Area of Trapeziums

Name the parallel sides, slant sides and height of a trapezium, see why its area is $\frac{1}{2}(a + b)h$ , and use the formula to find an area, a height or a parallel side.

Before you start

The area of a triangle is $\frac{1}{2} \times$ base $\times$ height, where the height is at right angles to the base. The area of a parallelogram is base $\times$ perpendicular height.
A trapezium is a quadrilateral with exactly one pair of parallel sides. In a figure, parallel sides are marked with arrows and equal sides with ticks.
An equation such as $14h = 84$ is solved by dividing both sides by 14: $h = 6$ .

The Parts of a Trapezium

The two sides of a trapezium that are parallel are its parallel sides. The other two sides, the non-parallel sides, are its slant sides. The parallel sides are usually of different lengths, and we call them $a$ and $b$ .

The height of a trapezium is the perpendicular distance between its parallel sides: the length of a line at right angles to both of them. It is drawn dashed, with a small square to show the right angle. It may be drawn from any point of one parallel side to the other parallel side, or to that side carried on (produced), and it always has the same length, because parallel lines stay the same distance apart.

Trapezium

A trapezium is a quadrilateral with exactly one pair of parallel sides.

The parallel sides ( $a$ and $b$ ) are marked with arrows. The other two sides are the slant sides.

The height $h$ is the perpendicular distance between the parallel sides. Usually it is not a slant side: both slant sides lean. Only in a right trapezium (below) is one slant side at right angles to the parallel sides, and then that slant side is also the height.

Two kinds of trapezium have names of their own. In an isosceles trapezium the two slant sides are equal (marked with ticks). In a right trapezium one of the two slant sides is at right angles to both parallel sides, so that slant side is the height.

The Area of a Trapezium

There are two ways to see where the formula for the area comes from. Both use a trapezium $ABCD$ with parallel sides $DC = a$ and $AB = b$ , and height $h$ .

First way: cut it along a diagonal. The diagonal $DB$ cuts the trapezium into two triangles. $\triangle ABD$ (blue) has base $AB = b$ . $\triangle BCD$ (amber) has base $DC = a$ . The height of each triangle is the distance between the parallel sides, $h$ . For $\triangle BCD$ it is measured from $B$ to the line $DC$ produced (carried on beyond $C$ ).

So the area of the trapezium is $\frac{1}{2}bh + \frac{1}{2}ah$ . Taking out the common factor $\frac{1}{2}h$ , this is $\frac{1}{2}(a + b)h$ .

Second way: two copies make a parallelogram. Make a second copy of the trapezium, give it a half turn, and fit it against the first along the slant side $BC$ . The parallel sides of the two copies now lie end to end: along the bottom, $b$ and then $a$ ; along the top, $a$ and then $b$ . The two copies together make a parallelogram with base $a + b$ and height $h$ .

The area of the parallelogram is base $\times$ height $= (a + b)h$ , and the trapezium is half of it, $\frac{1}{2}(a + b)h$ . Both ways give the same formula.

Area of a trapezium

Area of a trapezium $= \frac{1}{2} \times$ (sum of parallel sides) $\times$ height

$A = \frac{1}{2}(a + b)h$ , where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular height between them.

All three lengths must be in the same unit. The area is then in square units ( $\text{cm}^{2}$ or $\text{m}^{2}$ ).

Example 1: The area from the parallel sides and the height

Find the area of the trapezium.

Solution

The parallel sides (with the arrows) are 8 cm and 12 cm. The height between them is 6 cm.

Area $= \frac{1}{2}(a + b)h$

$= \frac{1}{2} \times (8 + 12) \times 6$

$= \frac{1}{2} \times 20 \times 6$

$= 60\text{ cm}^{2}$

The area of the trapezium is $60\text{ cm}^{2}$ .

The Height, Not the Slant Side

A figure often shows the slant sides as well as the height. The area uses only the parallel sides and the perpendicular height between them. A slant side that leans is longer than the height, because it runs across at a slant, so it is not used for the area. (A slant side is used only when it is at right angles to the parallel sides, in a right trapezium, below: then it is the height.)

The slant sides are needed for the perimeter, the distance all the way round: the sum of the four sides. A dashed height is not one of the four sides, so it is not added for the perimeter. (In a right trapezium the height is one of the slant sides, so it is on the perimeter, once, as that side.)

Example 2: Choosing the height

The figure shows an isosceles trapezium. Find (a) its area, (b) its perimeter.

Solution

(a) The parallel sides are 6 cm and 12 cm. The perpendicular height between them is 4 cm. The 5 cm sides are slant sides and are not used for the area.

Area $= \frac{1}{2}(a + b)h$

$= \frac{1}{2} \times (6 + 12) \times 4$

$= \frac{1}{2} \times 18 \times 4$

$= 36\text{ cm}^{2}$

(b) The perimeter is the sum of the four sides, slant sides included. Here the dashed height is not one of the sides, so it is not added.

Perimeter $= 6 + 12 + 5 + 5 = 28\text{ cm}$

(a) The area is $36\text{ cm}^{2}$ . (b) The perimeter is $28\text{ cm}$ .

In a right trapezium, one slant side is at right angles to both parallel sides, so that slant side is itself the height. It is marked with a right angle at each end. The other slant side leans, as in any trapezium.

Example 3: A right trapezium

Find the area of the right trapezium.

Solution

The parallel sides are 5 cm and 9 cm. The 6 cm slant side meets both of them at right angles, so it is the perpendicular height. The 7.2 cm slant side leans, so it is not used.

Area $= \frac{1}{2} \times (5 + 9) \times 6$

$= \frac{1}{2} \times 14 \times 6$

$= 42\text{ cm}^{2}$

The area of the right trapezium is $42\text{ cm}^{2}$ .

A trapezium may be turned so that its parallel sides are upright. The height is then the horizontal distance between them. Find the parallel sides by their arrows first, whichever way they point, and then the height between them.

Example 4: Parallel sides upright

Find the area of the trapezium.

Solution

The arrows are on the two upright sides, so these are the parallel sides: 7 cm and 4 cm. The height is the perpendicular distance between them, the horizontal dashed line: 5 cm.

Area $= \frac{1}{2} \times (7 + 4) \times 5$

$= \frac{1}{2} \times 11 \times 5$

$= 27.5\text{ cm}^{2}$

The area of the trapezium is $27.5\text{ cm}^{2}$ .

Finding the Height or a Parallel Side

When the area is given and one length is not, put the numbers you know into $A = \frac{1}{2}(a + b)h$ and solve the equation for the unknown length. Then check by putting your answer back into the formula.

Example 5: Finding the height

The trapezium has an area of $84\text{ cm}^{2}$ . Its parallel sides are 10 cm and 18 cm. Find its perpendicular height, $h$ cm.

Solution

$A = \frac{1}{2}(a + b)h$

$84 = \frac{1}{2} \times (10 + 18) \times h$

$84 = \frac{1}{2} \times 28 \times h$

$84 = 14h$

$6 = h$ (divide both sides by 14)

Check: $\frac{1}{2} \times (10 + 18) \times 6 = 14 \times 6 = 84$ , the area given.

The perpendicular height is 6 cm.

Example 6: Finding a parallel side

The area of the trapezium is $52\text{ cm}^{2}$ . Find $x$ .

Solution

$A = \frac{1}{2}(a + b)h$

$52 = \frac{1}{2} \times (x + 9) \times 8$

$52 = 4(x + 9)$ (since $\frac{1}{2} \times 8 = 4$ )

$13 = x + 9$ (divide both sides by 4)

$4 = x$ (subtract 9 from both sides)

Check: $\frac{1}{2} \times (4 + 9) \times 8 = \frac{1}{2} \times 13 \times 8 = 52$ , the area given.

$x = 4$ : the other parallel side is 4 cm long.

Sometimes the parallel sides are not given as numbers, but by how they are related: one is so many centimetres longer than the other, or the two are in a given ratio. Write both sides in terms of one unknown, form an equation from the area, and solve it. Then find both sides.

Example 7: Parallel sides given by a relation

One parallel side of a trapezium is 4 cm longer than the other. The perpendicular height is 6 cm and the area is $54\text{ cm}^{2}$ . Find the lengths of the two parallel sides.

Solution

Let the shorter parallel side be $x$ cm. Then the longer side is $(x + 4)$ cm.

$\frac{1}{2} \times (x + x + 4) \times 6 = 54$

$3(2x + 4) = 54$ (since $\frac{1}{2} \times 6 = 3$ )

$6x + 12 = 54$

$6x = 42$ (subtract 12 from both sides)

$x = 7$ (divide both sides by 6)

The parallel sides are 7 cm and $7 + 4 = 11$ cm.

Check: $\frac{1}{2} \times (7 + 11) \times 6 = \frac{1}{2} \times 18 \times 6 = 54$ , the area given.

The parallel sides are 7 cm and 11 cm long.

A Trapezium as a Rectangle and Triangles

Drawing the height from each end of the shorter parallel side cuts a trapezium into a rectangle and two right-angled triangles. (In a right trapezium, one height is enough: a rectangle and one triangle.) Adding the areas of the parts gives the same area as the formula. It is a good check, and a way to find the area if you forget the formula.

Example 8: Two methods, one area

In trapezium $ABCD$ , $DC \parallel AB$ , $DC = 5$ cm, $AB = 13$ cm and the height is 6 cm. The heights from $D$ and $C$ meet $AB$ at $E$ and $F$ , and $AE = 3$ cm.

Find the area of $ABCD$ (a) by adding the areas of the rectangle and the two triangles, (b) by the formula.

Solution

(a) $DEFC$ is a rectangle, so $EF = DC = 5$ cm. The rest of $AB$ is $FB = 13 - 3 - 5 = 5$ cm.

Area of $\triangle ADE = \frac{1}{2} \times 3 \times 6 = 9\text{ cm}^{2}$

Area of rectangle $DEFC = 5 \times 6 = 30\text{ cm}^{2}$

Area of $\triangle BCF = \frac{1}{2} \times 5 \times 6 = 15\text{ cm}^{2}$

Area of $ABCD = 9 + 30 + 15 = 54\text{ cm}^{2}$

(b) Area $= \frac{1}{2} \times (5 + 13) \times 6 = \frac{1}{2} \times 18 \times 6 = 54\text{ cm}^{2}$

Both methods give the area of $ABCD$ as $54\text{ cm}^{2}$ .

Trapeziums on a Grid

On a square grid, count the gaps between the dots along each parallel side and across from one parallel side to the other. A slant side that leans crosses the grid at a slant, so its length cannot be counted, and it is not needed for the area.

Example 9: Counting on a dot grid

The dots are 1 cm apart. Find the area of the trapezium.

Solution

The arrows are on the two horizontal sides, so they are the parallel sides. The top side runs across 3 gaps, so it is 3 cm. The bottom side runs across 7 gaps, so it is 7 cm. From the bottom side up to the top side is 4 gaps, so the height is 4 cm.

Area $= \frac{1}{2} \times (3 + 7) \times 4$

$= \frac{1}{2} \times 10 \times 4$

$= 20\text{ cm}^{2}$

The area of the trapezium is $20\text{ cm}^{2}$ .

Mixed Units and Problems in Context

In a real problem the lengths may be given in different units. Change them all to one unit before using the formula ( $80\text{ cm} = 0.8\text{ m}$ , $1.2\text{ m} = 120\text{ cm}$ ). The area is then in the square of that unit. A cost given per $\text{m}^{2}$ needs the area in $\text{m}^{2}$ , so work in metres.

Example 10: The cross-section of a monsoon drain

A monsoon drain beside a road in Tampines has a cross-section in the shape of a trapezium. It is 1.6 m wide at the top, 80 cm wide at the bottom and 90 cm deep. Find the area of the cross-section in $\text{m}^{2}$ .

Solution

The top and the bottom of the drain are the parallel sides. The depth is the perpendicular height between them. The area is wanted in $\text{m}^{2}$ , so change every length to metres first:

$80\text{ cm} = 0.8\text{ m}$ and $90\text{ cm} = 0.9\text{ m}$

Area $= \frac{1}{2} \times (1.6 + 0.8) \times 0.9$

$= \frac{1}{2} \times 2.4 \times 0.9$

$= 1.2 \times 0.9$

$= 1.08\text{ m}^{2}$

The area of the cross-section of the drain is $1.08\text{ m}^{2}$ .

Example 11: Painting the side wall of a staircase

Mrs Goh is having the side wall of the staircase in her house painted. The wall is a trapezium. Its bottom edge is level with the floor and 4 m long. Its two upright ends are 1.2 m and 3.6 m tall, and its top edge follows the slope of the stairs. The painter charges $3.50 per $\text{m}^{2}$ . Find the cost of painting the wall.

Solution

The two upright ends are both at right angles to the floor, so they are parallel: they are the parallel sides, 1.2 m and 3.6 m. The bottom edge is a slant side at right angles to both, so it is the height: the perpendicular distance between them, 4 m.

Area $= \frac{1}{2} \times (1.2 + 3.6) \times 4$

$= \frac{1}{2} \times 4.8 \times 4$

$= 9.6\text{ m}^{2}$

Cost: $9.6 \times 3.50 = 33.60$

The cost of painting the wall is $33.60.

Common mistakes

Using a slant side as the height. The height is the perpendicular distance between the parallel sides. A slant side that leans is always longer than the height. Only in a right trapezium is one slant side at right angles to the parallel sides, and then that slant side is the height.
Forgetting the $\frac{1}{2}$ . $(a + b)h$ is the area of the parallelogram made by two copies of the trapezium: twice the answer.
Multiplying the parallel sides together, or using only one of them. Add the parallel sides first, then multiply the sum by $\frac{1}{2}$ and by $h$ .
Taking the wrong pair of sides as the parallel sides when the trapezium is turned. Look for the arrows first.
Mixing units. $\frac{1}{2} \times (1.6 + 80) \times 90$ means nothing. Change every length to one unit before using the formula.
Stopping at $x$ . When the parallel sides are $x$ cm and $(x + 4)$ cm, the question wants both lengths: find $x$ , then $x + 4$ .
Writing an area in cm, or a length in $\text{cm}^{2}$ . Lengths are in cm or m; areas are in $\text{cm}^{2}$ or $\text{m}^{2}$ .

Practice

The questions get harder as you go. Try each one before you open its solution.

Practice 1: The area from the figure

Find the area of the trapezium.

ShowHide solution

The parallel sides are 9 cm and 15 cm, and the height is 8 cm.

Area $= \frac{1}{2} \times (9 + 15) \times 8$

$= \frac{1}{2} \times 24 \times 8$

$= 96\text{ cm}^{2}$

The area is $96\text{ cm}^{2}$ .

Practice 2: On a dot grid

The dots are 1 cm apart. Find the area of the trapezium.

ShowHide solution

The arrows are on the upright sides, so they are the parallel sides. The left side spans 2 gaps, so it is 2 cm. The right side spans 6 gaps, so it is 6 cm. Across from one to the other is 4 gaps, so the height is 4 cm.

Area $= \frac{1}{2} \times (2 + 6) \times 4 = \frac{1}{2} \times 8 \times 4 = 16\text{ cm}^{2}$

The area is $16\text{ cm}^{2}$ .

Practice 3: Two triangles

In trapezium $ABCD$ , $DC \parallel AB$ , $DC = 4$ cm, $AB = 10$ cm and the height is 6 cm. The diagonal $AC$ cuts the trapezium into two triangles.

(a) Find the area of $\triangle ABC$ and the area of $\triangle ACD$ .

(b) Hence find the area of $ABCD$ , and check your answer with the formula.

ShowHide solution

(a) $\triangle ABC$ has base $AB = 10$ cm. Its height is the distance from $C$ to $AB$ , 6 cm.

Area of $\triangle ABC = \frac{1}{2} \times 10 \times 6 = 30\text{ cm}^{2}$

$\triangle ACD$ has base $DC = 4$ cm. Its height is the distance from $A$ to the line $DC$ (produced). Since $DC \parallel AB$ , this is also the distance between the parallel sides, 6 cm.

Area of $\triangle ACD = \frac{1}{2} \times 4 \times 6 = 12\text{ cm}^{2}$

(b) Area of $ABCD = 30 + 12 = 42\text{ cm}^{2}$

Check with the formula: $\frac{1}{2} \times (4 + 10) \times 6 = \frac{1}{2} \times 14 \times 6 = 42\text{ cm}^{2}$ , the same.

(a) $30\text{ cm}^{2}$ and $12\text{ cm}^{2}$ . (b) The area of $ABCD$ is $42\text{ cm}^{2}$ .

Practice 4: The height among the sides

The figure shows an isosceles trapezium. Find (a) its area, (b) its perimeter.

ShowHide solution

(a) The parallel sides are 8 cm and 14 cm. The perpendicular height is 4 cm; the 5 cm sides are slant sides.

Area $= \frac{1}{2} \times (8 + 14) \times 4 = \frac{1}{2} \times 22 \times 4 = 44\text{ cm}^{2}$

(b) Perimeter $= 8 + 14 + 5 + 5 = 32\text{ cm}$

(a) The area is $44\text{ cm}^{2}$ . (b) The perimeter is $32\text{ cm}$ .

Practice 5: A right trapezium

Find (a) the area, (b) the perimeter of the right trapezium.

ShowHide solution

(a) The parallel sides are 7 cm and 10 cm. The 4 cm slant side is at right angles to both, so it is the perpendicular height. The 5 cm slant side leans, so it is not used for the area.

Area $= \frac{1}{2} \times (7 + 10) \times 4 = \frac{1}{2} \times 17 \times 4 = 34\text{ cm}^{2}$

(b) In a right trapezium the height is also a slant side, so here it is part of the perimeter, counted once.

Perimeter $= 7 + 10 + 4 + 5 = 26\text{ cm}$

(a) The area is $34\text{ cm}^{2}$ . (b) The perimeter is $26\text{ cm}$ .

Practice 6: Parallel sides upright

Find the area of the trapezium.

ShowHide solution

The arrows are on the upright sides, so the parallel sides are 12 cm and 6 cm. The perpendicular height is the horizontal distance between them, 4 cm. The two 5 cm sides are slant sides and are not used.

Area $= \frac{1}{2} \times (12 + 6) \times 4 = \frac{1}{2} \times 18 \times 4 = 36\text{ cm}^{2}$

The area is $36\text{ cm}^{2}$ .

Practice 7: A rectangle and a triangle

$ABCD$ is a right trapezium with $DC \parallel AB$ .

(a) Draw the height from $C$ to $AB$ , meeting $AB$ at $E$ . Find the area of $ABCD$ by adding the areas of the rectangle and the triangle.

(b) Check your answer with the formula.

ShowHide solution

(a) $AECD$ is a rectangle, so $AE = DC = 8$ cm and $CE = AD = 5$ cm. Then $EB = 12 - 8 = 4$ cm.

Area of rectangle $AECD = 8 \times 5 = 40\text{ cm}^{2}$

Area of $\triangle EBC = \frac{1}{2} \times 4 \times 5 = 10\text{ cm}^{2}$

Area of $ABCD = 40 + 10 = 50\text{ cm}^{2}$

(b) Area $= \frac{1}{2} \times (8 + 12) \times 5 = \frac{1}{2} \times 20 \times 5 = 50\text{ cm}^{2}$ , the same.

The area of $ABCD$ is $50\text{ cm}^{2}$ by both methods.

Practice 8: A plot in the eco-garden

A plot in a school's eco-garden is a trapezium with parallel sides 11 m and 14 m. Its area is $75\text{ m}^{2}$ . Find the perpendicular distance, $h$ m, between the parallel sides.

ShowHide solution

$75 = \frac{1}{2} \times (11 + 14) \times h$

$75 = \frac{1}{2} \times 25 \times h$

$75 = 12.5h$

$6 = h$ (divide both sides by 12.5)

Check: $\frac{1}{2} \times 25 \times 6 = 12.5 \times 6 = 75$ , the area given.

The parallel sides are 6 m apart.

Practice 9: Finding a parallel side

The area of the trapezium is $90\text{ cm}^{2}$ . Find $x$ .

ShowHide solution

$90 = \frac{1}{2} \times (x + 12) \times 10$

$90 = 5(x + 12)$ (since $\frac{1}{2} \times 10 = 5$ )

$18 = x + 12$ (divide both sides by 5)

$6 = x$ (subtract 12 from both sides)

Check: $\frac{1}{2} \times (6 + 12) \times 10 = \frac{1}{2} \times 18 \times 10 = 90$ , the area given.

$x = 6$ : the other parallel side is 6 cm long.

Practice 10: Parallel sides in a ratio

The parallel sides of a trapezium are in the ratio $2 : 3$ . The perpendicular height is 8 cm and the area is $100\text{ cm}^{2}$ . Find the lengths of the parallel sides.

ShowHide solution

The ratio $2 : 3$ means the parallel sides are $2x$ cm and $3x$ cm for some number $x$ .

$\frac{1}{2} \times (2x + 3x) \times 8 = 100$

$\frac{1}{2} \times 5x \times 8 = 100$

$20x = 100$

$x = 5$ (divide both sides by 20)

The parallel sides are $2 \times 5 = 10$ cm and $3 \times 5 = 15$ cm.

Check: $10 : 15 = 2 : 3$ , and $\frac{1}{2} \times (10 + 15) \times 8 = \frac{1}{2} \times 25 \times 8 = 100$ .

The parallel sides are 10 cm and 15 cm long.

Practice 11: Turfing a lawn

Siti's class looks after a lawn in a community garden in Bishan. The lawn is a trapezium with parallel sides 6 m and 450 cm, and the perpendicular distance between them is 4 m. Turf costs $5.50 per $\text{m}^{2}$ . Find the cost of turfing the whole lawn.

ShowHide solution

The cost is per $\text{m}^{2}$ , so work in metres: $450\text{ cm} = 4.5\text{ m}$ .

Area $= \frac{1}{2} \times (6 + 4.5) \times 4$

$= \frac{1}{2} \times 10.5 \times 4$

$= 21\text{ m}^{2}$

Cost: $21 \times 5.50 = 115.50$

The cost of turfing the lawn is $115.50.

Practice 12: The cross-section of a canal

A canal in Punggol has a cross-section in the shape of a trapezium. The canal is 3 m deep, its top is 4 m wider than its bottom, and the area of the cross-section is $42\text{ m}^{2}$ . Find the width of the canal at the bottom and at the top.

ShowHide solution

Let the width at the bottom be $x$ m. Then the width at the top is $(x + 4)$ m. These are the parallel sides, and the depth, 3 m, is the height.

$\frac{1}{2} \times (x + x + 4) \times 3 = 42$

$\frac{1}{2} \times (2x + 4) \times 3 = 42$

$(x + 2) \times 3 = 42$ (half of $2x + 4$ is $x + 2$ )

$3x + 6 = 42$

$3x = 36$ (subtract 6 from both sides)

$x = 12$ (divide both sides by 3)

The bottom is 12 m wide and the top is $12 + 4 = 16$ m wide.

Check: $\frac{1}{2} \times (12 + 16) \times 3 = \frac{1}{2} \times 28 \times 3 = 42$ , the area given.

The canal is 12 m wide at the bottom and 16 m wide at the top.

Key Points

A trapezium has exactly one pair of parallel sides, marked with arrows; the other two sides are its slant sides.
The height of a trapezium is the perpendicular distance between its parallel sides, not a slant side that leans.
Area of a trapezium $= \frac{1}{2} \times$ (sum of parallel sides) $\times$ height: $A = \frac{1}{2}(a + b)h$ .
Why: a diagonal cuts it into two triangles, $\frac{1}{2}ah + \frac{1}{2}bh$ ; or two copies make a parallelogram of base $a + b$ and height $h$ , and the trapezium is half of it.
In a right trapezium, one slant side is at right angles to the parallel sides, and that slant side is the height. When the parallel sides are upright, the height is horizontal.
A trapezium can also be cut into a rectangle and triangles; the parts add up to the same area.
To find the height or a parallel side, put the known values into the formula, solve the equation and check.
Parallel sides given by a relation are written in one unknown ( $x$ and $x + 4$ , or $2x$ and $3x$ ); find both sides at the end.
Change all lengths to one unit before using the formula; a cost per $\text{m}^{2}$ needs lengths in metres.
The perimeter is the sum of all four sides; a dashed height is not part of it (in a right trapezium the height is a slant side, and is counted once, as that side).