△Math Explorer · Grade 5
Lesson progress0 of 10 missions
Chapter 4 · Plane Geometry

Lesson 4.3 — Shared Bases and Shared Heights

第4讲 · 共享底与共享高

Triangles can look completely different and still have equal—or predictably related—areas. The secret is to compare the base and its perpendicular height before doing any arithmetic.

Same height? Compare bases. Same base? Compare heights.
Grade 5 enrichment 45–60 minutes Interactive SVG diagrams Autosaves in this browser
Mission 1

Area comparison starts with one formula

Not completed

For every triangle, the chosen base and its perpendicular height control the area:

Compare Triangle 1 with Triangle 2 in that order. A₁, b₁, and h₁ mean the area, base, and perpendicular height of Triangle 1; subscript 2 names Triangle 2. A ratio such as 3 : 8 means “first : second.”

Area of a triangle = ½ × base × perpendicular height

Same height

The factor ½ × height is the same, so areas change in the same ratio as the bases.

A₁ : A₂ = b₁ : b₂

Same base

The factor ½ × base is the same, so areas change in the same ratio as the heights.

A₁ : A₂ = h₁ : h₂

Same base and height

Both factors match, so the areas are equal—even when the triangles lean in different directions.

A₁ = A₂
Mission 2

Same height: compare the bases

Not completed

The two bases sit on the same lower line, while both top vertices lie on a line parallel to it. That guarantees one common perpendicular height.

Two triangles with a common height and adjustable bases Two triangles lie between the same pair of parallel horizontal lines. Sliders change their base lengths and common height. b₁ = 4 b₂ = 8 h = 6
4
8
6
12Area 1
24Area 2
1 : 2Area ratio
Watch what stays true: changing the common height changes both areas, but the area ratio always matches the base ratio.

These check questions use the measurements written below, regardless of your slider settings. Ratios are first triangle : second triangle.

cm²
Mission 3

Same base: compare the heights

Not completed

These triangles use the same base segment. Only the perpendicular distance from the top vertex to the base line changes.

Two triangles with the same base and adjustable heights Two triangles share one base. Their top vertices have independently adjustable perpendicular heights. shared base = 8 h₁ = 4 h₂ = 7
8
4
7
16Area 1
28Area 2
4 : 7Area ratio
A₁ = ½ × shared base × h₁
A₂ = ½ × shared base × h₂
A₁ : A₂ = h₁ : h₂

These check questions use the measurements written below, regardless of your slider settings. Ratios are first triangle : second triangle.

Mission 4

Slide the vertex without changing the area

Not completed

Keep the base fixed. Move the top vertex anywhere along a line parallel to the base. The triangle leans and changes shape, but its perpendicular height—and therefore its area—does not change.

A triangle whose top vertex slides along a line parallel to its base The base is 10 units and the height is 6 units. Moving the top vertex changes the shape but not the area. base = 10 moving vertex area = 30
center
Invariant: base = 10 and perpendicular height = 6, so every position gives ½ × 10 × 6 = 30 square units.
square units
Mission 5

Three shaded triangles, one base budget

Not completed
worked example

A rectangle 15 units wide and 10 units high contains three shaded triangles that do not overlap. Every shaded triangle reaches from the bottom edge to the top edge, so all three have height 10. Their three bases divide the rectangle’s full 15-unit bottom edge into separate segments with no gaps.

Three triangles inside a 15 by 10 rectangle The bases of three shaded triangles partition the full base of the rectangle. Their common height is 10. 4 5 6 combined base = 15 height = 10
4
5
Base 36
20Triangle 1 area
25Triangle 2 area
30Triangle 3 area
Total shaded area = ½b₁h + ½b₂h + ½b₃h
= ½(b₁ + b₂ + b₃)h
= ½ × 15 × 10
= 75 square units
Move the base sliders. Individual triangle areas change, but the total stays 75 because the base sum stays 15.
units
square units
Mission 6

The shaded triangles fill exactly half the rectangle

Not completed
Exercise 9

The rectangle is 12 cm by 8 cm. The shaded triangles do not overlap. All have perpendicular height 8 cm, and their bases divide the full 12-cm bottom edge into separate segments with no gaps.

Three shaded triangles inside a 12 by 8 rectangle The bases of the three shaded triangles partition the entire 12-centimetre base, and all heights equal the rectangle height of 8 centimetres. base sum = 12 cm 12 cm 8 cm
Rectangle area = 12 × 8 = 96 cm²
Shaded area = ½ × 12 × 8 = 48 cm²
Unshaded area = 96 − 48 = 48 cm²
Conclusion: the shaded and unshaded regions have equal area. The three shaded triangles together occupy exactly half the rectangle.
cm²
cm²
Mission 7

Use the whole perimeter as a combined base

Not completed
Adapted from Exercise 10 · valid 5–12–13 triangle

Join an interior point to the three vertices. The triangle becomes three smaller triangles. Each small triangle uses one side of the large triangle as its base, and the stated perpendicular distance from the point to every side is 2 cm.

P512132 cm2 cm2 cm

This right triangle has side lengths 5, 12, and 13 cm. The marked distances are perpendicular.

5
12
13

Fixed side lengths: 5 cm, 12 cm, and 13 cm. Their sum is 30 cm. Point P is the particular interior point 2 cm from each side; an arbitrary interior point would not have these equal distances.

Area = ½a×2 + ½b×2 + ½c×2
= ½(a+b+c)×2
= ½ × perimeter × 2
= ½ × 30 × 2 = 30 cm²
5Small triangle 1
12Small triangle 2
13Small triangle 3
cm²
Mission 8

Base–height ratio detective

Not completed

Read each ratio as first triangle : second triangle. Choose the relationship before calculating. All six answers can be found from shared bases, shared heights, or equal-area reasoning.

Case A

Same height; bases 4 and 7.

Case B

Same base; heights 9 and 3.

Case C

Same height and equal area. First base is 12.

Case D

Same base; areas 20 and 35.

Case E

Same height; area ratio 2 : 5. First base is 6.

Case F

One triangle changes while its area stays fixed: its base doubles.

Mission 9

Shared-base and shared-height workshop

Not completed

Solve at least 6 of the 8 questions correctly. All lengths below are in cm and areas are in cm². Give ratios in the order the triangles are mentioned. Enter only the number unless a selection box is shown.

1Two triangles have the same height and bases 6 and 10. What is the simplified area ratio?

2Three triangles share height 7. Their bases total 13. Find their total area.

3Two triangles share base 12. Their heights are 5 and 8. Find the larger area minus the smaller area.

4An 18-by-6 rectangle contains triangles whose bases partition the full length 18 and whose height is 6. Find the total triangle area.

5In Mission 6’s 12-by-8 rectangle, compare shaded and unshaded areas.

6A triangle has side lengths 8 cm, 15 cm, and 17 cm. An interior point is at perpendicular distance 3 cm from every side. Find the triangle’s area in cm².

7Two triangles share a height. Triangle 1 has area 24 and base 6. Triangle 2 has base 9. Find Triangle 2’s area.

8Two triangles share a base. Triangle 1 has area 30 and height 5. Triangle 2 has height 8. Find Triangle 2’s area.

Strategy checklist
  1. Choose the base for each triangle.
  2. Identify whether the height or base is shared.
  3. Use a ratio before calculating full areas when possible.
  4. For several common-height triangles, add their bases first.
  5. Check that every base segment is counted exactly once.
Worked workshop solutions — open after attempting

Find the first step that differs from your own work, then try the problem again.

  1. Same height means the area ratio equals 6:10, which simplifies to 3:5.
  2. Combine the bases: 13 × 7 ÷ 2 = 45.5.
  3. Same base 12: use the height difference 8 − 5 = 3. Area difference = 12 × 3 ÷ 2 = 18.
  4. Combined base 18, shared height 6: area = 54.
  5. Both regions have area 48; the shaded triangles use half the rectangle.
  6. Add the three small triangle areas: perimeter × common distance ÷ 2 = 40 × 3 ÷ 2 = 60.
  7. The base grows by a factor of 9 ÷ 6 = 1.5, so area becomes 24 × 1.5 = 36.
  8. The height grows by 8 ÷ 5, so the area becomes 30 × 8 ÷ 5 = 48.
Mission 10

Exit ticket

Not completed

Complete all five checks. The optional reflection is not automatically graded.

square units
cm²

Shared-Area Strategist

You have completed this lesson.

You can compare triangle areas through shared bases, shared heights, base-height ratios, and combined base budgets.

Lesson 4.3 completed

Teaching notes

The original places three shaded triangles in a 15-by-10 rectangle.Their bases together span 15, so the original combines them as 15 × 10 ÷ 2 = 75.

In the 12-by-8 rectangle, the shaded triangle bases span the full length 12, giving shaded area 48; the rectangle area is 96, so the unshaded area is also 48.Mission 7 adapts Exercise 10 using a valid 5–12–13 triangle: perimeter 30, equal distance 2, and area 30.The original instead states perimeter 30 and distance 3, formally giving 45, but no triangle has those measurements.

The same-base, same-height, ratio, and sliding-vertex missions are instructional scaffolds added to make the worked solution explicit and self-contained. A technical teacher note: the specific perimeter-30 and distance-3 measurements in Exercise 10 do not correspond to a realizable Euclidean triangle, although the worked example’s intended formal decomposition yields 45. The lesson uses realizable measurements and perpendicular constructions instead.

All diagrams are explanatory and are not scale drawings.