Area of Parallelograms

See why the area of a parallelogram is base × perpendicular height, and use it to find an area, a base or a height, on a grid, with mixed units, with an equation and in problems with cost.

Before you start

A parallelogram is a quadrilateral whose opposite sides are parallel; its opposite sides are also equal (properties of special quadrilaterals). In a drawing, parallel sides carry matching arrows; ticks mean equal lengths.
A rectangle has four right angles. Area of a rectangle = length × breadth: the rectangle below has area $8 \times 5 = 40$ cm². Area is measured in square units (cm², m²); a perimeter, the distance round the boundary, in cm or m.
An equation such as $7h = 63$ is solved by dividing both sides by 7: $h = 9$ .

Why the Area is Base × Height

Parallelogram $ABCD$ below has base $AB = 10$ cm. From $D$ , draw the line $DE$ at right angles to $AB$ . Its length, 5 cm, is the perpendicular height of the parallelogram: the distance straight across from the base $AB$ to the side opposite it, $DC$ .

Cut off the right-angled triangle $AED$ (amber) along the height, and slide it 10 cm to the right, along the base. Its side $AD$ lands exactly on $BC$ , because opposite sides of a parallelogram are equal and parallel. The triangle now fills the space $BFC$ at the other end:

The two pieces make the rectangle $EFCD$ . Nothing was added and nothing was lost, so the rectangle has the same area as the parallelogram. Its length is $EF = 10$ cm, the base of the parallelogram (every point of the triangle moved 10 cm, so $EF = AB$ ), and its breadth is $DE = 5$ cm, the height of the parallelogram.

So the area of the parallelogram is $10 \times 5 = 50$ cm²: its base times its perpendicular height. A parallelogram that leans further can be cut along heights into more pieces and rebuilt in the same way, so the rule holds for every parallelogram.

Area of a parallelogram

Area of a parallelogram = base × height

$A = bh$

where $b$ is the length of the base and $h$ is the perpendicular height: the perpendicular distance between the base and the side opposite it. Both lengths are in the same unit, and the area is in square units (cm², m²).

Example 1: Area from the base and the height

Parallelogram $ABCD$ has base $AB = 15$ cm and perpendicular height 8 cm. Find its area.

Solution

Area = base × height

$= 15 \times 8$

$= 120$ cm²

The area of the parallelogram is 120 cm².

Base and Perpendicular Height

The base may be any side of the parallelogram. The height that goes with it is measured at right angles (perpendicular) to the base, from the base to the side opposite. It is drawn as a dashed line with a small square at its foot.

The height is not the slant side. The perpendicular is the shortest way across from one parallel side to the other; a slant side crosses at an angle, so it is always longer than the height. (A side is only as long as the height when it stands upright, at right angles to the base: then the parallelogram is a rectangle.) When a figure gives both, the area uses the height and leaves the slant side alone.

Example 2: A figure that also gives the slant side

Parallelogram $ABCD$ has base $AB = 12$ cm, slant side $AD = 7$ cm and perpendicular height $DE = 5$ cm. Find its area.

Solution

The height must be perpendicular to the base $AB$ . $DE$ is perpendicular to $AB$ (it has the right-angle mark); $AD$ is a slant side, so it is not used for the area.

Area = base × height

$= 12 \times 5$

$= 60$ cm²

Using the slant side would give $12 \times 7 = 84$ cm², which is wrong.

The area of the parallelogram is 60 cm².

The height can be drawn from any point of the side opposite the base: the two sides are parallel, so the distance between them is the same all along.

When a parallelogram leans a long way, the height from a vertex lands beyond the end of the base. The base is then produced (extended) with a dashed line, and the height meets it there, outside the figure. It is still the height for the base, and the base is still only the side of the parallelogram, not the dashed part.

Example 3: The height outside the figure

In parallelogram $ABCD$ , $AB = 5$ cm and $AD = 10$ cm. The base $AB$ is produced to $F$ , and $CF$ , perpendicular to $AF$ , is 6 cm. Find the area of $ABCD$ .

Solution

$CF$ is perpendicular to the line of the base, so it is the height for the base $AB$ , even though it is outside the figure. The slant side $AD = 10$ cm is not the height.

The base is $AB = 5$ cm, not $AF$ : the dashed part $BF$ is not a side of the parallelogram.

Area = base × height

$= 5 \times 6$

$= 30$ cm²

The area of $ABCD$ is 30 cm².

Either Side as the Base

A parallelogram has two pairs of parallel sides, so it has two heights, one for each pair. The height that goes with a base is always the one perpendicular to that base. Either way gives the same area:

area = $AB$ × (height to $AB$ ) = $AD$ × (height to $AD$ ).

Example 4: Two bases, two heights

Parallelogram $ABCD$ has $AB = 12$ cm and $AD = 9$ cm. The perpendicular height from $D$ to $AB$ is $DE = 7.5$ cm, and the perpendicular height from $B$ to $AD$ is $BG = h$ cm. Find (a) the area of $ABCD$ , (b) the value of $h$ .

Figure for (a), with $AB$ as the base:

Figure for (b), with $AD$ as the base:

Solution

(a) Take $AB$ as the base. The height perpendicular to $AB$ is $DE$ .

Area = base × height

$= 12 \times 7.5$

$= 90$ cm²

(b) Now take $AD$ as the base. The height perpendicular to $AD$ is $BG = h$ cm, and the area is still 90 cm².

$9 \times h = 90$

Divide both sides by 9: $h = 10$

Check: $9 \times 10 = 90$ , as in (a). Each height is shorter than the slant side it goes with: $7.5 < 9$ and $10 < 12$ .

(a) The area of $ABCD$ is 90 cm². (b) $h = 10$ : the height to $AD$ is 10 cm.

Finding the Height or the Base

If the area and one of the two lengths are known, put them into $A = bh$ and solve the equation for the other length, dividing both sides by the length you know.

Example 5: The height, and then the base, from the area

(a) A parallelogram has area 96 cm² and base 12 cm. Find its perpendicular height, $h$ cm.

(b) A parallelogram has area 91 cm² and perpendicular height 7 cm. Find its base, $b$ cm.

Solution

(a) Area = base × height

$12 \times h = 96$

Divide both sides by 12: $h = 96 \div 12 = 8$

Check: $12 \times 8 = 96$ .

(b) Area = base × height

$b \times 7 = 91$

Divide both sides by 7: $b = 91 \div 7 = 13$

Check: $13 \times 7 = 91$ .

(a) The perpendicular height is 8 cm. (b) The base is 13 cm.

A height found in this way cannot be longer than the slant side it goes with (it equals it only when the parallelogram is a rectangle). If it is longer, no such parallelogram can be drawn.

Example 6: A parallelogram that cannot exist

Ravi says he has drawn a parallelogram with sides 10 cm and 6 cm and an area of 72 cm². Can he be right?

Solution

Take a 10 cm side as the base, with height $h$ cm: $10 \times h = 72$ , so $h = 7.2$ .

The 6 cm sides are the slant sides joining the two 10 cm sides. The height $DE$ is the shortest way across from $D$ to the base; the slant side $AD$ also goes across, at an angle, so it is longer:

So the height can be at most 6 cm, and it is 6 cm only when the slant sides stand upright, that is, when the parallelogram is a rectangle. A height of 7.2 cm is longer than the 6 cm side: impossible.

Taking a 6 cm side as the base instead needs a height of $72 \div 6 = 12$ cm, longer than the 10 cm side: impossible again.

The largest area a parallelogram with sides 10 cm and 6 cm can have is the rectangle's, $10 \times 6 = 60$ cm², and 72 cm² is more than that.

No. An area of 72 cm² needs a height of 7.2 cm between the 10 cm sides, but the height cannot be longer than the 6 cm slant side. Such a parallelogram cannot exist.

Area and Perimeter

The perimeter is the distance round the boundary, so it uses the slant sides and never the height. Opposite sides are equal, so the perimeter is twice the sum of two neighbouring sides: $P = 2(a + b)$ .

A rectangle and a parallelogram on the same base and between the same two parallel lines have the same height, so they have the same area. Their perimeters are not the same: the slant sides are longer than the upright ones.

Example 7: Same area, different perimeters

Figure A is a rectangle 10 cm by 5 cm. Figure B is a parallelogram with base 10 cm, slant sides 7 cm and perpendicular height 5 cm. Find the area and the perimeter of each.

Solution

A: area = length × breadth $= 10 \times 5 = 50$ cm²; perimeter $= 2(10 + 5) = 30$ cm.

B: area = base × height $= 10 \times 5 = 50$ cm² (the slant side is not used).

B: perimeter $= 2(10 + 7) = 34$ cm (the slant sides are on the boundary; the height is not).

The more a parallelogram on this base leans, the longer its slant sides and its perimeter; its area stays 50 cm² as long as its height is 5 cm.

Both have area 50 cm², but the perimeter of A is 30 cm and the perimeter of B is 34 cm.

Parallelograms on a Grid

On a square grid, a base along a grid line can be counted, and so can the height, straight across the grid at right angles to it. Count the base along its line and the height row by row; do not count along a slant side.

Example 8: Counting the base and the height

The dots are 1 cm apart, across and up. Find the area of parallelogram $ABCD$ .

Solution

The base $AB$ lies along a row of dots: from $A$ to $B$ is 5 spaces, so $AB = 5$ cm.

The side $DC$ is 4 rows above $AB$ , so the perpendicular height is 4 cm (the dashed line from $D$ , straight down the grid to $AB$ ).

Area = base × height

$= 5 \times 4$

$= 20$ cm²

The area of $ABCD$ is 20 cm².

The Rhombus

A rhombus is a parallelogram with four equal sides (marked with ticks). Its area is found in the same way, area = base × height, and any side can be the base.

Example 9: The area of a rhombus from its perimeter

A rhombus has perimeter 52 cm and perpendicular height 12 cm. Find its area.

Solution

The four sides are equal, so each side is $52 \div 4 = 13$ cm.

Area = base × height

$= 13 \times 12$

$= 156$ cm²

The height, 12 cm, is shorter than the 13 cm side, so such a rhombus can be drawn.

The area of the rhombus is 156 cm².

Mixed Units and Equations

The base and the height must be in the same unit before they are multiplied. Change one of the lengths (1.2 m = 120 cm, or 45 cm = 0.45 m); never multiply metres by centimetres.

Example 10: Lengths in metres and centimetres

A signboard in the shape of a parallelogram has base 1.2 m and perpendicular height 45 cm. Find its area in cm².

Solution

The answer is wanted in cm², so change the base to centimetres: $1.2$ m $= 1.2 \times 100 = 120$ cm.

Area = base × height

$= 120 \times 45$

$= 5400$ cm²

Multiplying $1.2 \times 45 = 54$ mixes metres with centimetres and gives no correct area.

The area of the signboard is 5400 cm².

When a length is given in terms of an unknown, the area gives an equation. Solve it as any linear equation is solved, doing the same to both sides, then check by putting the value back.

Example 11: An equation from the area

A parallelogram has base $(x + 3)$ cm, perpendicular height 4 cm and area 36 cm². Find $x$ and the length of the base.

Solution

Area = base × height

$4(x + 3) = 36$

Divide both sides by 4: $x + 3 = 9$

Subtract 3 from both sides: $x = 6$

The base is $6 + 3 = 9$ cm. Check: $9 \times 4 = 36$ .

$x = 6$ , and the base is 9 cm long.

Problems

In a problem, decide first what is asked. Covering a region (turf, tiles, paint) needs its area, base × height; going round its edge (a fence, a border) needs its perimeter, which uses the slant sides.

Example 12: Turfing and fencing a garden plot

A community garden plot in Tampines is a parallelogram $ABCD$ with $AB = 25$ m, $AD = 20$ m and perpendicular height 16 m.

(a) The plot is covered with turf costing $4 per m². Find the cost of the turf.

(b) A fence is put up along all four sides at $12.50 per metre. Find the cost of the fence.

Solution

(a) Area = base × height

$= 25 \times 16$

$= 400$ m²

Cost of the turf: $400 \times 4 = 1600$ , so the turf costs $1600.

(b) The fence goes along the sides, so it uses the slant side, 20 m, and not the height.

Perimeter $= 2(25 + 20) = 2 \times 45 = 90$ m

Cost of the fence: $90 \times 12.50 = 1125$ , so the fence costs $1125.

(a) The turf costs $1600. (b) The fence costs $1125.

Common mistakes

Using the slant side as the height. In Example 2, $12 \times 7 = 84$ cm² is wrong; the height is the perpendicular 5 cm, and the area is 60 cm².
Pairing a base with the wrong height. The height must be perpendicular to the side used as the base: $AB$ goes with the height to $AB$ , $AD$ with the height to $AD$ .
Halving. $\frac{1}{2} \times$ base × height is the area of a triangle; the area of a parallelogram is base × height, not halved.
Counting the dashed, produced part as part of the base when the height falls outside the figure.
Multiplying lengths in different units, such as $1.2 \times 45$ with metres and centimetres. Change one length first.
Thinking that figures with the same area have the same perimeter, or using the height in a perimeter.
Writing an area in cm instead of cm², or leaving out the unit.

Practice

Try each question on paper first, writing the formula, the numbers and the unit, then open its solution.

Practice 1: Floor tiles at a void deck

The floor of a void deck is laid with tiles in the shape of a parallelogram. Each tile has base 18 cm and perpendicular height 11 cm.

(a) Find the area of one tile. (b) Find the area covered by 50 tiles.

ShowHide solution

(a) Area = base × height

$= 18 \times 11$

$= 198$ cm²

(b) Area of 50 tiles $= 198 \times 50 = 9900$ cm²

(a) 198 cm² (b) 9900 cm²

Practice 2: Height or slant side?

Find the area of parallelogram $PQRS$ .

ShowHide solution

The dashed height, 8 cm, is perpendicular to the base $PQ$ ; $PS = 10$ cm is a slant side and is not used.

Area = base × height

$= 14 \times 8$

$= 112$ cm²

The area of $PQRS$ is 112 cm².

Practice 3: The height outside the figure

In parallelogram $ABCD$ , $AB = 4$ cm and $AD = 10$ cm. $AB$ is produced to $F$ , and $CF$ , perpendicular to $AF$ , is 8 cm. Find the area of $ABCD$ .

ShowHide solution

$CF$ is perpendicular to the line of $AB$ , so it is the height for the base $AB$ , though it lies outside the figure.

The base is $AB = 4$ cm (not $AF$ ), and the slant side $AD = 10$ cm is not used.

Area = base × height

$= 4 \times 8$

$= 32$ cm²

The area of $ABCD$ is 32 cm².

Practice 4: The base from the area

A parallelogram has area 135 cm² and perpendicular height 9 cm. Find its base, $b$ cm.

ShowHide solution

Area = base × height

$b \times 9 = 135$

Divide both sides by 9: $b = 135 \div 9 = 15$

Check: $15 \times 9 = 135$ .

The base is 15 cm.

Practice 5: Parallelograms on a grid

The dots are 1 cm apart, across and up. Find the area of each of the parallelograms P and Q.

ShowHide solution

P: its bottom side lies along a row, 4 spaces long, so the base is 4 cm. The top side is 4 rows higher, so the height is 4 cm. Area = base × height $= 4 \times 4 = 16$ cm².

Q: two of its sides are upright, each 3 spaces long; take one as the base, 3 cm. The height must be perpendicular to it, so it runs across the grid: the two upright sides are 4 columns apart, so the height is 4 cm. Area = base × height $= 3 \times 4 = 12$ cm².

The area of P is 16 cm² and the area of Q is 12 cm².

Practice 6: Either side as the base

In parallelogram $ABCD$ , $AB = 16$ cm and $AD = 12$ cm. The perpendicular from $B$ to $AD$ is $BG = 14$ cm. $AB$ is produced to $F$ , and $CF$ , perpendicular to $AF$ , is $h$ cm. Find (a) the area of $ABCD$ , (b) the value of $h$ .

ShowHide solution

(a) Take $AD$ as the base; its height is $BG$ .

Area = base × height

$= 12 \times 14$

$= 168$ cm²

(b) Take $AB$ as the base; its height is $CF = h$ cm (outside the figure). The area is still 168 cm².

$16 \times h = 168$

Divide both sides by 16: $h = 168 \div 16 = 10.5$

Check: $16 \times 10.5 = 168$ , and each height is shorter than the slant side it goes with: $10.5 < 12$ and $14 < 16$ .

(a) The area is 168 cm². (b) $h = 10.5$

Practice 7: Can it be drawn?

Mei Ling says she has drawn a parallelogram with sides 9 cm and 5 cm, in which the perpendicular height to a 9 cm side is 6 cm.

(a) Explain why she must have made a mistake.

(b) What is the largest area that a parallelogram with sides 9 cm and 5 cm can have?

ShowHide solution

(a) The height to a 9 cm side is the distance across to the other 9 cm side, and the 5 cm sides are the slant sides joining them. The perpendicular is the shortest way across, so the height is at most 5 cm. A height of 6 cm is longer than the 5 cm side, so the figure cannot be drawn.

(b) The height to a 9 cm side is largest, 5 cm, when the 5 cm sides stand upright, that is, when the parallelogram is a rectangle.

Largest area $= 9 \times 5 = 45$ cm²

(a) The height cannot be longer than the 5 cm slant side, and 6 cm is longer. (b) 45 cm², when the parallelogram is a rectangle.

Practice 8: Two flower beds

A school garden has two flower beds. Bed A is a rectangle 9 m by 6 m. Bed B is a parallelogram with base 9 m, slant sides 7.5 m and perpendicular height 6 m.

(a) Show that the two beds have the same area.

(b) Ali says the two beds need the same length of edging round them. Is he right? Give a reason.

ShowHide solution

(a) Bed A: area = length × breadth $= 9 \times 6 = 54$ m². Bed B: area = base × height $= 9 \times 6 = 54$ m². The areas are the same.

(b) Bed A: perimeter $= 2(9 + 6) = 30$ m.

Bed B: perimeter $= 2(9 + 7.5) = 33$ m (the slant sides, not the height).

$33 - 30 = 3$ , so bed B needs 3 m more edging.

(a) Both beds have area 54 m². (b) No: bed A needs 30 m of edging and bed B needs 33 m.

Practice 9: A rhombus of card

Siti cuts a rhombus from a sheet of card. Its perimeter is 40 cm and its perpendicular height is 8 cm. Find its area.

ShowHide solution

The four sides of a rhombus are equal, so each side is $40 \div 4 = 10$ cm.

Area = base × height

$= 10 \times 8$

$= 80$ cm²

The area of the rhombus is 80 cm².

Practice 10: Solving for $x$

A parallelogram has base $(2x - 1)$ cm, perpendicular height 6 cm and area 54 cm². Find $x$ and the length of the base.

ShowHide solution

Area = base × height

$6(2x - 1) = 54$

Divide both sides by 6: $2x - 1 = 9$

Add 1 to both sides: $2x = 10$

Divide both sides by 2: $x = 5$

The base is $2 \times 5 - 1 = 9$ cm. Check: $9 \times 6 = 54$ .

$x = 5$ , and the base is 9 cm long.

Practice 11: Perimeter and area together

Parallelogram $ABCD$ has perimeter 84 cm and area 240 cm², and $AB = 24$ cm.

Find (a) the length of $AD$ , (b) the perpendicular height $h$ cm to $AB$ , (c) the perpendicular height to $AD$ , correct to 3 significant figures.

ShowHide solution

(a) Perimeter $= 2(AB + AD)$

$2(24 + AD) = 84$

Divide both sides by 2: $24 + AD = 42$ , so $AD = 18$ cm.

(b) Area = base × height, with $AB$ as the base:

$24 \times h = 240$ , so $h = 240 \div 24 = 10$

$18 \times k = 240$ , so $k = 240 \div 18 = 13.333\ldots$

$k \approx 13.3$ (3 s.f.)

Check: both heights are shorter than the slant sides they go with ( $10 < 18$ and $13.3 < 24$ ).

(a) $AD = 18$ cm (b) 10 cm (c) $\approx 13.3$ cm (3 s.f.)

Practice 12: Resurfacing a car park

Each lot in an HDB car park is a parallelogram. Its base along the driveway is 250 cm, its painted slant sides are 5 m long, and its perpendicular height is 4.8 m.

(a) Find the area of one lot in m².

(b) 30 lots are resurfaced at a cost of $15 per m². Find the total cost.

ShowHide solution

(a) The area is wanted in m², so change the base to metres: $250$ cm $= 250 \div 100 = 2.5$ m.

Area = base × height (the 5 m slant side is not used)

$= 2.5 \times 4.8$

$= 12$ m²

(b) Area of 30 lots $= 30 \times 12 = 360$ m²

Cost: $360 \times 15 = 5400$ , so the total cost is $5400.

(a) 12 m² (b) $5400

Key Points

Area of a parallelogram = base × height, $A = bh$ : a right-angled triangle cut off one end and moved to the other makes a rectangle with the same base and height.
The height is the perpendicular distance between the base and the side opposite it. It is not the slant side, and it is never longer than the slant side (the two are equal only in a rectangle).
The height may fall outside the figure, on the base produced; the base is still only the side of the parallelogram.
Either side may be the base, each with its own height: $AB$ × (height to $AB$ ) = $AD$ × (height to $AD$ ).
To find a height or a base, put the area and the known length into $A = bh$ and solve; check that the height is not longer than its slant side.
Same base and same height give the same area, but not the same perimeter. Perimeter $P = 2(a + b)$ uses the slant sides.
A rhombus is a parallelogram with four equal sides: area = side × height.
Change the lengths to the same unit before multiplying; an area is in cm² or m².

Before you start

Why the Area is Base × Height

Area of a parallelogram

Example 1: Area from the base and the height

Base and Perpendicular Height

Example 2: A figure that also gives the slant side

Example 3: The height outside the figure

Either Side as the Base

Example 4: Two bases, two heights

Finding the Height or the Base

Example 5: The height, and then the base, from the area

Example 6: A parallelogram that cannot exist

Area and Perimeter

Example 7: Same area, different perimeters

Parallelograms on a Grid

Example 8: Counting the base and the height

The Rhombus

Example 9: The area of a rhombus from its perimeter

Mixed Units and Equations

Example 10: Lengths in metres and centimetres

Example 11: An equation from the area

Problems

Example 12: Turfing and fencing a garden plot

Common mistakes

Practice

Practice 1: Floor tiles at a void deck

Practice 2: Height or slant side?

Practice 3: The height outside the figure

Practice 4: The base from the area

Practice 5: Parallelograms on a grid

Practice 6: Either side as the base

Practice 7: Can it be drawn?

Practice 8: Two flower beds

Practice 9: A rhombus of card

Practice 10: Solving for xxx

Practice 11: Perimeter and area together

Practice 12: Resurfacing a car park

Key Points

Practice 10: Solving for $x$