Math Quest · Lesson 5.5
Mission progress0 of 10
Chapter 5 · Equal-Area Transformations

Use an Area Ledger: Add, Subtract, and Complete

用面积账本：相加、相减与补全

Complicated shaded regions become manageable when every known piece is recorded, every equality is justified, and the target is written as a clean area equation.

Name the whole. List the pieces. Balance the area budget.
Grade 5 enrichment10 missionsLayered SVG diagrams40–55 minutes
W

Start with a whole

Use the rectangle, parallelogram, trapezoid, half, or quarter whose area is known.

=

Record equal pieces

Parallel lines, diagonals, midpoints, and common-region subtraction can create equal entries.

−

Subtract complements

Remove nonoverlapping known pieces exactly once.

✓

Check the budget

All ledger entries must add back to the chosen whole.

Mission 1

Build an area ledger

Not completed

An area ledger is a list of nonoverlapping regions whose areas make one useful whole. It tells you what to add, what to subtract, and what remains.

All areas on this page are in square units and lengths in units unless cm or cm² is stated. A region name in a ledger entry stands for its area.

1What is the whole?Rectangle, trapezoid, half, quarter…
2What is known?Given areas and easy derived areas
3What is equal?Record why the equality is true
4What is the target?Write it as one area equation

Try the ledger machine

70
18
12
Whole: 70Remove: 18Remove: 12Target: ?
Target area4070 − 18 − 12
Mission 2

Keep useful halves and quarters ready

Not completed

Diagonals often create ready-made ledger entries. A diagonal of a rectangle or parallelogram makes two equal halves. Both diagonals make four equal-area triangles.

Rectangle divided by two diagonals Rectangle ABCD has both diagonals drawn and intersecting at O, creating four equal-area triangles. ABCDO

The rectangle’s area is 24 square units. The drawing is not necessarily to scale.

½
One diagonal
splits the rectangle into two equal-area triangles
24 ÷ 2 = 12
¼
Both diagonals
split it into four equal-area triangles
24 ÷ 4 = 6
Do not overgeneralize: an arbitrary quadrilateral is not guaranteed to be halved by a diagonal. We are using a rectangle or parallelogram.
Mission 3

A rectangle ledger reveals quadrilateral PMON

Not completed

Rectangle ABCD has area 24 square units. Its diagonals AC and BD meet at O. P lies on DC; AP meets BD at M, and BP meets AC at N. The combined area of triangles ADM and BCN is 7.8. We will not chase the shaded quadrilateral directly; we will account for it inside triangle ABP.

Rectangle area ledger with shaded quadrilateral PMON Rectangle ABCD has diagonals AC and DB. Point P is on the top side. Lines AP and PB meet the diagonals at M and N. Quadrilateral PMON is shaded. ABCDPMON ADMBCNPMON

Diagram not necessarily to scale. The points and lines reproduce the worked example relationships.

AOD + BOC=24 ÷ 2 = 12
ADM + BCN=7.8
AOM + NOB=12 − 7.8 = 4.2
Triangle ABP=24 ÷ 2 = 12
Triangle AOB=24 ÷ 4 = 6
Area(PMON) = 12 − 6 − 4.2 = 1.8

The key choice is the whole used for the final subtraction: triangle ABP, not the entire rectangle.

Mission 4

Build a fraction ledger in a parallelogram

Not completed

In parallelogram ABCD, E is the midpoint of BC, F lies on diagonal BD, and BF = 2DF. Given Area(BEF)=8, find the area of the parallelogram.

Parallelogram fraction ledger Parallelogram ABCD has diagonal BD. E is the midpoint of BC. F lies two thirds of the way from B to D. Triangle BEF is shaded. ABCDEF BEF = 8BDEBCD BF : FD = 2 : 1
1
BEF is 2/3 of BDE
They share the height from E to BD, and BF:BD=2:3.
BDE = 8 ÷ 2 × 3 = 12
2
BDE is 1/2 of BCD
E is the midpoint of BC.
BCD = 24
3
BCD is 1/2 of the parallelogram
Diagonal BD halves a parallelogram.
ABCD = 48
BEF = ⅙ of the whole, so 8 × 6 = 48.
Mission 5

Subtract equal remainders in a trapezoid

Not completed

Trapezoid ABCD has area 45, height 6, bottom base BC=10, and Area(AED)=5. The diagonals meet at E. Find Area(BCE).

Trapezoid ledger with equal remainders Trapezoid ABCD has parallel bases AD and BC. Diagonals AC and BD intersect at E, forming triangles AED, ABE, CDE, and BCE. ADBCE AED = 5ABECDEBCE
Ledger entryValueReason
Top base AD545 = ½(AD+10)×6
Triangle ABD15½×5×6
Triangle ACD15same base AD, equal height
ABE and CDE10 each15−5
Target BCE2045−5−10−10
45 = 5 + 10 + 10 + 20
Mission 6

Complete a rectangle, then subtract its complements

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. Area(ABC) = 14 square units. Find the area of shaded quadrilateral BCDE, which remains after removing triangles ABC and BFE.

Rectangle complement ledger Rectangle ABFD contains point C on top side AD and midpoint E on right side DF. The shaded quadrilateral BCDE is the rectangle minus triangles ABC and BFE. ABCDEF ABC = 14BFEshaded BCDE
Triangle ABD3×14 = 42
Rectangle ABFD2×42 = 84
Triangle BFE¼×84 = 21
Shaded area = 84 − 14 − 21 = 49

We complete the rectangle first because the shaded quadrilateral itself has no convenient direct formula from the given lengths.

Mission 7

Choose the ledger equation before calculating

Not completed

A correct area equation prevents double counting. Match each situation to the equation that records every nonoverlapping piece exactly once.

Rectangle PMON

Use half ABP, remove quarter AOB, then remove the remaining pair.

Trapezoid BCE

Remove center AED and two equal remainders from the whole.

Rectangle shaded region

Complete the rectangle, then remove the two complements.

Parallelogram whole

Triangle BEF is one sixth of the whole and has area 8.

Mission 8

Area-ledger error detective

Not completed

Decide whether each statement is valid. Use the ledger structure, not the appearance of the drawing.

Extra practice — put the relationships together

This additional practice combines area relationships with two different perpendicular heights. Work through the steps before the workshop; these extra checks do not add to the ten mission count.

Choose a second base, then use a 45° triangle

Adapted from original Exercise 7

The parallelogram has horizontal base 7 cm and sloping side 5 cm. The dashed 5.6 cm segment is perpendicular to that sloping side. The shaded right triangle has a 45° angle at its top-left vertex. Its vertical leg spans the full perpendicular height h between the parallelogram’s horizontal bases. Find its area.

Parallelogram with two perpendicular heights and a shaded right triangle with a 45 degree angle7 cm5 cm5.6 cmh45°

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

Use the sloping side and its matching perpendicular height to find the whole area. Use the horizontal base to recover h.

Worked solution — compare after trying
  1. With the sloping side as base, the whole area is 5 × 5.6 = 28.
  2. With the horizontal side as base, the same whole area is 7 × h. Thus h = 28 ÷ 7 = 4.
  3. The shaded triangle is right-angled and has a 45° angle. Its other acute angle is 45° too, so its perpendicular legs are equal. Both are h = 4.
  4. Its area is ½ × 4 × 4 = 8 cm². The two height measurements belong to different bases.
Teacher note — why these measurements were adapted

original Exercise 7 prints base 10, sloping side 8, perpendicular distance 9, and a 45° diagonal angle. The first three give vertical height 7.2, but a 45° diagonal then forces the horizontal offset to be 2.8, making the sloping side √(7.2² + 2.8²), not 8. The original measurements cannot all hold in one Euclidean parallelogram. This labeled adaptation uses consistent 3–4–5 geometry while preserving the intended two-base and 45° reasoning.

Mission 9

Area-ledger workshop

Not completed

Solve at least six of the eight. Areas are in square units. Write a ledger equation before doing arithmetic.

1

A rectangle has area 40. Inside a chosen half of area 20 are two non-overlapping pieces with areas 10 and 3. Find the target area left inside that half.

2

In Mission 3’s PMON rectangle, the whole is 24 and the given corner pair is 7.8. Find Area(PMON).

3

Triangle BEF has area 8 and is one sixth of a parallelogram. Find the parallelogram area.

4

A trapezoid has ledger 45 = 5 + 10 + 10 + target. Find the target.

5

A rectangle has area 84. Two nonoverlapping complements have areas 14 and 21. Find the remaining shaded area.

6

Two equal large triangles each have area 18 and share a central triangle of area 4. Find the sum of the two remainders.

7

A rectangle has area 96. Two nonoverlapping complements are one third and one quarter of the rectangle. Find the target remainder.

8

A target triangle is three eighths of a whole figure and has area 15. Find the whole area.

Workshop hint — choose your first step

Choose the useful whole and list each nonoverlapping part exactly once. If a fraction of the whole is known, work backward.

Worked workshop solutions — open after trying

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

  1. Use the half as the ledger whole: 20 − 10 − 3 = 7.
  2. The opposite half is 12. Remove 7.8: remaining pair = 4.2. Within ABP, target = 12 − 6 − 4.2 = 1.8.
  3. One sixth is 8, so the whole is 8 × 6 = 48.
  4. Subtract each piece exactly once: 45 − 5 − 10 − 10 = 20.
  5. The complements do not overlap, so 84 − 14 − 21 = 49.
  6. Each remainder is 18 − 4 = 14. Two remainders give 14 + 14 = 28.
  7. The nonoverlapping complement areas are 96 ÷ 3 = 32 and 96 ÷ 4 = 24. Target = 96 − 32 − 24 = 40.
  8. Three eighths is 15, so one eighth is 15 ÷ 3 = 5. The whole is 5 × 8 = 40.
Mission 10

Exit ticket

Not completed

Answer all five questions. Every scored answer is objectively checked.

Teaching notes

The ledger machine, equation-matching activity, error diagnoses, and additional practice questions are instructional scaffolds created for this page. All diagrams are redrawn and are not necessarily to scale.