MathematicsGrade 6–8

Parallelogram, triangle and trapezium by decomposing

Cut off a triangle and put it on the other side: the parallelogram becomes a rectangle. Triangles and trapeziums work the same way.

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What is it about?

You already know the area of a rectangle: length times width. Parallelograms, triangles and trapeziums can be turned into rectangles by cutting, moving and completing – the area does not change.

What matters is the base gg and the matching height hh: the perpendicular distance of the opposite corner (or side) from the base. If its foot lies outside, extend the base.

  • Parallelogram: cut off a triangle along the height and slide it to the other side: a rectangle gg long and hh wide, so A=g⋅hA = g \cdot h. If it is steeply slanted, one cut is not enough – then it is cut into several strips.
  • Triangle: a copy rotated by 180∘180^\circ about the midpoint MM of a side completes it to a parallelogram with the same base and height. The triangle is half of it, so A=g⋅h:2A = g \cdot h : 2.
  • Trapezium with parallel sides aa and cc: the rotated copy gives a parallelogram with base a+ca + c, so A=(a+c)⋅h:2A = (a + c) \cdot h : 2. Alternatively split it along a diagonal into two triangles: a⋅h:2+c⋅h:2a \cdot h : 2 + c \cdot h : 2.

Example: a=7 cma = 7\,\text{cm}, c=3 cmc = 3\,\text{cm}, h=4 cmh = 4\,\text{cm} gives A=(7+3)⋅4:2 cm2=20 cm2A = (7 + 3) \cdot 4 : 2\,\text{cm}^2 = 20\,\text{cm}^2.

One grid square is 1 cm1\,\text{cm} wide, so it is 1 cm21\,\text{cm}^2; “Count the squares” checks every result.

Try it

  1. Press “Cut and move” for this parallelogram. Which piece moves where? Why does it fit exactly?
  2. Then count the squares. Does the number match g⋅hg \cdot h?
  3. Cut up a steeply slanted parallelogram. Where is the foot of the height? Into how many pieces is it cut?
  4. Double the triangle and rearrange the parallelogram into a rectangle. Count the squares: which belong to the triangle, which to the copy – and what about the cut ones?
  5. Press “Slide the top” for the triangle or drag the top CC yourself. What changes about the shape, what about the area?
  6. Compare both ways for the trapezium: rotated copy and two triangles. Do they give the same result?
  7. Set c = 0 cm and c = a for the trapezium. Which shapes appear, and what becomes of the formula?

Tasks

Task 1: Parallelogram

Load task Find the area of the parallelogram. Then cut it up to check.

Show solution

g=5 cmg = 5\,\text{cm}, h=3 cmh = 3\,\text{cm}, so A=5 cm⋅3 cm=15 cm2A = 5\,\text{cm} \cdot 3\,\text{cm} = 15\,\text{cm}^2. The rectangle after moving consists of 1515 squares. Show

Task 2: Triangle

Load task Calculate the area of the triangle.

Show solution

A=8 cm⋅5 cm:2=40 cm2:2=20 cm2A = 8\,\text{cm} \cdot 5\,\text{cm} : 2 = 40\,\text{cm}^2 : 2 = 20\,\text{cm}^2. Two such triangles make a parallelogram of 40 cm240\,\text{cm}^2. Show

Task 3: An obtuse triangle

Load task The corners are A(0,0)A(0, 0), B(4,0)B(4, 0) and C(7,3)C(7, 3). Mia calculates 7⋅3:2=10.57 \cdot 3 : 2 = 10.5. What went wrong?

Show solution

The base is ABAB with g=4 cmg = 4\,\text{cm}, not 7 cm7\,\text{cm}. The height is h=3 cmh = 3\,\text{cm}; its foot (7,0)(7, 0) lies outside the base. So A=4⋅3:2 cm2=6 cm2A = 4 \cdot 3 : 2\,\text{cm}^2 = 6\,\text{cm}^2. Show

Task 4: Trapezium two ways

Load task Calculate the area with the trapezium formula and by splitting it into two triangles.

Show solution

Formula: A=(9+5)⋅4:2 cm2=28 cm2A = (9 + 5) \cdot 4 : 2\,\text{cm}^2 = 28\,\text{cm}^2. Two triangles: 9⋅4:2+5⋅4:2=18+10=289 \cdot 4 : 2 + 5 \cdot 4 : 2 = 18 + 10 = 28, so 28 cm228\,\text{cm}^2. Show

Task 5: Same size, different shape

Find parallelograms with a whole-number base and height (in cm) and an area of 24 cm224\,\text{cm}^2 (base at most 10 cm10\,\text{cm}, height at most 8 cm8\,\text{cm}). How many different parallelograms are there if you may also change the offset?

Show solution

g⋅h=24g \cdot h = 24 with g≤10g \le 10 and h≤8h \le 8: g=3g = 3, h=8h = 8; g=4g = 4, h=6h = 6; g=6g = 6, h=4h = 4; g=8g = 8, h=3h = 3. For each pair there are infinitely many parallelograms: sliding the top side parallel to the base keeps the area at 24 cm224\,\text{cm}^2. Show an example

Task 6: Working backwards

A trapezium has the parallel sides a=8 cma = 8\,\text{cm} and c=4 cmc = 4\,\text{cm} and an area of 30 cm230\,\text{cm}^2. How high is it?

Show solution

(8+4)⋅h:2=30(8 + 4) \cdot h : 2 = 30, so 6⋅h=306 \cdot h = 30 and h=5 cmh = 5\,\text{cm}. Show

Ideas for the classroom

  • Use: grade 6, area of parallelogram, triangle and trapezium derived by cutting and completing. Good as an introduction on the whiteboard: let the class guess first, then cut.
  • Order: parallelogram → rectangle, then triangle → parallelogram, finally the trapezium in two ways. With “Show the area right away” off, the result only appears after cutting or counting.
  • Shear: “Slide the top” shows that only base and height matter. This leads to the question why the slanted side appears in no formula.
  • Model: all corners lie on grid points, the base is horizontal. The height is drawn from the top left corner; for the trapezium hh is the distance between the two parallel sides.
  • Common stumbling blocks:
    • The slanted side is taken as the height – the height is perpendicular to the base.
    • For triangles “:2: 2” is forgotten; the doubled triangle is a reminder.
    • For obtuse triangles or steeply slanted parallelograms the foot of the height lies outside – the base must be extended, but its length stays gg.
    • For the trapezium, aa and cc are multiplied instead of added.