△Math Explorer · Grade 5
Lesson progress0 of 10 missions
Chapter 5 · Equal-Area Transformations

Lesson 5.2 — Turn Segment Ratios into Area Ratios

第5讲 · 线段比转化为面积比

A length ratio becomes an area ratio only when the triangles share the correct perpendicular height or base. Learn to translate part-to-part clues, part-to-whole clues, and chained ratios without being fooled by the drawing.

Translate the segment. Find the shared dimension. Transfer the ratio.
Grade 5 enrichment 55–70 minutes Interactive ratio diagrams Autosaves in this browser
Mission 1

Build the ratio bridge from the triangle formula

Not completed

The triangle formula explains exactly when a segment ratio may be copied into an area ratio. In every ratio, keep the triangles in the same order: first : second. A₁ and A₂ denote their areas; b denotes base length and h perpendicular height.

Triangle area = ½ × base × perpendicular height
b

Same perpendicular height

The factor ½ × height is common, so only the bases compare.

A₁ : A₂ = b₁ : b₂
h

Same base

The factor ½ × base is common, so only the heights compare.

A₁ : A₂ = h₁ : h₂
?

No shared dimension proved

A side ratio alone is not enough. First prove which base or perpendicular height is shared.

Do not copy a ratio by appearance.
Mission 2

Translate part-to-part clues into part-to-whole ratios

Not completed

Most ratio errors happen before any area calculation. A statement such as BE = 3 × AE compares two parts, not a part with the whole segment AB. Here E lies between A and B, so AB = AE + EB.

1
3
AE = 1
BE = 3
1 : 3AE : BE
4AB units
1 : 4AE : AB
3 : 4BE : AB

Change the two parts. The whole is always their sum.

original-style clue: if BE = 3AE, then AB = AE + BE = 4AE.
AE : BE = 1 : 3
AE : AB = 1 : (1 + 3) = 1 : 4
BE : AB = 3 : 4
Part-to-part is not part-to-whole. “Three times the other part” means four equal parts altogether.

The following questions use their written measurements, regardless of the slider settings.

Mission 3

Same-height machine: compare the bases

Not completed

These two triangles rest on the same line and have the same perpendicular height. Move the controls and watch their area ratio follow the base ratio.

Two triangles with equal perpendicular heights The triangles have adjustable bases and a shared adjustable height. Their area ratio equals their base ratio. base 3base 5 shared height 6

Both top vertices lie on one line parallel to the base line, so the perpendicular heights are equal.

3
5
6
3 : 5base ratio
9area A
15area B
3 : 5area ratio
Changing the common height changes both areas by the same factor. The ratio stays tied to the bases.

Use the fixed measurements written in these questions, not your current slider settings. Ratios compare the triangles in the order stated.

square units
Mission 4

Same-base machine: compare the perpendicular heights

Not completed

Now the base stays fixed while the third vertices sit at different perpendicular distances from the base line.

Two triangles sharing one base with different heights The triangles share a horizontal base. Their adjustable heights determine their area ratio. shared base 8height 4height 7

The triangles may overlap. The shared base is still the same segment.

8
4
7
4 : 7height ratio
16area A
28area B
4 : 7area ratio
With one shared base, the farther vertex creates the larger perpendicular height and therefore the larger area.

Use the fixed measurements written in these questions, not your current slider settings. Ratios compare the triangles in the order stated.

square units
Mission 5

Transfer one diagonal ratio through a trapezoid

Not completed

In right trapezoid ABCD, AD is parallel to BC and diagonals AC and BD meet at O. Area(BOC) = 6 square units and Area(DOC) = 3 square units. Two known triangle areas reveal a segment ratio on diagonal BD. That same segment ratio can then control a second pair of triangles.

Trapezoid divided by diagonals into four triangles The area of triangle BOC is 6 and the area of triangle DOC is 3. Equal-height triangle comparisons transfer the ratio BO to OD to the top pair of triangles. 63 AOB = 3 BO : OD = 2 : 1AOB : AOD = 2 : 1 ABCDO

Diagram not necessarily to scale. Every ratio is justified by a shared perpendicular height.

1 · Compare the lower pairBO : OD = 6 : 3 = 2 : 1

Triangles BOC and DOC have the same height from C to line BD.

2 · Transfer the segment ratioAOB : AOD = 2 : 1

Triangles AOB and AOD share the height from A to line BD.

3 · Use equal remaindersArea(AOB) = Area(COD) = 3

Triangles ADB and ADC share AD and have equal heights from the parallel line BC. Subtract their common triangle AOD to get Area(AOB) = Area(COD).

4 · FinishArea(AOD) = 3 ÷ 2 = 1.5

Total area = 6 + 3 + 3 + 1.5 = 13.5.

square units
square units
Mission 6

Turn side fractions into a rectangle-area chain

Not completed

Rectangle ABFD has width AB equal to two thirds of length AD. C lies on AD with AC = ⅓AD, and E is the midpoint of DF. Triangle ABC has area 14 square units. Find the area left after triangles ABC and BFE are removed from the rectangle.

Rectangle with one-third top segment and midpoint on the right side The shaded polygon is the rectangle after removing triangle ABC and triangle BFE. ⅓ AD⅔ ADequal ABC = 14BFE = ¼ rectangle ABCDEF

The shaded region is found from the whole rectangle, not by measuring its sloping sides.

AC : AD = 1 : 3 ⇒ Area(ABC) : Area(ABD) = 1 : 3
Area(ABD) = 3 × 14 = 42
Diagonal BD halves the rectangle ⇒ rectangle area = 2 × 42 = 84
DE = EF ⇒ Area(BFE) = ¼ × 84 = 21
84 − 14 − 21 = 49shaded area
The clue AC = ⅓AD compares bases of triangles ABC and ABD. The clue DE = EF makes E the midpoint of the rectangle’s right side.
square units
square units
square units
square units
Mission 7

Multiply two justified fractions in a parallelogram

Not completed

In parallelogram ABCD, E is the midpoint of BC. Point F lies on diagonal BD and BF = 2DF. Triangle BEF has area 8. Find the parallelogram’s area.

Parallelogram with midpoint E and ratio point F on diagonal BD E is the midpoint of BC and F divides BD in the ratio two to one. Triangle BEF is one sixth of the parallelogram. BE = ½BC BF : FD = 2 : 1Area = 8 ABCDEF

First use the midpoint fraction. Then use the diagonal fraction. Finally remember that diagonal BD halves the parallelogram.

MidpointBE : BC = 1 : 2

Triangles BDE and BCD have the same perpendicular height from D to BC, so Area(BDE) is half of Area(BCD).

Diagonal ratioBF : BD = 2 : 3

Triangles BEF and BED have the same perpendicular height from E to BD, so Area(BEF) is two thirds of Area(BDE).

Combine⅔ × ½ = ⅓

Triangle BEF is one third of triangle BCD.

Whole figure⅓ × ½ = ⅙

Triangle BCD is half the parallelogram, so BEF is one sixth of it.

8 × 6 = 48parallelogram area
square units
Mission 8

Ratio detective: decide what is actually justified

Not completed

Choose the conclusion that follows from the stated geometry—not from how the sketch looks.

1 · Same height, bases 2 : 5

The area ratio is:

2 · Same base, heights 3 : 4

The area ratio is:

3 · E lies between A and B; BE = 3AE

AE : AB is:

4 · D lies between A and C; AC = 4CD

AD : AC is:

5 · Equal areas only

Must the bases be equal?

6 · Two arbitrary side ratios

May we multiply them to get the area ratio without proving shared dimensions?

Mission 9

Segment-ratio workshop

Not completed

Reach at least 6 out of 8. Lengths below use the same unit, and areas are in square units. Begin each area comparison with a shared-base or shared-height reason.

1Same height: bases are 4 and 7. The base-4 triangle has area 20. Find the other area.
Hint

Scale 20 by 7 ÷ 4.

2Same base: heights are 3 and 8. The height-3 triangle has area 12. Find the other area.
Hint

Area follows the height ratio 3 : 8.

3E lies between A and B. If BE = 5AE, what is AE : AB?
Hint

AB contains 1 + 5 equal parts.

4Point D lies between A and C. If AC = 5CD, what is AD : AC?
Hint

AD = AC − CD.

5Use the trapezoid arrangement in Mission 5 with new measurements: Area(AOB) = 9 square units and BO = 3OD. Find the trapezoid’s total area.
Hint

First use the equal remainders to find COD. Then apply BO : OD to pairs that share a height.

6For Mission 6’s rectangle, with Area(ABC)=14, AC=⅓AD, and DE=EF, find the shaded area.
Hint

Recover the rectangle area from ABC’s base fraction. Use E’s midpoint condition to find the other removed triangle.

7For Mission 7’s parallelogram, Area(BEF)=8, E is midpoint of BC, and BF=2DF. Find the parallelogram area.
Hint

Use the midpoint and diagonal fractions one at a time, naming the shared height at each step.

8Point D divides BC with BD : DC = 2 : 3. Triangle ABC has area 50. Find Area(ACD).
Hint

Triangles ABD and ACD share the height from A; their bases total 5 parts.

Workshop hint — choose your first step

Translate every part-to-part clue into the needed part-to-whole ratio. Name the common base or height before transferring that ratio.

Worked workshop solutions — open after trying

Compare your first different step, then close this panel and try again.

  1. Same height: scale area by the base factor, 20 × 7 ÷ 4 = 35.
  2. Same base: scale area by the height factor, 12 × 8 ÷ 3 = 32.
  3. AB = AE + BE = AE + 5AE = 6AE, so AE : AB = 1 : 6.
  4. AC contains five CD-sized parts; AD contains four. AD : AC = 4 : 5.
  5. COD = AOB = 9. BO : OD = 3 : 1 gives AOD = 3 and BOC = 27. Whole = 9 + 9 + 3 + 27 = 48.
  6. ABC is one third of ABD, so ABD = 42 and rectangle = 84. BFE is one quarter of the rectangle, 21. Target = 84 − 14 − 21 = 49.
  7. BEF is (2/3) × (1/2) × (1/2) = 1/6 of the parallelogram. Whole = 8 × 6 = 48.
  8. DC takes three of the five BC parts. With the shared height from A, ACD = 50 × 3/5 = 30.
Mission 10

Exit ticket

Not completed

Complete all five objective checks. The optional reflection is not part of the score.

Segment-to-Area Ratio Navigator

You have completed this lesson.

You can translate segment ratios, justify shared dimensions, and transfer ratios through equal-area geometry.

Lesson 5.2 · Grade 5 Math Explorer

Teaching notes

This student-facing lesson is self-contained.Its central trapezoid ratio chain follows Chapter 5 Example 1 and Guided Practice 1.The rectangle fraction chain follows Example 6, and the midpoint-and-diagonal parallelogram chain follows Guided Practice 4.Additional practice items are newly written to isolate the same ratio principles.

All figures are redrawn as instructional SVGs and are not intended to be measured as scale drawings.