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Cut and add
Split the target into non-overlapping rectangles, triangles, parallelograms, or trapezoids.
whole area = piece 1 + piece 2 + …Lesson 4.2 — Cut Composite Figures into Useful Pieces
第4讲 · 把组合图形分割成有用的部分
A difficult-looking region often becomes easy after one thoughtful line. You will learn to split figures into non-overlapping pieces, add their areas, or surround a target and subtract what you do not need.
Draw one helpful line. Turn the unfamiliar into familiar figures.
Mission 1
Not completed
Meet the composite-figure toolbox
A composite figure is made from two or more familiar figures. The goal is not to invent a new formula. It is to transform the region into pieces whose areas you already know how to find.
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Surround and subtract
Place the target inside a familiar outer figure, then remove the parts that are not wanted.
target = outer area − unwanted area╱
Draw a diagonal
Connect two nonadjacent vertices so an irregular polygon becomes triangles or other familiar figures.
quadrilateral = triangle + triangle＋
Complete a figure
Add a temporary piece to create a rectangle or triangle, then account for the piece you added.
target = completed figure − added pieceMission 2
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A useful cut creates solvable pieces
There may be more than one good cut. A useful cut should create non-overlapping familiar figures whose dimensions are known or can be found.
The drawing is explanatory and not a scale drawing.
Vertical cut: the shape becomes a 7 × 8 rectangle and a 5 × 4 rectangle.
Vertical method: 7 × 8 + 5 × 4 = 56 + 20 = 76
Horizontal method: 12 × 4 + 7 × 4 = 48 + 28 = 76
Surround-and-subtract check: 12 × 8 − 5 × 4 = 96 − 20 = 76
Three correct methods agree. The diagonal is not impossible, but it creates awkward pieces with missing heights, so it is not the most useful choice here.
square units
Mission 3
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Draw the diagonal that unlocks an irregular quadrilateral
Quadrilateral AFCE is not a standard shape with one ready-made formula. The diagonal AC splits it into two triangles: △AFC and △CEA.
Choose a line. For this split, choose an interior diagonal that creates solvable pieces.
1
Look for nonadjacent verticesA and C are not joined by a side.
2
Draw ACThe quadrilateral becomes two triangles.
3
Plan the sumArea(AFCE) = Area(AFC) + Area(CEA).
Mission 4
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Use an area ledger to add the two triangles
The figure below supplies two base–height pairs, with all lengths in centimetres. The perpendicular heights land on extensions of the triangle bases, but they are still valid heights.
AB is perpendicular to the line containing FC. CD is perpendicular to the line containing AE.
| Piece | Base | Height | Area |
|---|---|---|---|
| △AFC | FC = 2 | AB = 6 | cm² |
| △CEA | AE = 5 | CD = 4 | cm² |
| Quadrilateral AFCE | cm² | ||
△AFC: ½ × FC × AB = ½ × 2 × 6
△CEA: ½ × AE × CD = ½ × 5 × 4
The two triangles do not overlap, so their areas are added.
Mission 5
Not completed
Surround and subtract: three nested squares
The outer ring and the tiny center square are shaded. Subtracting the entire middle square removes the center too, so the center must be added back once.
shadedunshaded
25outer area
9middle area removed
1center added back
17shaded area
Why add the center back? Because the subtraction of the 3 × 3 square removed every point inside it, including the shaded 1 × 1 center.
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Mission 6
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Subtract three corner triangles from a square
The target △BFE is surrounded by the square and three corner triangles. These four regions exactly fill the square without overlap.
Top-left corner: △ABE = ½ × 9 × 4 = 18
Right corner: FC = 9 − 2 = 7, so △BCF = ½ × 9 × 7 = 31.5
Bottom-left corner: DE = 9 − 4 = 5, so △DEF = ½ × 5 × 2 = 5
The subtraction works because the target and the three corner triangles form an exact partition of the square.
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Mission 7
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Perpendicular diagonals create four right triangles
When a quadrilateral’s two diagonals meet inside it at 90°, they divide it into four right triangles. Grouping their four areas produces a compact formula.
Use the sliders to change AO and BO in cm. CO = 4 − AO and DO = 5 − BO. The labels and areas update; the sketch stays fixed and is not to scale.
2.50area AOB
2.50area AOD
2.50area COD
2.50area COB
Four triangles: ½AO·BO + ½AO·DO + ½CO·DO + ½CO·BO
Factor by grouping: ½(AO + CO)(BO + DO)
Therefore: area = ½ × AC × BD
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Mission 8
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Strategy detective
Choose the most efficient first move. More than one method may sometimes work, but select the method that uses the given measurements most directly.
Case A · L-shaped floor
All horizontal and vertical lengths are known.
Case B · Square frame
An inner square opening is centered inside a larger square.
Case C · Irregular quadrilateral AFCE
Use the AFCE figure in Mission 4, with its two given base–height pairs.
Case D · Triangle inside a square
The three surrounding corner triangles have easy dimensions.
Case E · Perpendicular diagonals
Both diagonal lengths are known.
Case F · Two non-overlapping triangles
The base and height of each triangle are known.
Mission 9
Not completed
Composite-figure workshop
Solve at least 6 of the 8 questions correctly. All lengths below are in centimetres; give areas in cm². Enter only the number unless a selection box is shown.
1A 12 × 8 rectangle has a 5 × 4 corner removed. Find the remaining area.
2In quadrilateral AFCE, △AFC has base 2 and height 6; △CEA has base 5 and height 4. Find the quadrilateral area.
3Centered squares have side lengths 7, 5, and 1. The outer ring and center are shaded. Find the shaded area.
4In Mission 6’s square ABCD, side = 9 cm, E lies on AD with AE = 4 cm, and F lies on DC with DF = 2 cm. Find the area of △BFE.
5A quadrilateral has perpendicular diagonals 12 cm and 7 cm. Find its area.
6An irregular quadrilateral is split into two non-overlapping triangles. One has base 4 cm and perpendicular height 5 cm; the other has base 6 cm and perpendicular height 3 cm. Find the total area in cm².
7What is a segment joining two nonadjacent vertices of a polygon called?
8A 14 × 9 rectangle has a 4 × 3 rectangular notch removed. Find the remaining area.
Strategy checklist
- Identify the target region.
- Choose a cut or outer figure that creates familiar shapes.
- Make sure pieces do not overlap when you add them.
- When subtracting, account for every removed part exactly once.
- Write an area ledger before doing arithmetic.
Try the diagonal method with new measurements
Guided Practice 3
In ABCD, ∠B and ∠D are right angles. E lies on AD and F on BC. AE = 5 cm, AB = 10 cm, FC = 12 cm, and DC = 15 cm. Find the area of AFCE.
Use the marked dimensions and relationships; do not measure the drawing.
Hint — choose a first step
Use diagonal AC. For each new triangle, choose the labeled base and the height perpendicular to its line.
Worked solution — compare after trying
- △AFC has base FC = 12 and height AB = 10, so its area is 60.
- △CEA has base AE = 5 and height CD = 15, so its area is 37.5.
- The triangles share only their diagonal. Add: 60 + 37.5 = 97.5 cm².
Use a 45° angle to complete a triangle
Exercise 6
In ABCD, ∠B = ∠D = 90° and ∠BCD = 45°. BC = 7 cm and AD = 3 cm. Extend CD to meet the upward extension of AB at T, as shown. Find the area of ABCD.
Use the marked dimensions and relationships; do not measure the drawing.
Hint — choose a first step
A right triangle with a 45° angle has equal legs: its other acute angle is also 45°. Apply this to BCT and ADT.
Worked solution — compare after trying
- In right triangle BCT, angle C = 45°, so BT = BC = 7. Its area is 7×7÷2 = 24.5.
- Angle T is 45°. Triangle ADT is right at D, so DT = AD = 3. Its area is 3×3÷2 = 4.5.
- Remove the small triangle from the large one: 24.5 − 4.5 = 20 cm².
Find a shaded area in a stepped figure
Exercise 8
The outline has only horizontal and vertical sides. FG = DE = 4 cm, FE = 4 cm, HG = 1 cm, DC = 2 cm, and BC = 2 cm. The shaded region is AEDC. Find its area.
Use the marked dimensions and relationships; do not measure the drawing.
Hint — choose a first step
Draw AD. For ADE use the vertical base DE; for ADC use the horizontal base DC.
Worked solution — compare after trying
- The horizontal distance from A to DE is 1 + 4 = 5. Area ADE = ½ × 4 × 5 = 10.
- DC is 2 above A, and DC = 2. Area ADC = ½ × 2 × 2 = 2.
- Add the two pieces: 10 + 2 = 12 cm².
Worked workshop solutions — open after attempting
Find the first step that differs from your own work, then try the problem again.
- Surround and subtract: 12 × 8 − 5 × 4 = 76.
- Add the two triangle areas: 2 × 6 ÷ 2 + 5 × 4 ÷ 2 = 16.
- Remove the middle square and restore the center: 49 − 25 + 1 = 25.
- The square area is 81. Corner areas are 18, 31.5, and 5; 81 − 18 − 31.5 − 5 = 26.5.
- The right angles between the diagonals justify 12 × 7 ÷ 2 = 42.
- The pieces do not overlap: 4 × 5 ÷ 2 + 6 × 3 ÷ 2 = 19.
- A diagonal connects nonadjacent vertices.
- Outer area 14 × 9 = 126; remove 4 × 3 = 12. Remaining area = 114.
Mission 10
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Exit ticket
Complete all five checks. The optional reflection is not automatically graded.
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Composite-Figure Architect
You have completed this lesson.
You can choose useful auxiliary lines, split figures into non-overlapping pieces, add known areas, and subtract unwanted regions carefully.
Lesson 4.2 completed
Teaching notes
The original draws diagonal AC in quadrilateral AFCE, uses FC = 2 and external height AB = 6 to find area 6, and uses AE = 5 and external height CD = 4 to find area 10.The quadrilateral area is therefore 16 square centimetres.
Mission 7 derives the perpendicular-diagonal result from the exercise on the same page with AC = 4 and BD = 5.
Additional guided investigations cover Practice 3 and Exercises 6 and 8, with their diagrams and worked solutions. The L-shaped introduction, dynamic diagonal model, added numerical practice, feedback, and exit ticket are instructional scaffolds created to make the lesson self-contained. All diagrams are explanatory and are not scale drawings.