A diagonal does more than divide a shape. It creates a network of equal areas, ratios, halves, and sums that can carry information from one region to another.
Build the relationships first. Add the whole last.
Parallel lines and shared bases create equal triangles.
:
Ratio link
Triangles sharing one height have areas in the same ratio as their bases.
½
Half-area link
A diagonal divides every parallelogram into two equal-area triangles.
Σ
Sum link
The whole figure is the sum of all nonoverlapping regions.
Mission 1
Meet an area network
Not completed
In trapezoid ABCD, the parallel sides are AD and BC. Its diagonals meet at O, creating four smaller triangles. Each relationship below has a different reason.
All numerical areas on this page are in square units and lengths are in units, unless stated otherwise. Region names inside area equations stand for those regions’ areas.
top regionleft regionbottom regionright region
Diagram not necessarily to scale. The parallel marks, labels, and shared segments—not visual size—justify the relationships.
=
Left–right area equality
Triangles ADB and ADC share base AD and have equal heights from the parallel line BC. Remove their common triangle AOD: Area(AOB) = Area(COD).
:
Same-height ratio
BOC : COD = BO : OD because the two triangles share the height from C to BD.
Σ
Whole-area sum
Area(ABCD) is the sum of AOD, AOB, BOC, and COD.
Important: the four small triangles are not automatically equal. Only the relationships supported by parallel lines, shared bases, or shared heights may be used.
Mission 2
Follow the worked example network: 3, 6, and a hidden 1.5
Not completed
In the trapezoid below, Area(DOC) = 3 and Area(BOC) = 6. We will transfer information around the network until every region is known.
1
Equal opposite regions Area(AOB) = Area(DOC) = 3.
2
Read the diagonal ratio at C BO : OD = 6 : 3 = 2 : 1.
3
Transfer the ratio to vertex A Area(AOB) : Area(AOD) = 2 : 1, so Area(AOD) = 1.5.
4
Add the network 6 + 3 + 3 + 1.5 = 13.5.
Mission 3
Guided network: Area(AOB) = 9 and BO = 3·OD
Not completed
Use the same relationships, but begin with a different kind of information: one area and one segment ratio.
Opposite equality: Area(COD) = Area(AOB)
At A: Area(AOB) : Area(AOD) = BO : OD = 3 : 1
At C: Area(BOC) : Area(COD) = 3 : 1
Whole = 9 + 9 + 3 + 27
Mission 4
Build your own trapezoid network
Not completed
Let the equal left and right triangle areas, Area(AOB) and Area(COD), be y, and let BO : OD = r : 1. The network then determines all four regions.
Area(AOD)3
Area(AOB)6
Area(COD)6
Area(BOC)12
Whole trapezoid27
The drawing is not rescaled when r changes. Use the labels and relationships, not ruler measurements. If an area has a repeating decimal, round your answer to two decimal places.
Mission 5
A parallelogram diagonal transfers an area
Not completed
Triangle ABC has area 56 square units, twice the area of parallelogram DEFC. ED is parallel to AC. Find the area of shaded triangle AED.
1
Parallelogram DEFC has area 56 ÷ 2 = 28.
2
Diagonal EC halves the parallelogram, so Area(DEC) = 14.
3
ED ∥ AC. Triangles AED and CED share base ED and have equal heights.
4
Therefore the shaded triangle AED also has area 14.
Mission 6
Equal large triangles create equal remainders
Not completed
Trapezoid ABCD has AD parallel to BC, and diagonals AC and BD meet at E. Its area is 45 square units, its perpendicular height is 6 units, BC = 10 units, and the top triangle AED has area 5 square units. Find the area of triangle BCE.
1
Use the trapezoid formula: 45 = ½(AD + 10)×6, so AD = 5.
2
Area(ABD) = ½×5×6 = 15. Because AD ∥ BC, Area(ACD) is also 15.
3
Subtract the same central area 5: Area(ABE) = Area(CDE) = 10.
4
Area(BCE) = 45 − 5 − 10 − 10 = 20.
Mission 7
Practice network: 20 and 30 determine the whole
Not completed
In trapezoid ABCD, Area(AOD) = 20 and Area(ABO) = 30. Find the whole trapezoid area.
Area(COD) = Area(AOB) = 30
BO : OD = Area(AOB) : Area(AOD) = 30 : 20 = 3 : 2
Area(BOC) : Area(COD) = 3 : 2
Area(BOC) = 45, so whole area = 20 + 30 + 30 + 45 = 125
Mission 8
Area-network detective
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Decide whether each statement is always justified under the stated conditions.
Mission 9
Area-network workshop
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Enter only the final numerical answer. Lengths are in units and areas in square units. In Questions 1–4 and 7, use trapezoid ABCD with AD parallel to BC and diagonals meeting at O, as in Mission 1. Reach at least 6 of 8.
1
In a trapezoid network, Area(COD)=4 and BO:OD=2:1. Find the whole area.
2
Area(AOD)=5 and Area(AOB)=10. Find the whole trapezoid area.
3
Area(DOC)=3 and Area(BOC)=6. Find the whole area.
4
Area(AOB)=9 and BO=3·OD. Find the whole area.
5
Use Mission 5’s arrangement with new values: triangle ABC has area 72, twice the area of parallelogram DEFC. A diagonal halves the parallelogram, and that half transfers to the shaded triangle. Find the shaded area.
6
Use Mission 6’s trapezoid arrangement with new values: area 54, height 6, bottom base BC = 12, and top triangle AED area 6. Using the equal-remainder method, find the bottom triangle area.
7
Area(AOD)=12 and Area(AOB)=18. Find the whole trapezoid area.
8
A parallelogram has area 40. Its diagonal creates one half, and a shared-base parallel-line transfer makes the shaded triangle equal to that half. Find the shaded area.
Workshop hint — choose your first step
Find the equal opposite side regions first. Transfer a diagonal ratio to the other pair, then add the four areas.
Worked workshop solutions — open after trying
Compare your first different step, then close this panel and try again.
The opposite side regions are both 4. BO : OD = 2 : 1 gives top 2 and bottom 8. Whole = 2 + 4 + 4 + 8 = 18.
Side regions are both 10. The diagonal ratio is 10 : 5 = 2 : 1, so bottom = 20. Whole = 5 + 10 + 10 + 20 = 45.
Side regions are 3 each, and top is 3 ÷ 2 = 1.5. Whole = 6 + 3 + 3 + 1.5 = 13.5.
Side regions are 9 each; top = 9 ÷ 3 = 3 and bottom = 9 × 3 = 27. Sum = 48.
Parallelogram = 72 ÷ 2 = 36. Its diagonal half is 18; the parallel-line transfer preserves that area.
54 = (AD + 12) × 6 ÷ 2 gives AD = 6. ABD and ACD each have area 6 × 6 ÷ 2 = 18. Remove the common top 6: each side is 12. Bottom = 54 − 6 − 12 − 12 = 24.
Side regions are 18 each. BO : OD = 18 : 12 = 3 : 2, so bottom = 18 × 3/2 = 27. Whole = 12 + 18 + 18 + 27 = 75.
The diagonal half has area 40 ÷ 2 = 20. The shared-base, equal-height transfer gives shaded area 20.
Mission 10
Exit ticket
Not completed
Answer all five questions. Areas are in square units. For Questions 1–3, use trapezoid ABCD with AD parallel to BC and diagonals meeting at O, as in Mission 1. Every scored part is objectively checked.
Certificate unlocked
Area-Network Navigator
Lesson 5.4 checkpoints completed
You connected equality, ratio, diagonal, and sum relationships across trapezoids and parallelograms.
Teaching notes
The adjustable network machine, added numerical examples, error-diagnosis items, and objective assessments are instructional scaffolds created for this standalone page.
All diagrams are redrawn and are not necessarily to scale. Children should use only marked parallel lines, shared bases, shared heights, stated ratios, and stated areas.