Why can't I use the slanted side as the height of a parallelogram or trapezoid?

Because area counts unit squares stacked in rows, and a row's thickness is measured straight across at a right angle to the base. The slanted side leans, so it is longer than that straight-across distance and would give an area larger than the shape really is. Only a segment meeting the base at 90^ is a height.

Does it matter which side of a triangle I call the base?

No. Every base-and-height pair in the same triangle gives the same area. What matters is that the two measurements match: if you use the 8 cm side as your base, you must use the height drawn perpendicular to that 8 cm side, not a height drawn to a different side.

Should I decompose or subtract when I find the area of an irregular polygon?

Use whichever needs less work. Subtracting from a bounding rectangle is usually quickest when the shape looks like a rectangle with a bite removed. Decomposing works better when the figure has parts sticking out, or includes triangles and trapezoids. If you have time, do it both ways as a check.

What is the most common mistake on these problems?

Three show up constantly: dropping the 1/2 in the triangle or trapezoid formula, using a slanted side instead of the perpendicular height, and overlapping pieces when decomposing so part of the area gets counted twice. Labeling each piece and comparing your total to the surrounding rectangle catches all three.

M7MATH-8.1

Area of Polygons

Learn to find the area of triangles, parallelograms and trapezoids using perpendicular height, then break irregular polygons into familiar pieces.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Area of Polygons, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Area of Polygons free

What this lesson covers

Every area formula you will use in this lesson comes from one idea: a rectangle's area is base times height. Slide a parallelogram's corner over and it becomes a rectangle. Put two identical triangles together and you get a parallelogram. Turn a trapezoid upside down next to itself and you get a parallelogram too. That is why the formulas look so similar, and why they all depend on the same tricky measurement — the perpendicular height.

In this lesson you will practice choosing the right base-and-height pair, applying the three formulas, and handling shapes that have no formula at all. For those irregular polygons, you decompose the figure into rectangles, triangles and trapezoids, find each piece, and add. Sometimes it is faster to surround the shape with a big rectangle and subtract the missing corners. Both moves show up again when you find surface area later in this unit.

Perpendicular Height: The Measurement That Trips People Up

The base of a polygon is any side you choose to build from. The height is the perpendicular distance from that base to the opposite vertex or the opposite side. Perpendicular means the height makes a

90^\circ

angle with the base — it is the shortest, straight-across distance, not a slanted side.

This matters because figures are almost always drawn with an extra number on them. A parallelogram might show a bottom side of 10 cm, a slanted side of 7 cm, and a dashed segment of 6 cm inside marked with a right-angle symbol. Only the dashed 6 cm is the height that goes with the 10 cm base. Multiplying

10 \times 7

gives a number that is too big, because the slanted side is longer than the true straight-across distance.

A second point: base and height must be a matching pair. A triangle has three bases and three matching heights, and any pair gives the same area. If you use the 8 cm side as the base, you must use the height drawn perpendicular to that 8 cm side.

For an obtuse triangle, the height sometimes falls outside the triangle, drawn to an extension of the base. That still counts. The height is a distance, not a piece of the shape.

Finally, area is always in square units:

\mathrm{cm^2}

\mathrm{m^2}

, square feet. Writing a bare number, or writing cm instead of

\mathrm{cm^2}

, is the most common way a correct calculation ends up as an incomplete answer.

The Three Formulas and Why They Work

All three formulas come from the rectangle.

Shape	Formula	Where it comes from
Parallelogram	$A = bh$	Cut off the triangle on one end, slide it to the other end, and you have a rectangle with the same base and height
Triangle	$A = \frac{1}{2}bh$	Two copies of the triangle fit together into a parallelogram with base $b$ and height $h$ , so one triangle is half of it
Trapezoid	$A = \frac{1}{2}(b_1 + b_2)h$	Two copies form a parallelogram whose base is $b_1 + b_2$ and whose height is $h$

In the trapezoid formula,

b_1

and

b_2

are the two parallel sides, and

h

is the perpendicular distance between them. The non-parallel slanted sides never enter the calculation. A useful way to read the formula is that

\frac{b_1+b_2}{2}

is the average width of the trapezoid, so

A = (\text{average of the parallel sides}) \times h

Example: a trapezoid with parallel sides 5 m and 11 m and height 4 m has average width

\frac{5+11}{2} = 8

m, so

A = 8 \times 4 = 32\ \mathrm{m^2}

.

Watch the order of operations. In

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

you must add inside the parentheses first. Computing

\frac{1}{2}(5) + 11 \times 4

is a frequent slip. And in the triangle formula, do not forget the

\frac{1}{2}

— that single omission doubles every answer.

Decomposing Irregular Polygons

An irregular polygon has no formula, so you make your own. Decomposing means slicing the figure with straight cuts into non-overlapping pieces you do know: rectangles, triangles, trapezoids. Find each piece, then add.

The other method is subtracting. Draw the smallest rectangle that completely contains the shape, find its area, then subtract the corner pieces that are not part of the figure. For an L-shape or a rectangle with a bite taken out, subtracting is usually faster.

The real work is finding missing side lengths. Diagrams rarely label every segment, but opposite sides of a rectangle are equal, so you can reason. If the whole bottom is 10 cm and one part is 6 cm, the other part is

10 - 6 = 4

cm. If the full height is 8 cm and the notch is 3 cm tall, the short piece is 5 cm.

Three habits that prevent mistakes:

Draw your cut lines right on the figure and label each piece A, B, C. Then write one area statement per piece so you can check your work later.

After decomposing, make sure the pieces do not overlap. Double-counting a shared strip is the most common error in these problems, and it always makes the answer too large.

Check reasonableness. Your total should be less than the surrounding rectangle and more than the largest single piece. If you get a number bigger than the bounding rectangle, you either overlapped pieces or added when you should have subtracted.

Working Backwards from a Known Area

Many problems give you the area and ask for a missing length. The move is the same as solving any equation: substitute what you know, then undo the operations.

A triangle has area

48\ \mathrm{cm^2}

and base 12 cm. Then

48 = \frac{1}{2}(12)h

, so

48 = 6h

and

h = 8

cm.

A parallelogram has area

91\ \mathrm{m^2}

and height 7 m. Then

91 = 7b

, so

b = 13

m.

A trapezoid has area

54\ \mathrm{ft^2}

, height 6 ft, and one parallel side 5 ft. Then

54 = \frac{1}{2}(5 + b_2)(6) = 3(5 + b_2)

, so

18 = 5 + b_2

and

b_2 = 13

ft.

Notice the answer to a missing-length question is in plain units, not square units, because you are naming a distance. Mixing that up is a quick way to signal the wrong idea even when the arithmetic is right.

These problems also show up in real settings: a floor plan needs 150 square feet of tile and the room is 10 feet wide, so it must be 15 feet long. Or a triangular sail must cover 24 square meters with a 6 meter base, so the sail needs a height of 8 meters. When a problem mixes units — a base in centimeters and a height in millimeters — convert first so both measurements match, or your square units will be meaningless.

Key terms

Base.: Any side of a polygon chosen as the reference side for an area calculation. Triangles have three possible bases; a trapezoid's formula uses its two parallel sides.
Perpendicular height (altitude).: The straight-across distance from the base to the opposite vertex or opposite parallel side, meeting the base at a $90^\circ$ angle. It may fall outside the figure in an obtuse triangle.
Parallelogram.: A quadrilateral with both pairs of opposite sides parallel. Its area is $A = bh$ , where $h$ is measured perpendicular to the chosen base, not along the slanted side.
Trapezoid.: A quadrilateral with at least one pair of parallel sides, called $b_1$ and $b_2$ . Its area is $A = \frac{1}{2}(b_1+b_2)h$ .
Decomposing.: Cutting a figure into non-overlapping familiar shapes, finding each area, and adding the results.
Bounding rectangle method.: Enclosing an irregular figure in the smallest rectangle that contains it, then subtracting the areas of the leftover corner pieces.
Composite figure.: A shape built from two or more basic polygons joined together; its area equals the sum of its non-overlapping parts.
Square units.: The units of area, such as $\mathrm{cm^2}$ or square feet, counting how many unit squares cover the region.

Worked example

A community garden is an irregular polygon. Its main part is a rectangle 10 m wide and 6 m tall. A triangular planting bed sticks out from the right side of the rectangle: its base is the rectangle's 6 m right side, and its perpendicular height is 4 m. A 3 m by 2 m rectangular shed pad in the bottom-left corner is paved and is not part of the garden. Find the area of the garden.

Step 1: Name the pieces. Piece A is the main rectangle, piece B is the triangle on the right, and piece C is the shed pad that must be removed.

Step 2: Piece A.

A = bh = 10 \times 6 = 60\ \mathrm{m^2}

.

Step 3: Piece B. The base is the shared 6 m side and the perpendicular height is 4 m, measured straight out from that side.

A = \frac{1}{2}bh = \frac{1}{2}(6)(4) = 12\ \mathrm{m^2}

. Check that the 4 m is perpendicular to the 6 m side — it is, since it is measured straight out from the rectangle's edge.

Step 4: Piece C.

A = 3 \times 2 = 6\ \mathrm{m^2}

. This corner is inside the rectangle, so it gets subtracted, not added.

Step 5: Combine.

60 + 12 - 6 = 66\ \mathrm{m^2}

.

Step 6: Check for reasonableness. The whole figure fits inside a rectangle 14 m wide by 6 m tall, which is

84\ \mathrm{m^2}

, and the answer 66 is comfortably less than that but larger than the biggest single piece, 60. The final answer is

66\ \mathrm{m^2}

, written in square meters because it is an area.

Practice questions

A parallelogram is drawn with a bottom side of 10 cm, a slanted left side of 6 cm, and a dashed segment inside marked with a right-angle symbol measuring 5 cm from the bottom side up to the top side. What is the area of the parallelogram?

$30\ \mathrm{cm^2}$
$50\ \mathrm{cm^2}$
$60\ \mathrm{cm^2}$
$25\ \mathrm{cm^2}$

Answer: $50\ \mathrm{cm^2}$

The height must be perpendicular to the base, so the matching pair is the 10 cm base with the 5 cm dashed height:

A = bh = 10 \times 5 = 50\ \mathrm{cm^2}

. Using the 6 cm slanted side gives

60\ \mathrm{cm^2}

, which is too big because a slanted side is always longer than the straight-across height. The

25\ \mathrm{cm^2}

answer comes from using the triangle's

\frac{1}{2}

, which does not belong in a parallelogram formula.

A trapezoid-shaped tabletop has an area of 60 square inches. Its parallel sides measure 7 inches and 13 inches. Find the perpendicular height, and explain why the answer is in inches rather than square inches.

Answer: The height is 6 inches.

Substitute into

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

60 = \frac{1}{2}(7+13)h

. Add inside the parentheses first, giving

60 = \frac{1}{2}(20)h = 10h

. Divide both sides by 10 to get

h = 6

. The answer is in inches, not square inches, because height is a single distance — one measurement across the shape — while square inches count how many unit squares cover the surface. A quick check: average width

\frac{7+13}{2} = 10

inches, and

10 \times 6 = 60

square inches.

An L-shaped room is 12 feet wide and 9 feet tall on the outside. A rectangular section 5 feet wide and 4 feet tall is missing from the top-right corner. Find the floor area two different ways and show that the answers agree.

Answer: $88$ square feet

Subtraction method: the bounding rectangle is

12 \times 9 = 108

square feet, and the missing corner is

5 \times 4 = 20

square feet, so

108 - 20 = 88

square feet. Decomposition method: cut the L into a left rectangle

7

feet wide by

9

feet tall, since

12 - 5 = 7

, giving

63

square feet, plus a right rectangle

5

feet wide by

5

feet tall, since

9 - 4 = 5

, giving

25

square feet. Then

63 + 25 = 88

square feet. Both routes agree, which is a strong check. If your two methods disagree, look first for pieces that overlap or a missing side length you subtracted incorrectly.

FAQ

Why can't I use the slanted side as the height of a parallelogram or trapezoid?: Because area counts unit squares stacked in rows, and a row's thickness is measured straight across at a right angle to the base. The slanted side leans, so it is longer than that straight-across distance and would give an area larger than the shape really is. Only a segment meeting the base at $90^\circ$ is a height.
Does it matter which side of a triangle I call the base?: No. Every base-and-height pair in the same triangle gives the same area. What matters is that the two measurements match: if you use the 8 cm side as your base, you must use the height drawn perpendicular to that 8 cm side, not a height drawn to a different side.
Should I decompose or subtract when I find the area of an irregular polygon?: Use whichever needs less work. Subtracting from a bounding rectangle is usually quickest when the shape looks like a rectangle with a bite removed. Decomposing works better when the figure has parts sticking out, or includes triangles and trapezoids. If you have time, do it both ways as a check.
What is the most common mistake on these problems?: Three show up constantly: dropping the $\frac{1}{2}$ in the triangle or trapezoid formula, using a slanted side instead of the perpendicular height, and overlapping pieces when decomposing so part of the area gets counted twice. Labeling each piece and comparing your total to the surrounding rectangle catches all three.

Learn this with a teacher, not a page

The Crimsora tutor teaches Area of Polygons live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

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