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

Lesson 5.3 — Extend Sides and Scale Triangle Areas

第5讲 · 延长边与三角形面积倍数

Decode what an extension statement really means, scale one geometric direction at a time, and combine two justified factors without mistaking an added length for the new whole length.

Translate the extension. Scale one direction. Multiply the justified factors.
Grade 5 enrichment 55–70 minutes Interactive extension diagrams Autosaves in this browser
Mission 1

Decode the language of extending a side

Not completed

An extension statement describes the new piece added beyond an endpoint. It does not directly give the length of the complete enlarged segment.

1
A B D
1original AB
1added BD in AB-units
2whole AD in AB-units
Translation rule: if BD = k × AB, then AD = AB + BD = (1 + k) × AB.
“Extend AB by 1 original length” → BD = AB → AD = 2AB
“Extend AC by 2 original lengths” → CE = 2AC → AE = 3AC
Added factor k → whole factor 1 + k
Common trap: extending by two original lengths makes the whole three original lengths, not two.

These questions use the fixed values written below, regardless of your slider settings.

Mission 2

One extension creates one area factor

Not completed

Keep the top vertex fixed and extend the base along the same line. The perpendicular height stays the same, so the triangle’s area changes by exactly the base factor.

2
A triangle whose base is extended Triangle ABC is the original triangle. The base is extended through C to D. Both triangles use the same perpendicular height from A. ABCD 1 base unit2 added units
3×area of enlarged triangle ÷ area of original triangle
BD = BC + CD = 3BC
same perpendicular height from A
Area(ABD) = 3 × Area(ABC)
One base factor changes area once. No angle measurement or similarity argument is needed.

These questions use the fixed values written below, regardless of your slider settings.

Mission 3

Two independent scale factors multiply

Not completed

Scale one direction, then scale the other. Each justified change multiplies the current area, so the combined multiplier is the product—not the sum—of the two factors.

2
3
Original and scaled right triangles The enlarged triangle has a base and height scaled independently from the original triangle. original base ×2 height ×3
2first area factor
3second area factor
6combined area factor
6×2 × 3 = 6
Do not add the factors. A factor of 2 followed by a factor of 3 gives 2 × 3 = 6, not 2 + 3 = 5.

These questions use the fixed values written below, regardless of your slider settings.

Mission 4

Read the worked extension figure correctly

Not completed

Start with triangle ABC. Extend AB through B by one original AB length to D. Extend AC through C by two original AC lengths to E. The target is the area multiplier from triangle ABC to triangle ADE.

Triangle ABC inside extended triangle ADE D lies beyond B with BD equal to AB. E lies beyond C with CE equal to twice AC. ABCDE BD = ABCE = 2AC

Diagram not necessarily to scale. Use the stated equalities, not a ruler.

AD = 2ABone original AB was added
AE = 3ACtwo original AC lengths were added
?area multiplier to prove
Before doing any area reasoning, translate every extension statement into a whole-segment factor.
The diagram’s long sides may not look exactly 2 and 3 times as long. The written relationships control the mathematics.
Mission 5

Connect BE and prove the area factor is 6

Not completed

The extension factors become useful only after we choose area comparisons with a shared height. Reveal the proof one layer at a time.

Layered proof for the extended triangle Triangle ABC, triangle BCE, triangle DBE, and auxiliary segment BE show that triangle ADE is six times triangle ABC. ABCDE 1 unit2 units3 units
1
Scale along AC.
Triangles BCE and BCA share the perpendicular height from B to line AE. Their bases CE and CA are in ratio 2 : 1, so Area(BCE) = 2 × Area(ABC).
2
Combine the top pieces.
Area(ABE) = 1 + 2 = 3 original triangle areas.
3
Use AB = BD.
Triangles ABE and DBE share the height from E to line AD and have equal bases AB and BD, so their areas are equal.
4
Combine the two halves.
Area(ADE) = 3 + 3 = 6 × Area(ABC).
6×the large triangle’s area compared with the original
Mission 6

Guided practice: double the base and double the height

Not completed

Points A, B, and E lie in that order on one straight line. Points B, C, and D lie in that order on another. Given AE = 3AB and BD = 2BC, compare triangle DBE with triangle ABC.

Two triangles on intersecting lines with doubled bases and perpendicular heights BE is twice AB. Point D lies twice as far from B along the ray through C, so its perpendicular distance to line AE is twice that of C. ABCDE AB = 1 partBE = 2 parts h2h

Because C and D lie on the same ray from B, doubling BC doubles the perpendicular distance to line AE.

AE = 3AB → BE = AE − AB = 2AB
BD = 2BC → height from D to AE = 2 × height from C to AE
Area factor = base factor × height factor
Area(DBE) ÷ Area(ABC) = 2 × 2 = 4
Both the base and the corresponding perpendicular height double. That is why the area becomes four times as large.
Another proof using shared heights: Compare ABC with ABD on bases BC and BD, using the same height from A: the area doubles. Then compare ABD with EBD on bases AB and EB, using the same height from D: the area doubles again. Total factor: 2 × 2 = 4.
Mission 7

A second extension pattern also produces a factor of 6

Not completed

Extend AB through B by one original AB length to D. Extend BC through C by two original BC lengths to E. The original triangle is ABC; the enlarged triangle is ADE.

Triangle ABC extended to triangle ADE B is the midpoint of AD. E lies beyond C and CE equals twice BC. ABCDE BD = ABCE = 2BC
1
Extend BC by two lengths.
BE = BC + CE = 3BC.
2
Use the same height from A.
Area(ABE) = 3 × Area(ABC).
3
Use AB = BD.
Triangles ABE and DBE have equal bases AB and BD on the same line AD and share the perpendicular height from E. Therefore Area(DBE) = Area(ABE).
4
Add the two parts.
Area(ADE) = 3 + 3 = 6 original areas.
Mission 8

Extension and scaling error detective

Not completed

Choose the mathematically justified statement in each case.

1 · Extend by two

A, C, and E lie in that order on a straight line. If CE = 2AC, then AE equals:

2 · Combine factors

Area factors 2 and 3 combine to:

3 · A side looks twice as long

May we declare the area doubled from appearance alone?

4 · Similarity

In Mission 5, must the large and small triangles be similar (same angles and proportional sides) for the proof to work?

5 · One-length extension

“Extend AB through B to D by one original AB length” means:

6 · Two pieces

Two pieces cover a whole without gaps or overlap. Each has three times the original triangle’s area. The whole area is:

Mission 9

Extension-factor workshop

Not completed

Reach at least 6 out of 8. Areas are in square units. For an area factor or an expression using ÷, enter a single number: for example, enter 6 for “six times as large.” Translate added lengths into whole factors first.

1An original triangle has area 18. Its base is extended by two original base lengths while the height stays the same. Find the enlarged area.
Hint

The complete base is 3 times the original.

2From one vertex of a triangle, extend both sides outward along their rays. Add one original length to the first side and three original lengths to the second. Find the enlarged triangle’s area factor.
Hint

The whole factors are 2 and 4.

3In Missions 4–5’s figure, BD = AB and CE = 2AC. Find Area(ADE) ÷ Area(ABC).
Hint

The whole factors are 2 and 3.

4In Mission 6’s figure, AE = 3AB and BD = 2BC. Find Area(DBE) ÷ Area(ABC).
Hint

BE = 2AB and the height factor is 2.

5In Mission 7’s second extension pattern, BD = AB and CE = 2BC. Find Area(ADE) ÷ Area(ABC).
Hint

Triangle ABE is 3 originals, and triangle DBE is equal to it.

6An area multiplier is 15. One justified direction factor is 3. Find the other factor.
Hint

15 ÷ 3.

7From one vertex, extend the two triangle sides outward along their rays. Add four original lengths to one side and one original length to the other. Find the total area factor.
Hint

The whole factors are 5 and 2.

8The total area factor is 12. One whole-segment factor is 3. How many original lengths were added in the other direction?
Hint

The other whole factor is 4, so the added factor is one less.

Workshop hint — choose your first step

Write the new whole length as original + added length. Then multiply the justified whole factors.

Worked workshop solutions — open after trying

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

  1. Two added base lengths make three in total. The height stays fixed: 18 × 3 = 54.
  2. The whole direction factors are 1 + 1 = 2 and 1 + 3 = 4. The area factor is 2 × 4 = 8.
  3. AD = 2AB and AE = 3AC. Successive shared-height comparisons give the factor 2 × 3 = 6.
  4. BE = AE − AB = 2AB. BD = 2BC also doubles the perpendicular height to AE. Factor = 2 × 2 = 4.
  5. BE = 3BC, so ABE is three ABC areas. Since AB = BD, ABE and DBE have equal areas. Their sum ADE is six ABC areas.
  6. Each direction multiplies area. The missing factor is 15 ÷ 3 = 5.
  7. Use whole lengths: (1 + 4) × (1 + 1) = 5 × 2 = 10.
  8. The other whole factor is 12 ÷ 3 = 4. One part was already there, so 4 − 1 = 3 original lengths were added.
Mission 10

Exit ticket

Not completed

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

Triangle Area Scale Architect

You have completed this lesson.

You can decode side extensions, justify one-direction area factors, and multiply independent scale factors correctly.

Lesson 5.3 · Grade 5 Math Explorer

Teaching notes

This student-facing lesson is self-contained.The central worked investigation follows Chapter 5 Example 2, where AB is extended by one original length and AC by two original lengths.The guided 4-times comparison follows Guided Practice 2.The second 6-times pattern follows Exercise 5.

The interactive right-triangle scale machine and several practice questions are newly written scaffolds. All diagrams are instructional redrawings and are not intended to be measured as scale drawings.