancel, leaving 14 × 8 ÷ 2 = 56.
- Each whole loses the same overlap: 40 − 17 = 23.
- With the same height, one base unit corresponds to 12 ÷ 4 = 3 area units. Seven units give 7 × 3 = 21.
Mission 10
Not completed
Exit ticket
Complete all five checks. The optional reflection is not automatically graded.
square units
cm²
square units
cm²
Equal-Area Transformer
You have completed this lesson.
You can prove equal triangle areas through shared bases, perpendicular heights, parallel lines, midpoints, and common-region subtraction.
Lesson 5.1 completed
Teaching notes
Added original practice reproduces Exercise 2’s two parallel strips and derives the shaded total of 24 cm².
The original gives Area(ABC) = 56 and states that parallelogram DEFC has half that area.Diagonal EC halves the parallelogram, and DE ∥ AC makes triangles AED and CED equal in area, yielding a shaded area of 14.
D and E are midpoints, so the shaded triangle is one quarter of the whole.
The sliding-vertex, dynamic parallel-line, diagnostic, and additional practice activities are instructional scaffolds created to make the worked example relationships explicit and self-contained. All diagrams are explanatory and are not scale drawings.