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Area of parallelograms and trapezoids

A shape leans over, so it is no longer a rectangle. How much surface does it cover, and which lengths do you need to find out?

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Why this exists

A rectangle’s area is easy: length times width, because rows of unit squares fill it exactly. But plenty of real shapes lean. The lines of a slanted parking space make a parallelogram. The side of a ski ramp is wider at the bottom than at the top. A pane of stained glass is often cut on a slant so the pieces fit together.

Counting squares no longer works cleanly on a leaning shape, because its slanted edges cut through the squares. What you want is a way to turn a leaning shape back into a rectangle without changing how much surface it covers. That turns out to be easy, and it shows exactly which measurements matter. The surprise is that the slanted sides, the lengths that are easiest to measure with a ruler, are not the ones you need.

This lesson runs over two days. Day 1 is the parallelogram and what “height” means. Day 2 is the trapezoid, whose formula comes from the parallelogram’s.

Definition

A parallelogram is a four-sided shape whose opposite sides are parallel: two pairs of parallel sides. Its opposite sides are also equal in length (properties of parallelograms). A rectangle is a parallelogram whose corners are all right angles.

A trapezoid is a four-sided shape with at least one pair of parallel sides (trapezoids). The two parallel sides are its bases, written b_1 and b_2. The other two sides are its legs. Some books say “exactly one pair” instead; with the “at least one” wording used here, every parallelogram is also a trapezoid, and the formulas below agree about it.

For area, two words carry everything:

  • The base b of a parallelogram is whichever side you choose to measure along. In a trapezoid the bases are always the two parallel sides.
  • The height h is the perpendicular distance between the base and the side parallel to it: the length of a segment that crosses from one to the other meeting both at right angles. It is measured straight across, never along a slanted side.
b
h
b_1
b_2
h
Figure 1The height is the dashed distance straight across, at a right angle to the base. It is not the slanted side.

The height does not have to be drawn from a corner, or even inside the shape. Any segment that meets both parallel lines at right angles has the same length, because parallel lines stay the same distance apart all the way along. That is what “parallel” guarantees.

Intuition

Take a neat stack of paper, a rectangle seen from the side. Push the top of the stack sideways so it leans. The side view is now a parallelogram, but every sheet is still there and every sheet is still the same length. The stack’s base has not changed and neither has its height, measured straight up. So the leaning stack covers the same area as the upright one: base times height.

Notice what did change: the slanted side got longer as the stack leaned. That length tells you how far the stack leans, not how much paper is in it. This is why the height, and not the slanted side, belongs in the formula.

A trapezoid is a parallelogram’s close cousin. Its top and bottom are parallel but different lengths. If you swap the two lengths for their average, the middle width, you get a rectangle of the same area. The formula on Day 2 says exactly that.

Results

Area of a parallelogram. For a base b and the height h that goes with it,

A = bh

Area of a trapezoid. For parallel sides b_1 and b_2 and the height h between them,

A = \tfrac{1}{2}(b_1 + b_2)\,h

The height in both formulas is the perpendicular distance. The length of a slanted side never appears. Lengths must be in the same unit, and the area comes out in square units of that unit.

Why it is true

The parallelogram: cut and slide. In Figure 2, drop the dashed height from a top corner to the base. It cuts a shaded piece off the left end. Slide that piece along the base to the right end.

b
h
b
h
Figure 2Cut the shaded piece off one end and slide it to the other. Nothing is lost and nothing is added.

It fits exactly. The piece’s slanted edge was the parallelogram’s left side, and the gap at the right end is bounded by the right side, which is parallel to the left side and the same length. So the piece slots into the gap edge against edge. The new shape has square corners where the dashed cut was, so it is a rectangle. Its bottom is still b long, since the piece removed from one end is added back at the other, and its height is still h. The rectangle’s area is b \times h, and the parallelogram is made of the same pieces, so its area is bh too.

Sometimes a parallelogram leans so far that the height dropped from the top left corner lands past the right end of the base, outside the shape, so there is no piece to cut off. The same idea still works by adding and taking away. Under the overhanging top right end there is an empty right-angled triangle; fill it in, so the shape gets a square corner there. Then cut along the height from the top left corner, which takes away a right-angled triangle at the left end. The two triangles are the same size and shape (each is as wide as the overhang and h tall), so the area has not changed, and what is left is a rectangle b long and h tall. However far it leans, a parallelogram covers the same area as the rectangle with its base and height.

The trapezoid: two copies. Take a trapezoid with parallel sides b_1 and b_2 and height h. Make an exact copy, turn it upside down, and place it beside the first so that its copy of the right-hand slanted side lies exactly against the original’s right-hand slanted side.

b_1
b_2
b_2
b_1
h
Figure 3Two copies of a trapezoid, one turned upside down, always make a parallelogram.

Turning the copy over swaps its top and bottom, so along the bottom of the new shape you meet b_1 then b_2, and along the top b_2 then b_1. Both rows are straight, because each base of the copy lies on the same line as a base of the original, and both have length b_1 + b_2. The top and bottom are parallel and equal, and so are the two outer slanted sides, so the new shape is a parallelogram with base b_1 + b_2 and height h. Its area is (b_1 + b_2)h. It is made of two equal trapezoids, so one trapezoid has half of that:

A = \tfrac{1}{2}(b_1 + b_2)\,h

Read the formula as \displaystyle \frac{b_1 + b_2}{2} \times h and it says what the Intuition promised: the average of the two bases, times the height. The same two-copies idea, used on a triangle, is how the area of a triangle is found.

Worked examples

Example 1

Day 1: a slanted parking space

Aziz paints the lines for slanted parking spaces. Each space is a parallelogram (Figure 4). Its front edge, along the aisle, is 3 m long. Its slanted side lines are 5.5 m long. The distance straight across from the aisle line to the curb, at a right angle to both, is 5 m. How much ground does one space cover?

3\text{ m}
5.5\text{ m}
5\text{ m}
Figure 4One slanted parking space. Only two of the three lengths belong in the area.
Solution

Take the front edge as the base, b = 3 m. The height must be measured at a right angle to that base, and the 5 m distance is exactly that, so h = 5 m. The 5.5 m slanted side is the length of a line Aziz paints, but it is not the distance across:

A = bh = 3 \times 5 = 15

One space covers 15\text{ m}^2.

Using the slanted side would give 3 \times 5.5 = 16.5, which is too much. The slanted side is longer than the distance across because it leans, and leaning adds length to the side without adding any ground inside.

Example 2

Day 1: two sides, two heights

A pane of stained glass that Eira is cutting is a parallelogram with sides 10 cm and 8 cm (Figure 5). The height to the 10 cm side is 6 cm, and the height to the 8 cm side is 7.5 cm. Find the area using each side as the base.

10\text{ cm}
8\text{ cm}
6\text{ cm}
10\text{ cm}
8\text{ cm}
7.5\text{ cm}
Figure 5Each side has its own height. Either pair gives the area.
Solution

With the 10 cm side as base, the matching height is the distance between the two 10 cm sides, 6 cm:

A = 10 \times 6 = 60

With the 8 cm side as base, the matching height is the distance between the two 8 cm sides, 7.5 cm:

A = 8 \times 7.5 = 60

The pane has area 60\text{ cm}^2 either way, as it must, since it is one pane. What would be wrong is pairing a side with the other side’s height: 10 \times 7.5 = 75 multiplies the 10 cm side by a distance measured at right angles to a different side, and it describes no rectangle the pane can be rearranged into.

Example 3

Day 1: fractional lengths

Belen cuts a small parallelogram of glass with base 4\tfrac{1}{2} in and height 2\tfrac{2}{3} in. What is its area?

Solution

The formula is the same whatever kind of number the lengths are. Write each mixed number as a fraction, then multiply the fractions:

A = 4\tfrac{1}{2} \times 2\tfrac{2}{3} = \frac{9}{2} \times \frac{8}{3} = \frac{72}{6} = 12

The piece has area 12\text{ in}^2.

Check. The answer should sit between 4 \times 2 = 8 and 5 \times 3 = 15, since each length lies between those whole numbers. It does.

Example 4

Day 1: finding a missing height

Briar has a parallelogram pane with area 63\text{ in}^2 and base 9 in. How far apart are the base and the side opposite it? Could the pane’s slanted side be 6 in long?

Solution

The formula A = bh ties the three numbers together, so put in the two that are known:

\begin{aligned} 63 &= 9h \\ h &= 63 \div 9 = 7 \end{aligned}

Dividing both sides by 9 undoes the multiplication by 9. The base and its opposite side are 7 in apart.

The slanted side cannot be 6 in. It runs from the base to the opposite side, so it has to cover at least that 7 in distance, and a slanted path between two parallel lines is longer than the straight-across one. The slanted side must be at least 7 in, and it is exactly 7 in only if it stands straight up, which makes the pane a rectangle.

Example 5

Day 2: the side of a ski ramp

Yarrow is building a small ski ramp and needs plywood for one side of it. The side is a trapezoid (Figure 6): 6 m along the snow, 2 m along the flat top, with slanted edges of 2.5 m, and the top is 1.5 m above the snow. Find its area.

6\text{ m}
2\text{ m}
1.5\text{ m}
2.5\text{ m}
2.5\text{ m}
Figure 6The side of Yarrow’s ramp. The top and bottom are parallel.
Solution

The top and bottom are parallel, so they are the bases: b_1 = 6 and b_2 = 2. The height is the straight-up distance between them, h = 1.5. The 2.5 m slanted edges are not needed:

A = \tfrac{1}{2}(6 + 2)(1.5) = \tfrac{1}{2}(8)(1.5) = 4 \times 1.5 = 6

The side needs 6\text{ m}^2 of plywood.

Check by cutting. Two dashed heights, one from each end of the top, split the side into a middle rectangle and two right-angled end pieces. The middle rectangle is 2 by 1.5, area 3. The bottom has 6 - 2 = 4 m left over, and the ramp is symmetric, so each end piece is 2 m along the snow and 1.5 m tall. Turn one end piece over and lay it against the other and the two make a 2 by 1.5 rectangle, area 3. The total is 3 + 3 = 6.

Example 6

Day 2: a trapezoid standing on its side

Eira’s window panel is a trapezoid whose parallel sides stand upright (Figure 7): 18.5 cm on the left and 11.5 cm on the right. The bottom edge is 9 cm long and meets both upright sides at right angles. Find the area of the panel.

18.5\text{ cm}
11.5\text{ cm}
9\text{ cm}
Figure 7The parallel sides stand upright here, so the height runs across the bottom.
Solution

The bases are the parallel sides, wherever they sit. Here they are upright, so b_1 = 18.5 and b_2 = 11.5. The height is the distance between them, measured at right angles to both, which is across, not up. The bottom edge meets both at right angles, so it is that distance: h = 9.

A = \tfrac{1}{2}(18.5 + 11.5)(9) = \tfrac{1}{2}(30)(9) = 15 \times 9 = 135

The panel has area 135\text{ cm}^2.

Here the “height” is a horizontal length. The word means the distance between the bases, whichever way the shape is turned.

Example 7

Day 2: working backwards to the height

Next to a parking lot, Aziz is laying turf on a strip of ground shaped like a trapezoid. The two parallel edges are 10 m and 14 m long, and the strip has area 96\text{ m}^2. How far apart are the parallel edges?

Solution

Put what is known into the formula and simplify the part that is all numbers:

\begin{aligned} 96 &= \tfrac{1}{2}(10 + 14)\,h \\ 96 &= \tfrac{1}{2}(24)\,h \\ 96 &= 12h \\ h &= 96 \div 12 = 8 \end{aligned}

The parallel edges are 8 m apart.

The line 96 = 12h has a meaning of its own. The average of the two edges is 12 m, so the strip covers the same area as a rectangle 12 m wide, and a 12 m wide rectangle with area 96\text{ m}^2 is 8 m long.

Check. \tfrac{1}{2}(10 + 14)(8) = 12 \times 8 = 96.

Boundaries

The height must be perpendicular. A slanted side is longer than the distance it crosses, so putting it in the formula gives too large an area. The only exception is a side that already meets the base at a right angle, as in a rectangle or in Example 6.

Base and height must belong together. In a parallelogram the height is measured at right angles to the chosen base, between that base and its opposite side. A height measured from a different side gives a wrong product (Example 2).

The trapezoid formula needs the parallel sides. The two lengths added are the parallel sides, never the legs. A four-sided shape with no parallel sides is neither a parallelogram nor a trapezoid, and neither formula applies; cut it into pieces you know instead, as in finding the area of any polygon.

One unit throughout. A base in meters and a height in centimeters must be put in the same unit before multiplying. The area is in square units, such as \text{m}^2 or \text{cm}^2, and changing those is its own job: converting square units.

Side lengths alone do not fix the area. A parallelogram with sides 10 and 8 can lean a little or a lot, and its area changes as it leans: 80 when it stands upright as a rectangle, and less and less the further it leans. Without a height or some other measurement, the area cannot be found.

Where this goes next

The parallelogram formula is the key to the next shapes. A triangle is half of a parallelogram, and that gives the area of a triangle as half of base times height. With rectangles, parallelograms, trapezoids and triangles in hand, any shape with straight sides can be measured by cutting it into those pieces, and shapes with holes or cut-outs by adding and subtracting areas.

The cut-and-rearrange idea goes further than straight sides. A circle cut into thin wedges and laid side by side makes something very close to a parallelogram, and that is how the area of a circle is found. The stack of paper from the Intuition grows into Cavalieri’s principle: two solids built from slices of matching size, slice for slice, have the same volume, however the slices are pushed sideways.

Exercises

Day 1: area of a parallelogram

  1. A parallelogram has base 11 m and height 4 m. Find its area.
  2. A parallelogram has base 15 cm, a slanted side of 7 cm, and height 6 cm to the 15 cm side. Find its area.
  3. A parallelogram has base 2.5 m and height 1.2 m. Find its area.
  4. A parallelogram has base 3\tfrac{3}{4} in and height 2\tfrac{2}{5} in. Find its area.
  5. A parallelogram on a coordinate grid has corners (2, 1), (9, 1), (11, 6) and (4, 6). Find its base, its height and its area.
  6. Aziz paints a row of 8 slanted parking spaces. Each is a parallelogram 2.75 m along the aisle and 4.8 m straight across from the aisle line to the curb. How much ground do the 8 spaces cover?
  7. A parallelogram has sides 20 cm and 12 cm. The height to the 20 cm side is 9 cm. Find the area, then find the height to the 12 cm side.

Day 1: working backwards

  1. A parallelogram has area 91\text{ ft}^2 and height 7 ft. How long is its base?
  2. A parallelogram has area 37.5\text{ cm}^2 and base 7.5 cm. Find its height.
  3. Eira cuts 6 identical parallelogram pieces of glass, each with base 5 cm and height 3.5 cm. How much glass do the pieces use in total?

Day 2: area of a trapezoid

  1. A trapezoid has parallel sides 5 m and 9 m, and height 6 m. Find its area.
  2. A trapezoid has parallel sides 4 cm and 10 cm, legs of 5 cm each, and height 4 cm. Find its area.
  3. A trapezoid has parallel sides 2.4 m and 3.6 m, and height 1.8 m. Find its area.
  4. A trapezoid has parallel sides 2\tfrac{1}{2} ft and 4\tfrac{1}{4} ft, and height 3 ft. Find its area.
  5. Briar’s ski ramp has trapezoid sides: 9 ft along the snow, 4 ft along the top, with the top 2\tfrac{1}{2} ft above the snow. How much plywood covers both sides of the ramp?
  6. A trapezoid on a coordinate grid has corners (1, 1), (1, 7), (6, 5) and (6, 2). Which sides are parallel? Find its area.
  7. A trapezoid has area 60\text{ m}^2 and parallel sides 7 m and 13 m. Find its height.
  8. Belen’s parking lot is a trapezoid whose parallel sides are 40 m and 64 m, and they are 25 m apart. Paving costs 12 dollars per square meter. Find the area and the cost of paving it.

Reasoning

  1. Explain why a parallelogram with sides 6 cm and 10 cm cannot have an area of 70\text{ cm}^2.
  2. A rectangular frame 8 in by 5 in has hinged corners. It is pushed over until the long sides are 3 in apart. Find the area inside the frame before and after, and explain why it changed when no side changed length.
  3. Show that the trapezoid formula gives bh for a parallelogram, when the two parallel sides both have length b.
  4. Two slanted parking spaces are painted between the same two parallel lines, the aisle line and the curb, and both are 3.2 m along the aisle, but one leans much more than the other. Explain why they cover the same area.
  5. A trapezoid has parallel sides 6 cm and 10 cm and height 3 cm. Find its area, and the width of a rectangle 3 cm tall with the same area.
  6. In the two-copies argument for the trapezoid, explain why the copy has to be turned upside down, and why the trapezoid’s area is half the parallelogram’s.
  7. A trapezoid has parallel sides 3 m and 5 m and height 2 m. Find its area. Then double every length and find the area again. How many times larger is it, and why?
Answers
  1. 44\text{ m}^2. A = 11 \times 4 = 44.
  2. 90\text{ cm}^2. A = 15 \times 6 = 90. The 7 cm slanted side is not needed.
  3. 3\text{ m}^2. A = 2.5 \times 1.2 = 3.
  4. 9\text{ in}^2. \displaystyle \frac{15}{4} \times \frac{12}{5} = \frac{180}{20} = 9.
  5. Base 7, height 5, area 35 square units. The bottom runs from x = 2 to x = 9, so it is 9 - 2 = 7 long. The top and bottom are horizontal, so the height is the vertical distance 6 - 1 = 5, and A = 7 \times 5 = 35.
  6. 105.6\text{ m}^2. One space is 2.75 \times 4.8 = 13.2\text{ m}^2, and 8 \times 13.2 = 105.6.
  7. Area 180\text{ cm}^2; height to the 12 cm side 15 cm. A = 20 \times 9 = 180. With the 12 cm side as base, 180 = 12h, so h = 180 \div 12 = 15.
  8. 13 ft. 91 = b \times 7, so b = 91 \div 7 = 13.
  9. 5 cm. 37.5 = 7.5h, so h = 37.5 \div 7.5 = 5.
  10. 105\text{ cm}^2. Each piece is 5 \times 3.5 = 17.5\text{ cm}^2, and 6 \times 17.5 = 105.
  11. 42\text{ m}^2. \tfrac{1}{2}(5 + 9)(6) = 7 \times 6 = 42.
  12. 28\text{ cm}^2. \tfrac{1}{2}(4 + 10)(4) = 7 \times 4 = 28. The 5 cm legs are not needed.
  13. 5.4\text{ m}^2. \tfrac{1}{2}(2.4 + 3.6)(1.8) = 3 \times 1.8 = 5.4.
  14. 10\tfrac{1}{8}\text{ ft}^2. The bases add to \displaystyle 6\tfrac{3}{4} = \frac{27}{4}, and \displaystyle \tfrac{1}{2} \times \frac{27}{4} \times 3 = \frac{81}{8} = 10\tfrac{1}{8}.
  15. 32.5\text{ ft}^2. One side is \tfrac{1}{2}(9 + 4)(2.5) = 6.5 \times 2.5 = 16.25\text{ ft}^2, and two sides need 2 \times 16.25 = 32.5.
  16. The two upright sides, on x = 1 and x = 6, are parallel; area 22.5 square units. Their lengths are 7 - 1 = 6 and 5 - 2 = 3, and the distance between them is 6 - 1 = 5, so A = \tfrac{1}{2}(6 + 3)(5) = 22.5.
  17. 6 m. 60 = \tfrac{1}{2}(7 + 13)h = 10h, so h = 60 \div 10 = 6.
  18. Area 1{,}300\text{ m}^2; cost 15{,}600 dollars. \tfrac{1}{2}(40 + 64)(25) = 52 \times 25 = 1{,}300, and 1{,}300 \times 12 = 15{,}600.
  19. Take the 10 cm side as the base. The height to it is the distance across to the opposite side, and the 6 cm slanted side spans that distance, so the height is at most 6 cm. The area is at most 10 \times 6 = 60\text{ cm}^2, which is less than 70.
  20. Before, 40\text{ in}^2; after, 24\text{ in}^2. As a rectangle, 8 \times 5 = 40. Pushed over, the base is still 8 but the height is now 3, so 8 \times 3 = 24. The sides kept their lengths, but area depends on the distance between the sides, and leaning shrinks it.
  21. With b_1 = b_2 = b, \tfrac{1}{2}(b + b)h = \tfrac{1}{2}(2b)h = bh, the parallelogram formula.
  22. Both have base 3.2 m. Both heights are the distance between the aisle line and the curb, and parallel lines are the same distance apart everywhere, so the heights are equal. Equal base and equal height give equal area; the lean changes only the slanted sides.
  23. Area 24\text{ cm}^2; width 8 cm. \tfrac{1}{2}(6 + 10)(3) = 8 \times 3 = 24, and a rectangle 3 cm tall needs width 24 \div 3 = 8, the average of 6 and 10.
  24. Turned over, the copy’s long base sits on top and its short base below, so the bottom row is b_1 + b_2 and the top row is b_2 + b_1: equal and parallel, which makes a parallelogram. Without the turn, the two short bases would sit together on top and the long ones below, and (unless the two bases are equal) the top and bottom would be different lengths, so the shape would not be a parallelogram. The parallelogram is exactly two equal trapezoids, so one trapezoid is half its area.
  25. 8\text{ m}^2, then 32\text{ m}^2: 4 times larger. \tfrac{1}{2}(3 + 5)(2) = 8 and \tfrac{1}{2}(6 + 10)(4) = 32. Doubling the bases doubles their average, and doubling the height doubles it again, so the area is multiplied by 2 \times 2 = 4.
kg1128 · updated 2026-10-13 · CC-BY-SA-4.0