Maths · 6 Mensuration

6.2 Area and perimeter

Before you start

Metric units of length (mm, cm, m, km) and converting between them, e.g. $90$ cm $= 0.9$ m.
Multiplying and dividing decimals, with and without a calculator.
Pythagoras' theorem, $a^{2} + b^{2} = c^{2}$ , for finding a missing side of a right-angled triangle.
Solving a simple linear equation such as $5 x + 15 = 55$ .

Syllabus objectives covered (1)

Carry out calculations involving the perimeter and area of a rectangle, triangle, parallelogram and trapezium.

Part 1 of 7

Perimeter: the distance round the outside

The perimeter of a shape is the total length of its boundary: the distance you would walk if you went once round the edge. To find it, add the lengths of every side.
Perimeter is a length, so its unit is a length unit: cm, m, km. It is never cm² or m².
A rectangle has two lengths and two widths. So its perimeter is twice the sum of one length and one width.
$P = 2 (l + w)$
For a rectangle $8$ cm by $5$ cm: $P = 2 (8 + 5) = 2 \times 13 = 26$ cm.
The trap is a side with no label on it. Before you add, count the sides of the shape and check that you have that many numbers. A parallelogram has four sides even if only two are labelled, because opposite sides are equal.
No perimeter rule is printed on the list of formulas. You do not need one: perimeter is always the sum of the sides.

Part 2 of 7

Area: counting squares, and the rectangle

Area measures how much flat surface a shape covers. It is counted in unit squares: $1$ cm² is a square $1$ cm by $1$ cm.
A rectangle $8$ cm long and $5$ cm wide holds $5$ rows of $8$ one-centimetre squares. That is $5 \times 8 = 40$ squares, so its area is $40$ cm².
Rows times squares in a row is length times width. That is where the formula comes from.
$A = l \times w$
The unit is squared because you multiplied a length by a length: cm $\times$ cm $=$ cm². Write the squared unit on every area answer.
A square is a rectangle with all four sides equal. For side $s$ : area $s^{2}$ , perimeter $4 s$ .
Both lengths must be in the same unit before you multiply. For a strip $3$ m by $40$ cm, change $40$ cm to $0.4$ m first: $3 \times 0.4 = 1.2$ m². Multiplying $3 \times 40$ gives $120$ , which is neither m² nor cm².
The rectangle area formula is not on the list of formulas. You must recall it.

Part 3 of 7

Triangle: half of a rectangle

Draw a rectangle round a triangle, with the same base and the same height. The triangle fills exactly half of that rectangle.
To see why, draw the height from the top vertex down to the base. It cuts the rectangle into two smaller rectangles. Each one is cut in half by a side of the triangle, so the triangle has half of each.
So the area of a triangle is half of base times height.
$A = \frac{1}{2} \times b \times h$
Here $h$ is the perpendicular height: the distance straight from the base to the opposite vertex, meeting the base at a right angle. It is not a sloping side of the triangle.
Any side can be the base. The height you use must be perpendicular to the side you chose.
In a right-angled triangle, the two sides that meet at the right angle are a base and its perpendicular height. Sides $6$ cm and $8$ cm at the right angle give $\frac{1}{2} \times 6 \times 8 = 24$ cm².
In a triangle with an obtuse angle, the perpendicular height can fall outside the triangle. The base is still only the side of the triangle, not the side plus the extension you drew to meet the height.
graph diagram — not drawn yet
An obtuse-angled triangle ABC with base AB horizontal and labelled 6 cm, vertex C up and to the right beyond B. AB extended to the right as a dashed line to a point N directly below C. A vertical dashed line CN labelled h = 4 cm with a right-angle mark at N.
The height meets the extended base outside the triangle, but the base in the formula is still AB = 6 cm, so the area is 12 cm².
The area of a triangle is the one formula in this subtopic that is printed on the list of formulas on papers 2 and 4. You still need to know which length is $h$ , because the list does not pick it out of a diagram for you.
The perimeter of a triangle is the sum of its three sides. If a right-angled triangle has a side missing, find it with Pythagoras first.

Part 4 of 7

Parallelogram: base times perpendicular height

A parallelogram has two pairs of parallel sides. Opposite sides are equal in length.
Cut the right-angled triangle off one end, along the perpendicular height, and slide it to the other end. The pieces make a rectangle with the same base and the same height. Nothing was added or taken away, so the area is the same.
graph diagram — not drawn yet
Two diagrams side by side. Left: a parallelogram with horizontal base b, a dashed vertical height h dropped from the top-left vertex with a right-angle mark, and the small right-angled triangle to the left of the dashed line shaded. Right: the same shaded triangle moved to the right-hand end, so the shape is now a rectangle of width b and height h.
The parallelogram and the rectangle have the same base, the same height and the same area.
So the area is base times perpendicular height. There is no half, because the shape is a whole rectangle rearranged, not half of one.
$A = b \times h$
Here $h$ is the perpendicular distance between the two parallel sides. A parallelogram with base $9$ cm, slant side $5$ cm and perpendicular height $4$ cm has area $9 \times 4 = 36$ cm², not $9 \times 5 = 45$ cm².
The slant side is always longer than the height. It is the hypotenuse of the right-angled triangle you cut off, and the height is one of its shorter sides. So an area worked out with the slant side always comes out too big.
The perimeter uses the slant sides, because they are part of the boundary: $2 (9 + 5) = 28$ cm. The height is not on the boundary at all. So the area uses the height and the perimeter uses the slant side.
The parallelogram area formula is not on the list of formulas. Recall it from the cut-and-slide picture.

Part 5 of 7

Trapezium: the mean of the parallel sides

A trapezium has exactly one pair of parallel sides. Call their lengths $a$ and $b$ . The perpendicular distance between them is the height $h$ .
Take two identical trapezia and turn one of them upside down. Put them side by side and they make a parallelogram. Its base is $a + b$ and its height is $h$ .
graph diagram — not drawn yet
A trapezium with parallel horizontal sides, top labelled a and bottom labelled b, and a dashed perpendicular height h. Next to it a second identical trapezium rotated half a turn, placed so the two together form a parallelogram whose bottom edge is labelled b + a and whose top edge is labelled a + b.
Two copies of the trapezium make a parallelogram of base a + b and height h, so one trapezium is half of it.
One trapezium is half of that parallelogram, so its area is half of $(a + b) \times h$ .
$A = \frac{1}{2} (a + b) h$
Another way to read it: $\frac{1}{2} (a + b)$ is the mean of the two parallel sides. The trapezium has the same area as a rectangle that wide and $h$ tall.
For parallel sides $6$ cm and $10$ cm, $4$ cm apart: $\frac{1}{2} \times (6 + 10) \times 4 = \frac{1}{2} \times 16 \times 4 = 32$ cm².
Check it a second way. A diagonal splits the trapezium into two triangles, both of height $4$ cm, with bases $6$ cm and $10$ cm: $\frac{1}{2} \times 6 \times 4 + \frac{1}{2} \times 10 \times 4 = 12 + 20 = 32$ cm². Same answer.
Add the parallel sides first. If you type $0.5 \times 6 + 10 \times 4$ into a calculator without the bracket, it halves only the $6$ and gives $43$ . Use brackets, or add $6 + 10$ before anything else.
The two sloping sides play no part in the area. You need them only for the perimeter.
The trapezium area formula is not on the list of formulas. Recall it, or rebuild it from the two-copies picture.

Part 6 of 7

Working backwards from an area

Sometimes the area is given and a length is missing. Write the area formula, put in every value you know, and solve the equation for the length you do not know.
A rectangle has area $45$ cm² and length $9$ cm. Then $9 \times w = 45$ . Divide both sides by $9$ , because $9$ is what multiplies $w$ : $w = 5$ cm.
A triangle has area $24$ cm² and base $8$ cm. Then $\frac{1}{2} \times 8 \times h = 24$ , which is $4 h = 24$ , so $h = 6$ cm.
$\frac{1}{2} \times 8 \times h = 24 \Rightarrow 4 h = 24 \Rightarrow h = 6$
The trap here is the half. If you write $8 h = 24$ you get $h = 3$ , which is half the true height. Simplify $\frac{1}{2} \times 8$ to $4$ before you divide.
A trapezium has area $60$ cm² and parallel sides $5$ cm and $7$ cm. Then $\frac{1}{2} \times 12 \times h = 60$ , which is $6 h = 60$ , so $h = 10$ cm.
Check a backwards answer by putting it back into the formula: $\frac{1}{2} \times (5 + 7) \times 10 = 60$ . It gives the area you started with.
The lengths can be algebraic, for example sides $x$ and $x + 6$ . The method does not change: substitute into the formula and solve the equation you get.

Part 7 of 7

Compound shapes: split, or subtract

A compound shape is made of simpler shapes joined together. Compound shapes are syllabus subtopic 6.5, not 6.2. They are here as practice, because they use only the four shapes above. You find the area of a compound shape in one of two ways.
Split: cut the shape into rectangles, triangles or trapezia, find each area, and add.
Subtract: draw the large rectangle that encloses the shape, find its area, and take away the area of the piece that is missing.
Both methods give the same answer. Use the one that needs fewer missing lengths, and use the other one as a check.
graph diagram — not drawn yet
An L-shaped floor plan made from a 10 m by 8 m rectangle with a 4 m by 3 m rectangle removed from the top-right corner. Outside edges labelled anticlockwise starting with the bottom edge: bottom 10 m, right side 5 m, inner horizontal step 4 m, inner vertical step 3 m, top 6 m, left side 8 m. A dashed horizontal line 5 m above the bottom splits the shape into a 10 m by 5 m rectangle and a 6 m by 3 m rectangle.
The dashed split line is used for the area only. It is inside the shape, so it is not part of the perimeter.
Before you split, find any missing lengths. In a shape whose sides are all horizontal or vertical, the horizontal edges along the top add up to the horizontal edges along the bottom. The same is true for the vertical edges on the left and right.
Write each piece's length and width on the diagram as you find them, and write each piece's area as a product, such as $10 \times 5$ . Method marks are for a valid method, so the examiner needs to see which numbers you multiplied.
For the perimeter of a compound shape, add only the outside edges. A line you drew to split the shape is inside it, so it is not part of the boundary.
A useful check: if a rectangular piece is cut from a corner of a rectangle, the perimeter does not change. The two edges of the cut replace the two pieces of edge that were removed. This is not true for a notch cut into the middle of a side, which adds two extra edges.

Worked examples

Example 1. A parallelogram has a base of $12$ cm and slant sides of $7$ cm. The perpendicular height from the base to the opposite side is $5.5$ cm. Work out (a) the area and (b) the perimeter of the parallelogram.

Show the working

Decide which lengths each answer needs before you calculate. The area needs the base and the perpendicular height. The perimeter needs the base and the slant side.
(a) Area is base times perpendicular height. The $7$ cm is not used, because it is not perpendicular to the base.
$A = 12 \times 5.5 = 66 {cm}^{2}$
(b) The perimeter has two sides of $12$ cm and two of $7$ cm. The $5.5$ cm is not used, because it is not on the boundary.
$P = 2 (12 + 7) = 2 \times 19 = 38 cm$
Check the units: the area is in cm² and the perimeter in cm.

Answer: (a) $66$ cm² (b) $38$ cm

Example 2. Trapezium $A B C D$ has $A B$ parallel to $D C$ . $A B = 15$ cm, $D C = 9$ cm, and side $A D = 8$ cm is perpendicular to both $A B$ and $D C$ . Work out (a) the area and (b) the perimeter of the trapezium.

Show the working

Sketch it: $A B$ along the bottom, $A D$ straight up from $A$ , and $D C$ along the top from $D$ . Side $B C$ slopes down from $C$ to $B$ .
(a) The parallel sides are $15$ cm and $9$ cm. $A D$ is perpendicular to both, so it is the height. Add the parallel sides first.
$A = \frac{1}{2} \times (15 + 9) \times 8 = \frac{1}{2} \times 24 \times 8 = 96 {cm}^{2}$
(b) The perimeter needs $B C$ , which is not given. Draw a vertical line down from $C$ to $A B$ . It makes a right-angled triangle with height $8$ cm.
The bottom of that triangle is the part of $A B$ that sticks out beyond $D C$ : $15 - 9 = 6$ cm. $B C$ is the hypotenuse.
$B C = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 cm$
Add all four outside edges.
$P = 15 + 10 + 9 + 8 = 42 cm$

Answer: (a) $96$ cm² (b) $42$ cm

Example 3. A floor is the shape of a $10$ m by $8$ m rectangle with a $4$ m by $3$ m rectangle removed from one corner. Work out (a) the area and (b) the perimeter of the floor.

Show the working

(a) Use the subtract method: the whole rectangle minus the missing corner.
$A = 10 \times 8 - 4 \times 3 = 80 - 12 = 68 m^{2}$
Check with the split method. Cut horizontally where the corner starts. The bottom piece is the full width, $10$ m, and $8 - 3 = 5$ m tall. The top piece is $10 - 4 = 6$ m wide and $3$ m tall.
$10 \times 5 + 6 \times 3 = 50 + 18 = 68 m^{2}$
(b) The L-shape has six outside edges. Two of them are not given, so find them first. The short vertical edge is $8 - 3 = 5$ m. The short horizontal edge is $10 - 4 = 6$ m.
Go round the shape and add all six edges once each.
$P = 10 + 5 + 4 + 3 + 6 + 8 = 36 m$
Check: the piece was cut from a corner, so the perimeter should equal the full rectangle's, $2 (10 + 8) = 36$ m. It does.

Answer: (a) $68$ m² (b) $36$ m

Example 4. A trapezium has an area of $75$ cm². Its parallel sides are $7$ cm and $13$ cm long. Work out the perpendicular distance between the parallel sides.

Show the working

Write the trapezium formula with every known value in it. The unknown is $h$ .
$\frac{1}{2} \times (7 + 13) \times h = 75$
Simplify the left side: $7 + 13 = 20$ , and half of $20$ is $10$ .
$10 h = 75$
Divide both sides by $10$ , because $10$ is what multiplies $h$ .
$h = 7.5$
Check by putting it back in: $\frac{1}{2} \times 20 \times 7.5 = 75$ . It gives the area you started with.

Answer: $7.5$ cm

Where people go wrong

Mistake: Multiplying the base by the slant side to find the area of a parallelogram or triangle.

Instead: Use the perpendicular height: the distance at right angles to the base.

The slant side is the hypotenuse of a right-angled triangle that has the height as a shorter side, so it is always longer than the height and gives an area that is too big.

Mistake: Leaving out the $\frac{1}{2}$ for a triangle or trapezium, or halving only one of the parallel sides.

Instead: Triangle: $\frac{1}{2} \times b \times h$ . Trapezium: add $a + b$ first, then halve, then multiply by $h$ .

A triangle is half a rectangle, and a trapezium is half a parallelogram. Without the half, the answer is exactly double. Without the bracket, only one side is halved.

Mistake: Multiplying lengths in different units, such as metres by centimetres.

Instead: Change every length to the same unit before you multiply.

$3$ m $\times 40$ cm is not $120$ of anything. As $3 \times 0.4$ it is $1.2$ m²; as $300 \times 40$ it is $12 000$ cm². Both are the same area.

Mistake: Including the split line in the perimeter of a compound shape, or leaving out an unlabelled edge.

Instead: Add only the outside edges, and work out any edge that has no label before you add.

Perimeter is the boundary. A line drawn inside the shape is not on it, and an edge with no number is still on it.

Mistake: Writing the unit of an area as cm, or the unit of a perimeter as cm².

Instead: Area: squared units (cm², m²). Perimeter: length units (cm, m).

Area is a length times a length. Perimeter is lengths added together.

Check yourself

A rectangle is $7.5$ cm long and $4$ cm wide. Work out its area and its perimeter.
Answer
Area $7.5 \times 4 = 30$ cm². Perimeter $2 (7.5 + 4) = 23$ cm.
A triangle has a base of $9$ cm, a perpendicular height of $6$ cm, and two other sides of $7.5$ cm each. Work out its area and its perimeter.
Answer
Area $\frac{1}{2} \times 9 \times 6 = 27$ cm². Perimeter $9 + 7.5 + 7.5 = 24$ cm.
A parallelogram has a base of $14$ m, slant sides of $7.5$ m and a perpendicular height of $6$ m. Work out its area and its perimeter.
Answer
Area $14 \times 6 = 84$ m². Perimeter $2 (14 + 7.5) = 43$ m.
A trapezium has parallel sides of $5.2$ cm and $8.8$ cm, and the perpendicular distance between them is $4.5$ cm. Work out its area.
Answer
$\frac{1}{2} \times (5.2 + 8.8) \times 4.5 = \frac{1}{2} \times 14 \times 4.5 = 31.5$ cm².
A rectangle has area $54$ cm² and length $9$ cm. Work out its perimeter.
Answer
Width $54 \div 9 = 6$ cm, so perimeter $2 (9 + 6) = 30$ cm.
A rectangular board is $2.4$ m long and $85$ cm wide. Work out its area in m².
Answer
$85$ cm $= 0.85$ m, so area $2.4 \times 0.85 = 2.04$ m².
A triangle has area $36$ cm² and perpendicular height $8$ cm. Work out the length of its base.
Answer
$\frac{1}{2} \times b \times 8 = 36$ , so $4 b = 36$ and $b = 9$ cm.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A parallelogram $A B C D$ has $A B = 11$ cm and $A D = 6$ cm. The perpendicular distance between $A B$ and $D C$ is $4.8$ cm.

(a) Work out the perimeter of the parallelogram. 2 marks No calculator

Mark scheme

M1 11 + 6 + 11 + 6 or 2 × (11 + 6) oe
A1 34 [cm]

(b) Work out the area of the parallelogram. 2 marks No calculator

Mark scheme

M1 11 × 4.8
A1 52.8 [cm²]

(c) Explain why the area is not $11 \times 6$ . 1 mark No calculator

Mark scheme

B1 AD (6 cm) is a slant side, not perpendicular to AB, so it is not the height oe

Question 2. A trapezium has parallel sides of $10$ cm and $22$ cm. Its two sloping sides are each $10$ cm long, and the trapezium is symmetrical.

(a) Show that the perpendicular distance between the parallel sides is $8$ cm. 3 marks No calculator

Mark scheme

M1 (22 − 10) ÷ 2 = 6 for the part of the longer side beyond each end of the shorter side
M1 10² − 6² oe, with the sloping side as the hypotenuse
A1 √64 = 8 with no errors seen; nfww

(b) Calculate the area of the trapezium. 2 marks No calculator

Mark scheme

M1 ½ × (10 + 22) × 8 oe
A1 128 [cm²]

(c) Write down the perimeter of the trapezium. 1 mark No calculator

Mark scheme

B1 52 [cm]

Question 3. A wall is a rectangle $4.2$ m wide and $2.5$ m high. It contains a rectangular window $1.2$ m wide and $90$ cm high. The wall, but not the window, is to be painted.

(a) Calculate the area to be painted. Give your answer in m². 3 marks

Mark scheme

B1 90 cm = 0.9 m soi, or all lengths changed to cm
M1 4.2 × 2.5 − 1.2 × 0.9 oe, FT their conversion
A1 9.42 [m²] cao

(b) One tin of paint covers $4$ m². Work out the least number of tins needed to paint the wall once. 2 marks

Mark scheme

M1 their (a) ÷ 4, e.g. 9.42 ÷ 4 = 2.355
A1 3, or FT their (a) ÷ 4 rounded up to the next whole number

Question 4. A trapezium has parallel sides of length $x$ cm and $(x + 6)$ cm. The perpendicular distance between them is $5$ cm. The area of the trapezium is $55$ cm².

(a) Find the value of $x$ . 3 marks No calculator

Mark scheme

M1 ½ × (x + x + 6) × 5 = 55 oe
M1 a correct first step towards solving, e.g. 2x + 6 = 22 or 5x + 15 = 55
A1 x = 8

(b) Write down the length of the longer parallel side. 1 mark No calculator

Mark scheme

B1 14 [cm], FT their x + 6

How this is examined

Finding a missing length when the area is given: A Find or Work out part that gives the area of a triangle, rectangle or trapezium and asks for a side or the height, sometimes with the sides in terms of $x$ , worth about 3 marks The area formula written with the given area put in, e.g. $\frac{1}{2} (x + x + 6) \times 5 = 55$ , is the method; a correct next step such as $5 x + 15 = 55$ shows the solving; the value of the length is the answer.
Perimeter when some edges are not labelled: A Work out or Calculate part on a compound shape or a trapezium, where an edge comes from subtracting known lengths or from Pythagoras, worth about 2 to 3 marks Each missing edge worked out and shown ( $8 - 3 = 5$ , or $\sqrt{6^{2} + 8^{2}} = 10$ ), then every outside edge added once, with no inside split lines, and a length unit on the answer.
Area of a trapezium from the two parallel sides and the distance between them: A 2-mark Calculate or Work out on a drawn trapezium, where the sloping sides may be labelled as well as the height $\frac{1}{2} \times (a + b) \times h$ written with the two parallel sides and the perpendicular height in it shows the method; the sloping sides must not appear. You must recall this formula, because it is not on the list of formulas.
Using the perpendicular height, not the slant side: A Calculate on a parallelogram or triangle with both a slant side and a height labelled, sometimes followed by a 1-mark Explain Base $\times$ perpendicular height for a parallelogram, $\frac{1}{2} \times$ base $\times$ perpendicular height for a triangle. For the Explain: a statement that the slant side is not perpendicular to the base, so it is not the height.
Area of a compound shape, by splitting it or by subtracting a missing piece: A Calculate or Work out part on an L-shape or a rectangle with a triangle or trapezium attached, with some lengths left for you to find, worth about 3 to 4 marks. This is syllabus 6.5 (compound shapes), but every piece is a 6.2 shape The areas of the pieces written out with their numbers, e.g. $10 \times 5 + 6 \times 3$ or $10 \times 8 - 4 \times 3$ , shows the method; each piece must use lengths from the diagram or ones you worked out from it. The final value with a squared unit ( $68$ m²) is the answer.

Words to know

Perimeter: The total length of the boundary of a shape. Measured in length units such as cm or m.
Area: The amount of flat surface a shape covers. Measured in squared units such as cm² or m².
Perpendicular height: The distance from a base to the opposite vertex or side, measured at right angles to the base.
Parallelogram: A quadrilateral with two pairs of parallel sides. Area $= b \times h$ .
Trapezium: A quadrilateral with exactly one pair of parallel sides. Area $= \frac{1}{2} (a + b) h$ , where $a$ and $b$ are the parallel sides.
Compound shape: A shape made by joining simpler shapes. Find its area by splitting it into pieces, or by subtracting a missing piece from a larger shape.

All Maths lessons Next: 6.3 Circles, arcs and sectors