Perimeter and Area · Lesson 5 of 7
Finding Triangle Areas by Rearranging
“Discover triangle area by cutting rectangles and use triangles to measure more complicated regions.”
• Explain why a rectangle diagonal produces two equal-area triangles. • Connect base and perpendicular height to half a rectangle. • Show why different triangles can have the same area. • Find gridded polygon areas using rectangles and triangles.
A diagonal shares a rectangle equally
A rectangle formula can help us measure triangles. Draw a line from one corner of a rectangle to the opposite corner and cut along it. This diagonal creates two triangles. Turn one over or rotate it and check that it covers the other exactly.
A line segment joining two non-adjacent vertices of a polygon. In a rectangle it joins opposite corners.
The two matching triangles have equal areas. Together they fill the original rectangle without gaps or overlaps, so each has half its area. Try several long and short rectangles, and also a square. Their dimensions change the numerical area, but not the equal sharing.
Problem
A 6 cm by 4 cm rectangle is cut along a diagonal. Find each triangle’s area.
- 1.Rectangle area = 6 × 4 = 24 cm².
- 2.The two triangles match and together cover all 24 cm².
- 3.Each triangle therefore has area 24 ÷ 2 = 12 cm².
- 4.The cut does not remove area; it separates the same region into two equal parts.
Rearranging the two halves
The two triangles can be moved into a new figure. Rearranging does not change their combined area if they meet without gaps or overlaps. A useful arrangement makes one larger triangle with twice the rectangle’s width and the same perpendicular height. It demonstrates why a triangle does not use the full rectangle product.
Choose a triangle side as its base. Its height is the perpendicular distance from the opposite vertex to the line containing that base; perpendicular means meeting at a right angle. In the pictured rearrangement, the original width is half the new base. Its area is therefore half the base multiplied by the height.
The base is a chosen triangle side. The corresponding height measures the straight distance to the opposite vertex at a right angle to the line of the base.
Problem
A triangle has base 8 cm and perpendicular height 5 cm. Find its area.
- 1.A rectangle with the same base and height would have area 8 × 5 = 40 cm².
- 2.The triangle occupies half that amount: 40 ÷ 2 = 20 cm².
- 3.Its area is 20 cm². We used the perpendicular height, not a sloping side length.
A sloping side is usually not the perpendicular height. Height must meet the base line at a right angle. Also remember to take half the base-height product; that full product describes the matching rectangle.
Different-looking triangles with the same area
A triangle does not need to look like half of one whole rectangle to have half its area. We can split it into two smaller triangular halves of two rectangles. Adding those halves connects a new triangle back to the same original rectangle. This explains the red and blue triangles in the grid.
ABCD is 5 units wide and 4 units high, with area 20 square units. The blue triangle BAD is a diagonal half, so it has area 10. Point E lies on the top side and F is directly below it on AB. AF is 3 units and FB is 2 units; both smaller rectangles are 4 units high.
Problem
Find the red triangle ABE in the diagram and compare it with blue triangle BAD.
- 1.Rectangle AFED has area 3 × 4 = 12 square units. Triangle AEF is its diagonal half, with area 6.
- 2.Rectangle FBCE has area 2 × 4 = 8 square units. Triangle BEF is its diagonal half, with area 4.
- 3.AEF and BEF exactly fill ABE, so ABE has area 6 + 4 = 10 square units.
- 4.This equals BAD. Both triangles have base AB = 5 and perpendicular height 4, despite their different shapes.
If E slides along the top side, the two smaller rectangle widths still add to 5. Half of the first rectangle plus half of the second is half of the whole rectangle. Thus the area remains 10 square units. Equal base and equal perpendicular height give equal triangle areas here; matching shape is not required.
Draw a rectangle on grid paper and cut it along a diagonal. Check that the halves match, then rearrange them into the wider triangle shown. Next draw several triangles with the same bottom base and top vertices along one line parallel to that base. Use grid squares or two rectangle halves to check their areas.
Triangles help measure irregular polygons
Sloping edges often leave triangular pieces beside rectangular regions. Draw a helpful dividing line, calculate each simple part, then add them. Another method is to start with an enclosing rectangle and subtract missing triangles. Both work when all pieces together account for the region exactly.
Problem
Find the area of figure a. Each grid square is 1 square unit.
- 1.It fits in a rectangle 4 units wide and 7 units high, whose area is 28.
- 2.Two missing corner triangles each have base 4 and height 1, so each has area ½ × 4 × 1 = 2.
- 3.The polygon area is 28 − 2 − 2 = 24 square units.
Problem
Find the area of figure b.
- 1.Its enclosing rectangle is 4 units wide and 10 units high, with area 40.
- 2.The upper missing triangle has base 4 and height 3, so area 6. The lower missing triangle has base 4 and height 2, so area 4.
- 3.The polygon area is 40 − 6 − 4 = 30 square units.
Problem
Find the area of figure c.
- 1.The enclosing rectangle is 6 units wide and 12 units high, with area 72.
- 2.At the top are two missing right triangles, each with legs 3 and 2, so their combined area is 3 + 3 = 6.
- 3.The lower-left missing triangle has legs 3 and 2, with area 3. The lower-right missing triangle has legs 3 and 10, with area 15.
- 4.Total missing area is 6 + 3 + 15 = 24. Polygon area = 72 − 24 = 48 square units.
Problem
Find the area of figure d.
- 1.The enclosing rectangle is 4 by 5, so its area is 20.
- 2.The top notch is a triangle with base 4 and perpendicular depth 2, so its area is 4.
- 3.Subtract the notch: 20 − 4 = 16 square units.
- 4.You can also split the notch into two right triangles of areas 1 and 3.
Problem
Find the area of figure e.
- 1.Draw a horizontal line joining the leftmost and rightmost vertices. Its length is 4 units.
- 2.The upper triangle has height 2, so area ½ × 4 × 2 = 4.
- 3.The lower triangle has height 4, so area ½ × 4 × 4 = 8.
- 4.Add the two non-overlapping pieces: 4 + 8 = 12 square units.
Quiz
A rectangle of area 30 cm² is cut along a diagonal. What is each triangle’s area?
Which measurement must accompany a chosen base for triangle area?
A triangle has base 10 cm and perpendicular height 6 cm. What is its area?
In the red-and-blue diagram, what is triangle ABE’s area?
Why does moving E along DC leave triangle ABE’s area unchanged?
A polygon fills a 40-unit² rectangle except for triangles of areas 6 and 4. What is its area?
Practice Problems
- Cut an 8 cm by 5 cm rectangle along a diagonal and predict each triangle’s area before checking.
- Find the area of a triangle with base 12 m and perpendicular height 7 m.
- Draw three different triangles with base 6 grid units and height 4 units. Compare their areas.
- In rectangle ABCD of width 8 and height 3, let AF = 5 and FB = 3. Find the two triangular areas and their total.
- Redraw any two of figures a–e, show your dividing lines and verify their areas by a different decomposition.
- Explain why using a sloping side instead of the perpendicular height can give an incorrect triangle area.
Key Takeaways
• A rectangle diagonal makes two matching equal-area triangles. • Triangle area is half the product of base and perpendicular height. • Rearranging pieces without gaps or overlaps preserves area. • Different triangle shapes can have the same base, height and area. • Use rectangle-and-triangle decompositions, or subtract missing triangles, to measure gridded polygons.