Chapter 6 — Measuring Space: Perimeter and Area

The perimeter of a closed plane shape is the total length of its boundary, the distance around it. If you walk once round the edge of a field, the distance you walk is its perimeter. Perimeter is a length, so it is measured in units of length such as centimetres (cm), metres (m) or kilometres (km).

For a shape made of straight sides, the perimeter is the sum of the lengths of all its sides.

Composite shapes. When a shape is made by joining simpler shapes, the perimeter is the length of the outer boundary only. The sides where two parts are joined are inside the shape and are not counted. For example, two squares of side 4 cm placed side by side make a rectangle 8 cm by 4 cm. The perimeter is 2(8 + 4) = 24 cm, and not 16 + 16 = 32 cm, because the shared side of 4 cm on each square is not part of the outer boundary.

Using the units. If a question gives some lengths in metres and some in centimetres, convert them to the same unit first. A rectangular garden 15 m by 8 m needs a fence along its whole boundary, so we need 2(15 + 8) = 46 m of fencing. If the fencing costs ₹25 per metre, the total cost is 46 × 25 = ₹1150.

Perimeter and area are different. A rectangle 8 cm by 2 cm and a rectangle 5 cm by 5 cm have the same perimeter, of 20 cm and 20 cm. But their areas are 16 cm² and 25 cm², which are different. Two shapes can have the same perimeter and different areas, and the reverse is also possible. Never assume that one number tells you the other.

In the next topics we meet the shape for which the perimeter is not found by simply adding sides, the circle, and the number π that is hidden inside every circle.

The perimeter of a circle is given a special name: the circumference. Because a circle has no straight sides, we need another way to find it.

An experiment. Take several round objects, such as a bangle, a bottle cap, a plate and a coin. Wrap a thread once around each, mark the length, and measure it against a ruler. Measure the distance across the circle through its centre, the diameter, as well. Then divide the circumference by the diameter for each object. Whatever the size of the circle, the answer comes out close to 3.14. The ratio of the circumference to the diameter is the same for every circle. This constant is called π (pi).

C ÷ d = π, so C = πd = 2πr, where r is the radius and d = 2r.

The value of π. The number π is about 3.14159265… For calculations, we usually use 3.14 or 22/7. The fraction 22/7 is 3.142857…, which is close but not equal to π. In this book, you must use the value stated in the question.

π is irrational. No fraction p/q of two whole numbers equals π exactly, and its decimal expansion never ends and never repeats. That is, π is an irrational number. This was proved by Johann Lambert in 1761. The fractions 22/7 and 355/113 are only approximations, and the digits of π go on forever without a pattern.

A history of π. People have tried to find π accurately for thousands of years:

• The Greek mathematician Archimedes (about 250 BCE) trapped π between 223/71 and 22/7 by fitting polygons inside and outside a circle.
• The Indian mathematician Aryabhata (499 CE) gave a value equal to 62832/20000 = 3.1416 for the ratio of the circumference of a circle to its diameter.
• The Chinese mathematician Zu Chongzhi (5th century CE) found the fraction 355/113 = 3.1415929…, which is accurate to six decimal places.

Which approximation is better? 22/7 = 3.142857 differs from π by about 0.0013, while 355/113 differs by only 0.0000003. The second is far closer.

Using the formula. A circle of radius 7 cm has circumference 2 × (22/7) × 7 = 44 cm. A wheel of diameter 70 cm makes one turn over a distance of π × 70 = 220 cm, using 22/7. So to travel 22 m (2200 cm), it must turn 2200 ÷ 220 = 10 times.

An arc is a part of the circumference of a circle. If you cut a slice of a round cake, the curved edge of the slice is an arc. The two radii that bound the slice, and the arc between them, make a sector. The angle between the radii at the centre is called the central angle or the angle of the sector, written θ (theta).

The idea. The whole circle goes round through 360°, and its circumference is 2πr. An arc that is cut off by a central angle of θ is the fraction θ ÷ 360 of the whole circle. So its length is the same fraction of the whole circumference:

Length of arc = (θ ÷ 360) × 2πr

Examples. For a half-circle, θ = 180°, and the length is (180 ÷ 360) × 2πr = πr. For a quarter of a circle, θ = 90°, the arc is (1/4) × 2πr = πr ÷ 2. For a circle of radius 14 cm and θ = 90°, the arc length is (90 ÷ 360) × 2 × (22/7) × 14 = (1/4) × 88 = 22 cm.

Check this with another: for a circle of radius 21 cm and θ = 60°, the arc length is (60 ÷ 360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm.

Perimeter of a sector. The boundary of a sector is made of the arc and the two radii. So the perimeter of a sector is arc length + 2r. For the sector of radius 14 cm and angle 90°, the perimeter is 22 + 28 = 50 cm.

Finding the angle. If we know the arc length, we can find the angle. An arc of length 11 cm on a circle of radius 14 cm: 11 = (θ ÷ 360) × 88, so θ ÷ 360 = 1/8 and θ = 45°.

Real-life uses. Arc length tells us how far a point on a rotating wheel or on the minute hand of a clock moves. The tip of a minute hand 14 cm long travels one full circle of 88 cm in an hour. In 15 minutes, which is a quarter turn, it travels 22 cm. This is also how engineers find the length of a curved bend in a road or a rail track.

Units and the value of π. Keep the radius in the unit in which the answer is needed. When the radius is a multiple of 7, using π = 22/7 makes the arithmetic neat.

Perimeter seems simple: walk round the edge and measure. Yet some perimeter questions surprise even careful thinkers. Working through them sharpens two habits: break a curved boundary into arcs you know, and check what your intuition tells you with algebra.

Puzzle 1: two overlapping circles. Two circles of radius r are placed so that each passes through the centre of the other. What is the perimeter of the combined shape (the outside boundary only)? Let the centres be A and B and let the circles cross at C and D. Then AB, AC and BC are all radii, so ΔABC is equilateral and ∠CAB = 60°. In the same way ∠DAB = 60°, so ∠CAD = 120°. The arc of circle A that lies inside circle B subtends 120°, which is one third of the circle. The outer arc of circle A is therefore two thirds of its circumference, 2/3 × 2πr. Circle B gives the same. The perimeter is 2 × (2/3) × 2πr = 8πr/3. For r = 6 cm this is 16π ≈ 50.3 cm.

Puzzle 2: one big semicircle or three small ones? Points P and Q are joined by two paths. Path 1 is a semicircle on PQ as diameter. Path 2 is made of three semicircles on PQ whose diameters split PQ into three parts. Which path is longer? Many people guess path 2, because it wiggles. Let the radii be a for the big semicircle and b, c, d for the small ones. A semicircle of radius r has length πr, so path 1 is πa and path 2 is π(b + c + d). The small diameters fill PQ exactly: 2b + 2c + 2d = 2a, so b + c + d = a. The two paths are equal! This holds however many semicircles you use.

Puzzle 3: a flower. On each side of a square of side 14 cm, draw a semicircle outwards. The perimeter of the flower is four semicircles of radius 7 cm: 4 × π × 7 = 28π ≈ 28 × 22/7 = 88 cm.

A paradox: the staircase. Join opposite corners of a square of side 1 m by a staircase path going right, up, right, up, … with equal small steps. Whatever the step size, the total of the horizontal pieces is 1 m and the total of the vertical pieces is 1 m, so the staircase is always 2 m long. As the steps get smaller, the staircase looks more and more like the diagonal, whose length is √2 ≈ 1.414 m. Yet its length stays at 2 m. Shapes that look closer and closer need not have lengths that get closer. The lesson: perimeter cannot be judged by eye; it must be measured or calculated.

Everyday perimeter. A wheel of diameter 70 cm travels one circumference, π × 70 = 220 cm (with π ≈ 22/7), in one turn. To cover 11 km = 11,00,000 cm it makes 11,00,000 ÷ 220 = 5000 turns.

The area of a shape is the amount of flat surface it covers. It is measured in square units, such as cm² or m². One square centimetre is the area of a square with a side of 1 cm. We now recall the areas of the three basic shapes and see how they are linked.

Rectangle. A rectangle l units long and b units wide can be filled with l × b unit squares, so area = l × b. A rectangle of 12 cm by 5 cm has area 60 cm². A square of side a has area a².

Parallelogram. Take a parallelogram with base b and height h, where the height is the perpendicular distance between the two parallel sides, and not the slanting side. Cut off a right-angled triangle from one end, slide it to the other end, and the parallelogram becomes a rectangle with the same base b and the same height h. So the area is the same: area = b × h. A parallelogram with base 10 cm and height 6 cm has area 60 cm².

Triangle. Two identical triangles fit together to make a parallelogram, with the same base and the same height. So one triangle has half the area of the parallelogram: area = ½ × b × h. A triangle with base 14 cm and height 9 cm has area ½ × 14 × 9 = 63 cm². Any side can be taken as the base, as long as you use the height that is perpendicular to it.

Same base, same height. Triangles with the same base and lying between the same pair of parallel lines have the same height, so they have the same area. Shapes can look very different and still have equal areas.

A median divides a triangle into two equal areas. A median joins a vertex to the midpoint of the opposite side. The two triangles that it makes have equal bases, because the midpoint splits the side into two equal parts, and the same height, which is the perpendicular distance from the vertex to that side. So they have equal areas. In a triangle of area 60 cm², each triangle made by a median has an area of 30 cm². Three medians together divide the triangle into six pieces of equal area.

Using the height. If you know the area and the base, you can find the height. A triangle of area 84 cm² with base 14 cm has height h = 2 × 84 ÷ 14 = 12 cm.

Units. Convert lengths to the same unit before multiplying. A rectangle 2 m by 50 cm should be treated as 200 cm by 50 cm = 10 000 cm², or as 2 m by 0.5 m = 1 m². Remember that 1 m² = 10 000 cm².

Ancient mathematicians in India, Greece and elsewhere wanted to find a square with the same area as a given rectangle. This problem is called squaring the rectangle. In India, it appears in the Sulbasutras, ancient texts on the geometry of altars written by authors such as Baudhayana, who lived around 800 BCE or earlier. Priests had to build altars of a prescribed area in different shapes, so converting a rectangle to a square was a practical need.

The method. Let the rectangle have length l and breadth b, with l greater than b. Here are the steps:

• Step 1. Cut a square of side b from one end of the rectangle. A strip of length (l − b) and breadth b remains.
• Step 2. Cut this strip across its length, in the middle, into two equal pieces. Each piece is (l − b)/2 long and b wide.
• Step 3. Attach one piece to each of two adjacent sides of the b by b square, with the side of length b along the square. The pieces and the square together form an L-shaped figure called a gnomon, which fits inside a larger square of side b + (l − b)/2 = (l + b)/2.
• Step 4. The larger square is not quite full, because a small square of side (l − b)/2 is missing in the corner. If we add this small square, we get the complete large square. So the area of the rectangle equals the area of the large square minus the area of the small corner square.

In symbols: l × b = ((l + b)/2)² − ((l − b)/2)². This is the identity (x + y)(x − y) = x² − y² in a new form, which you met in the chapter on algebraic identities.

Converting a difference of two squares into one square. Baudhayana used the theorem that is today named after Pythagoras (but is called the Baudhayana theorem in many Indian texts). Draw a right-angled triangle whose hypotenuse is (l + b)/2 and whose one leg is (l − b)/2. By the theorem, the square on the other leg has area ((l + b)/2)² − ((l − b)/2)², which is exactly l × b. So the side of the required square is √(lb), and it equals the other leg of that triangle.

A numerical example. Take a rectangle 8 units by 2 units, with area 16. Then (l + b)/2 = 5 and (l − b)/2 = 3. A right triangle with hypotenuse 5 and one leg 3 has the other leg 4, because 3² + 4² = 5². So the side of the square is 4, and 4² = 16 matches the area of the rectangle.

Why this matters. The construction shows that you do not need to take a square root with a calculator: a ruler, a set of cuts and a right-angled triangle are enough. It also shows how the ideas of area, algebraic identities and the Pythagoras theorem are linked. A reminder: the Greek geometers solved the same problem with a different construction.

Try it. Make a paper rectangle of 9 cm by 4 cm. Using the method, show that the square has side 6 cm, since (9 + 4)/2 = 6.5 and (9 − 4)/2 = 2.5, and 6.5² − 2.5² = 42.25 − 6.25 = 36 = 6².

Suppose you know the three sides of a triangle but not its height, such as a triangular plot of land measured with a tape. How do you find its area? Heron’s formula does exactly this. It is named after Heron of Alexandria (about 60 CE), though it was probably known earlier.

If the sides of the triangle are a, b and c, let s be the semi-perimeter, that is, half the perimeter: s = (a + b + c) ÷ 2. Then:

Area = √[ s(s − a)(s − b)(s − c) ]

Example 1. A triangle has sides 13 cm, 14 cm and 15 cm. Then s = (13 + 14 + 15) ÷ 2 = 21. So s − a = 8, s − b = 7 and s − c = 6. The area is √(21 × 8 × 7 × 6) = √7056 = 84 cm². We can check this independently by finding the height on the side of 14 cm: h = 2 × 84 ÷ 14 = 12 cm. The foot of the perpendicular splits the base into 5 and 9, and 5² + 12² = 13², 9² + 12² = 15². ✓.

Example 2. The 3-4-5 triangle has s = 6, and its area is √(6 × 3 × 2 × 1) = √36 = 6 cm². This agrees with the right-triangle area ½ × 3 × 4 = 6.

Example 3: an equilateral triangle. Each side is a, so s = 3a/2, and s − a = a/2 for each. The area is √(3a/2 × a/2 × a/2 × a/2) = √(3a⁴/16) = (√3/4)a². For a = 2, the area is √3 ≈ 1.73.

Using the formula. Follow these steps: find s, subtract each side from s to get three numbers, multiply the four numbers, and take the square root. Always check that the three sides can really form a triangle; the sum of any two sides must be greater than the third.

Quadrilaterals. Heron’s formula can also find the area of a four-sided plot. Split the figure into two triangles by drawing a diagonal, find the area of each with Heron’s formula, and add the results. For a quadrilateral with a diagonal of 5 m and triangles of sides (3, 4, 5) and (5, 5, 6), the first triangle has area 6 m², the second has s = 8 and area √(8 × 3 × 3 × 2) = √144 = 12 m², and the total is 18 m².

Why it is useful. Surveyors measuring land can easily measure lengths with a tape, but heights are hard to measure. Heron’s formula needs only the three lengths.

How can we find the area of a shape with a curved boundary? The idea, which Archimedes and Indian mathematicians used, is to cut the circle into many thin pieces and rearrange them into a shape we know.

Deriving the formula. Divide a circle of radius r into a number of equal sectors, like the slices of a pizza. Arrange them side by side, pointing alternately up and down, as in a zigzag. The result looks like a parallelogram with curved top and bottom edges. The more slices we cut, the thinner they become, and the closer the shape comes to a rectangle.

Look at that rectangle. Its two long sides are made of the arcs of the slices. Half the arcs are on the top, and half are on the bottom, so each long side has a length equal to half the circumference, which is (1/2) × 2πr = πr. Its short sides are the radius r of the circle. The area of the rectangle is therefore πr × r, which is πr².

Area of a circle = πr²

Since the cutting and rearranging does not change the area, this is the area of the circle.

Examples. A circle of radius 7 cm has area (22/7) × 7 × 7 = 154 cm². A circle of diameter 20 cm has radius 10 cm, and area 3.14 × 100 = 314 cm². A circular garden with circumference 44 m has radius r such that 2πr = 44, so r = 7 m and the area is 154 m².

Mind the difference. Circumference is a length, 2πr, in units such as cm, while area is πr², in square units such as cm². If you double the radius, the circumference doubles, but the area becomes four times as large, because r is squared. A pizza of 30 cm diameter has four times the area of a 15 cm pizza, not twice.

An area of a ring. The area between two concentric circles with radii R and r is πR² − πr² = π(R² − r²). A circular path 2 m wide around a pond of radius 5 m has area π(7² − 5²) = π × 24, which is about 75.4 m² with π = 3.14.

Historical approximations. The ancient Egyptians, in the Rhind Papyrus, took the area of a circle to be that of a square with side 8/9 of the diameter. That gives an area of (8d/9)², which corresponds to π ≈ 256/81 ≈ 3.16. It is surprisingly close. Such approximations show that people in many cultures worked on this problem, and each generation made it more exact.

A sector is the region of a circle enclosed by two radii and the arc between them. A segment is the region enclosed by a chord and the arc that it cuts off. A slice of cake is a sector, and the cap of a circle cut off by a straight line is a segment.

Area of a sector. A sector with a central angle of θ degrees is the fraction θ ÷ 360 of the whole circle, so its area is the same fraction of the area of the circle:

Area of a sector = (θ ÷ 360) × πr²

Example. A sector of angle 60° in a circle of radius 21 cm has area (60 ÷ 360) × (22/7) × 21 × 21 = (1/6) × 1386 = 231 cm². For a semicircle, θ = 180°, and the area is (1/2)πr². For a quarter-circle, θ = 90°, and the area is (1/4)πr².

A shortcut uses the arc length. Since the arc length is l = (θ ÷ 360) × 2πr, the area of the sector can also be written as (1/2) × l × r. Check with the example above: the arc length is 22 cm, and (1/2) × 22 × 21 = 231 cm². ✓.

Area of a segment. The segment is the sector minus the triangle formed by the two radii and the chord:

Area of segment = area of sector − area of triangle

Example. A circle of radius 14 cm and a chord that subtends an angle of 60° at the centre. The sector has area (1/6) × (22/7) × 196 = 102.67 cm². The triangle formed by the two radii and the chord is equilateral, because the angle at the centre is 60° and the two radii are equal. Its area is (√3/4) × 14² = 49√3 ≈ 84.87 cm². So the segment’s area is 102.67 − 84.87 = about 17.8 cm².

For a different angle. If the central angle is 90°, the triangle is a right triangle with legs r and r, and its area is ½ r². The segment is then (1/4)πr² − (1/2)r², which for r = 14 cm is 154 − 98 = 56 cm².

Practical shapes. These ideas solve problems about the area swept by a wiper, the part of a field a cow can graze when tethered at a corner by a rope, and the area of a decorative window shaped like a sector. In such problems, first identify the central angle and the radius, then pick the sector formula, and finally add or subtract any triangles.

Heron’s formula gives the area of a triangle from its sides. Is there a similar formula for a four-sided figure? The Indian mathematician Brahmagupta (628 CE) found one, for an important class of quadrilaterals. It applies when the four vertices all lie on one circle. Such a quadrilateral is called a cyclic quadrilateral.

Let the sides of a cyclic quadrilateral be a, b, c and d, and let s be its semi-perimeter, s = (a + b + c + d) ÷ 2. Then:

Area = √[ (s − a)(s − b)(s − c)(s − d) ]

Compare this with Heron’s formula. It has the same pattern, with one more factor.

A worked example. A cyclic quadrilateral has sides 25, 39, 52 and 60 units. Here s = (25 + 39 + 52 + 60) ÷ 2 = 88. Then s − a = 63, s − b = 49, s − c = 36 and s − d = 28. The area is √(63 × 49 × 36 × 28) = √(3 111 696) = 1764 square units.

Checking by splitting. Is 1764 right? Notice that 25² + 60² = 625 + 3600 = 4225 = 65², and 39² + 52² = 1521 + 2704 = 4225 = 65². So the diagonal of length 65 splits the figure into two right-angled triangles, one with legs 25 and 60, and the other with legs 39 and 52. Their areas are ½ × 25 × 60 = 750 and ½ × 39 × 52 = 1014. The total is 750 + 1014 = 1764. ✓. The two methods agree. (Both triangles have their right angles at opposite corners on a circle with the diagonal as its diameter, so the quadrilateral is indeed cyclic.)

A special case: a rectangle. A rectangle is a cyclic quadrilateral. With sides a, b, a, b, we have s = a + b and the factors are b, a, b, a, so the area is √(a²b²) = ab. This is the familiar area of a rectangle.

The condition matters. Brahmagupta’s formula is true only for cyclic quadrilaterals. A quadrilateral with given sides is not fixed in shape: you can push its corners and its area changes. The cyclic arrangement is the one with the largest area for those four side lengths. For a general quadrilateral, the formula needs an extra term that depends on the angles.

Brahmagupta’s place in history. Brahmagupta worked in Ujjain and wrote the Brahmasphutasiddhanta. His work also describes the use of zero and negative numbers in arithmetic. The formula for cyclic quadrilaterals is among his best-known geometric results.

In mathematics, a generalisation is a statement that covers many cases at once, and a special case is one of those cases. Seeing that one result is a special case of another connects ideas and often explains why the first result looks the way it does. Heron’s formula and Brahmagupta’s formula are a good example.

Look at the two formulas side by side:

A triangle as a collapsed quadrilateral. Imagine a cyclic quadrilateral in which one side, d, is made smaller and smaller, until it shrinks to a single point, d = 0. Then the quadrilateral becomes a triangle with sides a, b and c, and the four corners reduce to three. A triangle always lies on a circle, so it is a valid cyclic case.

Now put d = 0 in Brahmagupta’s formula. Then s = (a + b + c + 0) ÷ 2 = (a + b + c) ÷ 2, the semi-perimeter of the triangle, and the factor (s − d) becomes s − 0 = s. The formula becomes √[ (s − a)(s − b)(s − c) × s ], which is exactly Heron’s formula. So Heron’s formula is the special case of Brahmagupta’s formula in which one side is zero, and Brahmagupta’s formula is a generalisation of Heron’s.

A numerical check. Take a triangle with sides 3, 4 and 5, and treat it as a quadrilateral with sides 3, 4, 5 and 0. Then s = 6, and the four factors are 6 − 3 = 3, 6 − 4 = 2, 6 − 5 = 1 and 6 − 0 = 6. The product is 3 × 2 × 1 × 6 = 36, whose square root is 6. This is the area of the 3-4-5 triangle.

Another check. Take the 13-14-15 triangle. With d = 0, s = 21, and the factors are 8, 7, 6 and 21. The product is 7056 and the square root is 84, which is the triangle’s area that we found earlier.

Why generalise? A single general formula replaces several. Once you know Brahmagupta’s formula, you also know Heron’s without learning it separately. Mathematicians look for such links, because they show the underlying structure.

Other examples of special cases. A square is a special case of a rectangle, because it is a rectangle with equal sides. A rectangle is a special case of a parallelogram. The area formula for a parallelogram, b × h, gives the rectangle formula when the sides are perpendicular. A circle is a special case of an ellipse. Recognising such relationships helps you to remember fewer things and understand more.

Thinking like a mathematician. To test whether one formula is a special case of another, set one of the quantities to a particular value, such as zero or a pair of equal values, and check whether the simpler formula appears.

Most real problems combine several shapes. A systematic approach makes them manageable. Follow this plan:

1. Draw the figure and write the given lengths on it. A clear sketch prevents most errors.

2. Decide what is asked: a perimeter (a length) or an area (a square unit), or a cost that depends on one of them.

3. Break the figure into simple parts such as rectangles, triangles, circles, sectors and semicircles. You may either add the areas of parts, or subtract the area of a part that has been removed.

4. Use the right formula for each part and keep the units consistent.

5. Combine and check that the answer is sensible, for example that a part is not bigger than the whole.

Example 1: a rectangle with a semicircle. A window has the shape of a rectangle 10 m long and 6 m wide with a semicircle on one of its 6 m sides. The area is the rectangle (10 × 6 = 60 m²) plus the semicircle of diameter 6 m (radius 3 m): ½ × 3.14 × 9 = 14.13 m². The total area is 74.13 m². For the perimeter, take the outer boundary: two sides of 10 m, one side of 6 m, and the semicircular arc, which is half the circumference of the circle: 3.14 × 3 = 9.42 m. The perimeter is 10 + 10 + 6 + 9.42 = 35.42 m. Notice that the 6 m side where the semicircle is attached is not part of the boundary.

Example 2: a path around a rectangular garden. A garden 40 m by 30 m has a path 2 m wide running all round the outside. The outer rectangle is 44 m by 34 m, with area 1496 m². The garden’s own area is 1200 m². So the area of the path is 1496 − 1200 = 296 m².

Example 3: the cost of a surface. A circular pond of radius 7 m has a circular path 1 m wide around it. The path’s area is π(8² − 7²) = π × 15 = (22/7) × 15 ≈ 47.14 m². If it is paved at ₹120 per m², the cost is about ₹5657.

Example 4: a field with a triangle and a quadrilateral. A triangular field has sides 40 m, 41 m and 9 m. Here s = 45, and the area is √(45 × 5 × 4 × 36) = √32 400 = 180 m². (Notice that 9² + 40² = 41², so it is a right triangle, and ½ × 9 × 40 = 180 as well.)

Common errors to avoid. Mixing units, such as metres and centimetres. Writing a perimeter in square units or an area in plain units. Using the diameter where the radius is needed. Using the slant height instead of the perpendicular height. Counting joined sides in the perimeter. Check each of these before you finish.

A calculator may be used for the final arithmetic, but the setting up of the problem is the real skill. Practise by sketching, labelling, and writing out the formulas before substituting numbers.