Area

Ask how long a rope is and you are measuring in one direction, so the answer is in metres. Ask how much cloth a tailor needs and you are measuring in two directions at once, so the answer is in square metres. Area is the amount of flat space a shape covers, and we measure it by asking one simple question: how many unit squares fit inside it?

A unit square is a square whose side is one unit. A square of side 1 cm has an area of one square centimetre, written 1 cm². A square of side 1 m has an area of 1 m². That is why every area answer carries a small 2 above the unit. If you write an area as 30 cm instead of 30 cm², you have written a length, and the answer is not correct even though the number is.

Draw a rectangle 5 cm long and 3 cm wide on squared paper and fill it with 1 cm squares. You get 3 rows with 5 squares in each row, so 3 × 5 = 15 squares, and the area is 15 cm². Nothing in that argument depended on the numbers 5 and 3. Any rectangle of length l and breadth b holds b rows of l squares, so its area is l × b. A square is simply a rectangle whose sides are equal, so a square of side s has area s × s, written s².

Example 1. A square tile has side 9 cm, so its area is 9 × 9 = 81 cm². A rectangular table top measures 4.5 m by 3.2 m, so its area is 4.5 × 3.2 = 14.4 m².

Example 2. Working backwards is just as useful. A rectangular sheet has area 84 cm² and length 12 cm. Since area = length × breadth, the breadth is 84 ÷ 12 = 7 cm. Check it: 12 × 7 = 84. Correct.

Area and perimeter are different measurements and they do not rise and fall together. Take every rectangle with whole-number sides and perimeter 20 cm. The sides can be 1 and 9, 2 and 8, 3 and 7, 4 and 6, or 5 and 5, because in each pair the two numbers add to 10 and the perimeter is twice that. Their areas are 9, 16, 21, 24 and 25 cm². The perimeter never changed, yet the area almost tripled as the rectangle became more square. Turn it around and fix the area at 24 cm² instead: the rectangle could be 1 by 24, 2 by 12, 3 by 8 or 4 by 6, with perimeters 50, 28, 22 and 20 cm. Same area, wildly different perimeters. Keep this in mind when a question asks about fencing, which is a perimeter, and you find yourself reaching for an area formula.

Not every shape has straight edges. Put a leaf or your palm on squared paper, draw around it and count. The usual agreement is to count a fully covered square as 1, a square more than half covered as 1, a square about half covered as one half, and a square less than half covered as 0. If a leaf covers 32 full squares and 14 half squares on 1 cm paper, its area is about 32 + 7 = 39 cm². That is an estimate, not an exact value, and it is honest to say so. The finer the squared paper, the closer the estimate gets, and that idea of closing in with smaller and smaller pieces is exactly what we will use later to find the area of a circle.

One habit is worth building from the very start: estimate before you calculate. A rectangle roughly 10 cm by 6 cm has an area of roughly 60 cm², so if your working produces 6 cm² or 600 cm² you know something has gone wrong before you write the answer down. Most slips in this chapter are not deep misunderstandings. They are a factor of ten, a forgotten half, or a unit left unconverted, and a quick estimate catches all three.

You already know that 1 cm = 10 mm. It is tempting to jump from there to 1 cm² = 10 mm², and that is wrong. Draw a square of side 1 cm and rule it into millimetre squares: you get 10 rows of 10, which is 100 small squares. So 1 cm² = 100 mm². The length factor was 10; the area factor is 10 × 10 = 100.

The same reasoning gives every other conversion. Since 1 m = 100 cm, a square metre is 100 rows of 100 centimetre squares, so 1 m² = 100 × 100 = 10,000 cm². Since 1 m = 1000 mm, one square metre is 1000 × 1000 = ten lakh square millimetres. Since 1 km = 1000 m, one square kilometre is 1000 × 1000 = ten lakh square metres. The whole rule fits in three words: square the factor.

Large pieces of land are measured in hectares. One hectare is the area of a square 100 m by 100 m, so 1 hectare = 10,000 m². Since a square kilometre is ten lakh square metres, one square kilometre works out to 10,00,000 ÷ 10,000 = 100 hectares. You will also meet the are, which is a square 10 m by 10 m, so 1 are = 100 m² and 1 hectare = 100 ares. The word hectare literally means a hundred ares, which makes it easy to remember.

Example 1. Convert 2.5 m² into cm². Multiply by 10,000: 2.5 × 10,000 = 25,000 cm².

Example 2. Convert 45,000 cm² into m². You are going the other way now, so divide: 45,000 ÷ 10,000 = 4.5 m².

Example 3. A field is 250 m long and 160 m wide, so its area is 250 × 160 = 40,000 m². To read that in hectares, divide by 10,000: the field is 4 hectares.

Example 4. A room is 5 m by 4 m and is to be covered with square tiles of side 20 cm. Work in one unit throughout. In centimetres the room is 500 cm by 400 cm, which is 2,00,000 cm², and one tile is 20 × 20 = 400 cm². The number of tiles is 2,00,000 ÷ 400 = 500. Do the whole thing in metres instead and you get the same answer: the room is 20 m², one tile is 0.2 × 0.2 = 0.04 m², and 20 ÷ 0.04 = 500. Either route works. Mixing the two does not.

That last point is the one to hold on to. A question that gives the room in metres and the tile in centimetres is not being unfair; it is checking whether you convert before you divide. Convert everything to a single unit as your first written step, keep the unit written beside every number while you work, and this whole family of errors disappears.

Example 5. A plot is 3.6 hectares, which is 3.6 × 10,000 = 36,000 m². If the plot is rectangular and 240 m long, its width is 36,000 ÷ 240 = 150 m. Check: 240 × 150 = 36,000. Correct.

A rough sense of scale helps you spot nonsense straight away. A classroom floor is a few tens of square metres. A cricket ground is around a hectare or two. A large farm is measured in hectares and a district in square kilometres. If a calculation tells you a bedroom is 400 m², or that a village covers 5 m², the answer is wrong however neat the working looks above it.

Start with a parallelogram, a quadrilateral whose opposite sides are parallel. Cut a paper parallelogram along the perpendicular dropped from one top corner down to the base. You now hold a right-angled triangle and a leftover piece. Slide that triangle across to the other end and it fits exactly, because the two slanting sides are equal and parallel. What you are left with is a rectangle with the same base and the same height as the parallelogram you started with. Nothing was added and nothing was thrown away, so the two areas must be equal. A rectangle's area is base × height, and so:

Area of a parallelogram = base × height.

The word height is doing serious work in that sentence. The height is the perpendicular distance from the base to the side parallel to it. It is not the slanting side. In a parallelogram with base 12 cm, height 7 cm and slanting side 8 cm, the area is 12 × 7 = 84 cm². The 8 cm is in the question to see whether you reach for the right number. Multiply 12 × 8 and you get 96 cm², which is not merely wrong but impossible, because the slanting side is always longer than the perpendicular height, so that answer is bigger than the true area every single time.

A parallelogram has two different bases and each has its own height, but both pairs give the same area. Take the parallelogram above, area 84 cm², whose adjacent side is that 8 cm side. Using it as the base, 8 × h = 84, so h = 84 ÷ 8 = 10.5 cm. That is a genuinely useful move: given one base-height pair and a second side, you can always work out the second height without measuring anything.

Now the triangle. Cut out two identical copies of any triangle you like. Turn one of them upside down and slide it against the other along a matching side. The two always join into a parallelogram, with the same base as the triangle and the same height. Two triangles make one parallelogram of area b × h, so one triangle is half of that:

Area of a triangle = 1/2 × base × height.

This works for every triangle, whether it is acute, obtuse or right-angled. For a right-angled triangle it is especially easy, because the two shorter sides are already perpendicular to each other, so either one can serve as the base while the other is the height.

Example 1. A triangle has base 10 cm and height 6 cm, so its area is 1/2 × 10 × 6 = 30 cm². Notice that the rectangle 10 cm by 6 cm has area 60 cm², exactly twice as much. That doubling is the check built into the formula.

Example 2. A right-angled triangle has shorter sides 9 cm and 12 cm, so its area is 1/2 × 9 × 12 = 54 cm². By Pythagoras its longest side is √(9² + 12²) = √(81 + 144) = √225 = 15 cm. Now use that 15 cm side as the base: 1/2 × 15 × h = 54, so 7.5h = 54 and h = 54 ÷ 7.5 = 7.2 cm. The area came out the same from both directions, as it had to.

Example 3. Working backwards, a triangle has area 96 cm² and base 16 cm. From 1/2 × 16 × h = 96 we get 8h = 96, so h = 12 cm. When you rearrange the triangle formula, the safest form to remember is height = 2 × area ÷ base, because the 2 is very easy to lose.

One last idea ties this section together. Fix a segment as the base and draw a line parallel to it. Every triangle with that base and its apex anywhere at all on the parallel line has the same height, so every one of them has the same area, however stretched or lopsided it looks. Triangles on the same base and between the same parallels are equal in area, and the same is true of parallelograms on the same base and between the same parallels. It is a surprising fact the first time you meet it, and it explains why a diagonal splits any parallelogram into two triangles of exactly equal area.

Example 4. A plot of land is shaped like a parallelogram with base 40 m and height 25 m, so its area is 40 × 25 = 1000 m². Laying grass over it at ₹12 per square metre costs 1000 × 12 = ₹12,000.

A trapezium is a quadrilateral with one pair of parallel sides. Call those parallel sides a and b, and call the perpendicular distance between them h. There are two ways to find its area and both are worth seeing, because a formula you can rebuild is a formula you can never quite forget.

The first way copies the triangle trick. Make two identical trapeziums, turn one of them through half a turn and push it against the other. The slanting sides match up and you get a parallelogram whose base is a + b and whose height is h. That parallelogram has area (a + b) × h, and it is built from two trapeziums, so one trapezium is half of it.

The second way needs no cutting at all. Draw a diagonal. It splits the trapezium into two triangles which share the same height h: one stands on base a and the other on base b. Adding them gives 1/2 × a × h + 1/2 × b × h = 1/2 × (a + b) × h. Same answer by a completely different route, and getting the same result twice is strong evidence that the formula is right.

Area of a trapezium = 1/2 × (sum of the parallel sides) × (distance between them).

There is a nicer way to say that. The quantity 1/2 × (a + b) is just the average of the two parallel sides, so a trapezium covers the same area as a rectangle whose width is that average and whose height is h. Reading it that way also hands you a free check: if a and b happen to be equal, the shape is really a parallelogram and the formula collapses to b × h, exactly as it should.

Example 1. A trapezium has parallel sides 12 cm and 8 cm with 5 cm between them. Area = 1/2 × (12 + 8) × 5 = 1/2 × 20 × 5 = 50 cm².

Example 2. A trapezium-shaped field has parallel edges 25 m and 15 m, standing 12 m apart. Area = 1/2 × 40 × 12 = 240 m².

Example 3. Backwards for the height. A trapezium has area 180 cm² and parallel sides 16 cm and 14 cm. Then 1/2 × 30 × h = 180, so 15h = 180 and h = 12 cm.

Example 4. Backwards for a missing side. A trapezium has area 84 cm², height 7 cm and one parallel side 15 cm. Then 1/2 × (15 + b) × 7 = 84. Multiply both sides by 2 to get (15 + b) × 7 = 168, then divide by 7 to get 15 + b = 24, so b = 9 cm. Check: 1/2 × 24 × 7 = 84. Correct.

Now the rhombus, a parallelogram whose four sides are all equal. Its diagonals cut each other in half and meet at right angles, and that is the key to its area. Draw the rectangle that just encloses the rhombus with its sides parallel to the diagonals. That rectangle is d₁ long and d₂ wide, and the rhombus takes up exactly half of it, because the four corner triangles left over pair up to fill the other half.

If you prefer arithmetic to pictures, the diagonals split the rhombus into four right-angled triangles, each with shorter sides d₁/2 and d₂/2. Each of those has area 1/2 × (d₁/2) × (d₂/2), which is d₁d₂/8, and four of them come to d₁d₂/2. The picture and the arithmetic agree.

Area of a rhombus = 1/2 × d₁ × d₂, where d₁ and d₂ are the diagonals. A rhombus is still a parallelogram, so base × height works on it too, and having both rules lets you find a height you were never given.

Example 5. A rhombus has diagonals 16 cm and 12 cm, so its area is 1/2 × 16 × 12 = 96 cm². Half of each diagonal is 8 cm and 6 cm, and those two are the shorter sides of a right-angled triangle whose longest side is a side of the rhombus, so the side is √(64 + 36) = √100 = 10 cm. Since area = base × height, the height on that 10 cm side is 96 ÷ 10 = 9.6 cm.

The half-the-product-of-the-diagonals rule is not only for rhombuses. It holds for any quadrilateral whose diagonals cross at right angles, which includes the kite and the square. A kite with diagonals 20 cm and 9 cm has area 1/2 × 20 × 9 = 90 cm². A square has two equal diagonals, so a square of diagonal 10 cm has area 1/2 × 10 × 10 = 50 cm².

Finally, a quadrilateral with no special properties whatsoever can still be handled. Draw one diagonal of length d and drop perpendiculars onto it from the two remaining corners, of lengths h₁ and h₂. The quadrilateral is now two triangles standing on the same base d, so its area is 1/2 × d × h₁ + 1/2 × d × h₂ = 1/2 × d × (h₁ + h₂). Surveyors use exactly this to measure an oddly shaped field: one long tape stretched along a diagonal and two short ones laid across it.

Every shape so far had straight edges, so cutting and rearranging gave an exact answer. A circle has no straight edges at all, so we need a slightly different idea, and it is the same one that let us estimate a leaf on squared paper: close in on the answer.

Draw a circle of radius r and cut it into a large number of thin sectors, like slices of a cake. Lay the slices out in a row, alternately pointing up and pointing down so that they interlock. What you build is almost a rectangle. Its height is the radius r, because that is the length of each slice from the centre outwards. Its width is made from the curved tips, and half of them lie along the top edge while the other half lie along the bottom, so the width is half the circumference, which is 1/2 × 2 × pi × r = pi × r. The bumpy edges are the only inaccuracy, and the more slices you cut the flatter they become. So the area of a circle is (pi × r) × r:

Area of a circle = pi × r².

The number pi is what you get when you divide any circle's circumference by its diameter. It is a little more than 3, it never ends and never repeats, so in schoolwork we replace it by 22/7 or by 3.14. Use 22/7 when the radius is a multiple of 7, because the 7 cancels and the arithmetic stays clean. That is exactly why so many questions choose radii like 7, 14 and 21.

Example 1. A circle has radius 7 cm. Taking pi = 22/7, its area is (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm². Its circumference, for comparison, is 2 × (22/7) × 7 = 44 cm. One is a length in centimetres and the other an area in square centimetres, and they are not interchangeable.

Example 2. A circular ground has radius 21 m, so its area is (22/7) × 441 = 22 × 63 = 1386 m².

Example 3. A circular lawn has diameter 14 m, so the radius is 14 ÷ 2 = 7 m and the area is 154 m². Reading a diameter as though it were a radius doubles the radius, and doubling the radius multiplies the area by four, so this is worth a moment's care every single time.

Example 4. Backwards. A circle has area 616 cm². From (22/7) × r² = 616 we get r² = 616 × 7 ÷ 22 = 4312 ÷ 22 = 196, so r = √196 = 14 cm. Check: (22/7) × 196 = 22 × 28 = 616. Correct.

A ring, which is what a circular path or a washer looks like, is a circle with a smaller circle taken out of the middle. If the outer radius is R and the inner radius is r, the ring's area is pi × R² - pi × r², which is neater written as pi × (R² - r²). A circular pond of radius 7 m with a path around it out to a radius of 14 m has a path area of (22/7) × (196 - 49) = (22/7) × 147 = 22 × 21 = 462 m². Notice that you subtract the squares. Squaring the difference instead would give (14 - 7)² = 49, which is nowhere near the right answer.

Now a sector, which is the slice bounded by two radii and the arc between them. A whole circle corresponds to a full turn of 360 degrees, so a sector whose angle at the centre is A degrees is simply the fraction A/360 of the circle. That single sentence gives you everything you need: area of a sector = (A ÷ 360) × pi × r², and length of its arc = (A ÷ 360) × 2 × pi × r.

Example 5. A quarter circle is a sector of 90 degrees, and 90/360 = 1/4. For a radius of 14 cm the whole circle has area (22/7) × 196 = 616 cm², so the quarter has area 616 ÷ 4 = 154 cm². A semicircle is a sector of 180 degrees, so it is simply half the circle.

Example 6. A sector of 60 degrees is cut from a circle of radius 21 cm. Since 60/360 = 1/6 and the whole circle has area 1386 cm², the sector has area 1386 ÷ 6 = 231 cm². Its arc is one sixth of the circumference: the circumference is 2 × (22/7) × 21 = 132 cm, so the arc is 132 ÷ 6 = 22 cm.

Be careful with the word perimeter here. The perimeter of a sector is not just its arc; it is the arc plus the two straight radii. For the sector in Example 6 that comes to 22 + 21 + 21 = 64 cm. Slices of pizza, hand fans, the glass swept clean by a windscreen wiper and the ground watered by a rotating sprinkler are all sectors, and in every one of them the curved edge and the straight edges are different things measured in different ways.

Very few real surfaces are a single textbook shape. A floor plan has an L-shaped room, a park has a path running around it, a field has two roads crossing it, a window has a semicircular top. Every one of these is a composite figure, and there are only two moves you ever need.

The first move is to split. Draw one or two extra lines to cut the figure into rectangles, triangles, trapeziums and pieces of circles, work out each piece and add them up. The second move is to surround and subtract. Draw the smallest simple shape that contains the whole figure, find its area, then take away the bits that are not part of the figure. Some shapes are easier one way and some the other, and doing a figure both ways is the best check there is, because the two answers have to agree.

Example 1. An L-shaped room fits inside a rectangle 8 m by 6 m, with a rectangle 3 m by 2 m missing from one corner. Subtracting: 8 × 6 = 48 m² and 3 × 2 = 6 m², so the area is 48 - 6 = 42 m². Splitting instead: the full-width strip along the bottom is 8 m by 4 m = 32 m², and the strip above it is (8 - 3) = 5 m wide by 2 m = 10 m². Adding gives 32 + 10 = 42 m². The two methods agree, which is exactly what you want to see.

Example 2. A house-shaped figure is a rectangle 10 m wide and 6 m high with a triangular roof of base 10 m and height 4 m. Its area is 10 × 6 + 1/2 × 10 × 4 = 60 + 20 = 80 m².

Example 3. A running-track infield is a rectangle 40 m by 14 m with a semicircle attached to each of the shorter ends. Each semicircle has diameter 14 m and therefore radius 7 m, and the two together make one full circle of area (22/7) × 49 = 154 m². The total area is 40 × 14 + 154 = 560 + 154 = 714 m². The boundary is separate work: the two straight sides give 80 m, and the two curved ends together make one full circumference, 2 × (22/7) × 7 = 44 m, so the boundary measures 80 + 44 = 124 m.

Example 4, a path outside. A rectangular garden 30 m by 20 m has a 2 m wide path running all around it on the outside. The path adds 2 m at each end of each direction, so the outer rectangle is (30 + 4) by (20 + 4), that is 34 m by 24 m = 816 m². The garden itself is 600 m², so the path covers 816 - 600 = 216 m². Gravelling it at ₹45 per square metre costs 216 × 45 = ₹9,720. The number to watch is that 4: a 2 m path widens the rectangle by 2 m on the left and another 2 m on the right.

Example 5, a path inside. A hall 22 m by 18 m has a 1.5 m wide verandah running along the walls on the inside. This time the path eats into the rectangle, so the inner floor is (22 - 3) by (18 - 3), that is 19 m by 15 m = 285 m². The hall is 22 × 18 = 396 m², so the verandah covers 396 - 285 = 111 m².

Example 6, crossing roads. A rectangular park is 60 m by 40 m and two roads 3 m wide cross it at right angles, one parallel to the length and one parallel to the breadth. The road along the length covers 60 × 3 = 180 m² and the road across covers 40 × 3 = 120 m². Adding those gives 300 m², but the small square where the roads cross has been counted twice, so subtract it once: it is 3 m by 3 m = 9 m². The roads cover 180 + 120 - 9 = 291 m², and the grass left over is 2400 - 291 = 2109 m². Forgetting to remove the crossing square is an easy slip, and it always makes the road area come out too big.

Example 7, cutting a circle out. The largest possible circle is cut from a square sheet of side 14 cm. The circle touches all four sides, so its diameter is 14 cm and its radius 7 cm. The square is 196 cm² and the circle is (22/7) × 49 = 154 cm², so the leftover material is 196 - 154 = 42 cm². Here is a pleasant twist: if instead you cut a quarter circle of radius 7 cm from each of the four corners, you have removed four quarters, which is again one whole circle, and the leftover is once more 42 cm².

Two habits make composite questions much safer. First, redraw the figure and mark every length you know on it, including the ones you work out along the way, because most of these questions are lost to a length that was never written down. Second, before you start, decide whether the question wants area, which covers things like tiling, painting and turfing, or perimeter, which covers fencing, edging and running around the outside. The words in the question tell you which, and choosing the wrong one costs you far more than an arithmetic slip ever will.

Area is one of the few school topics you will use whether or not you ever study mathematics again. Tiles, paint, turf, carpet, cloth, glass and land are all sold by area, so the calculation ends in rupees.

Tiling. The recipe never changes: find the floor area, find the area of one tile, divide, then multiply by the price. A room is 6.5 m by 4 m, so its floor is 6.5 × 4 = 26 m². The tiles are 50 cm by 50 cm, which is 0.5 m by 0.5 m = 0.25 m² each. The number of tiles is 26 ÷ 0.25 = 104, and at ₹65 a tile that comes to 104 × 65 = ₹6,760. In practice a mason orders a few extra for cutting and breakage, which is a useful reminder that the mathematics gives you the minimum, not the shopping list.

Painting. A wall is 9 m long and 3.5 m high, so its area is 9 × 3.5 = 31.5 m². It has a door 2 m by 1.2 m = 2.4 m² and two windows each 1.5 m by 1 m, giving 3 m² in all. You do not paint doors and windows, so the painted area is 31.5 - 2.4 - 3 = 26.1 m². At ₹28 per square metre the paint costs 26.1 × 28 = ₹730.80.

Land. A farmer's plot is 2.5 hectares, which is 2.5 × 10,000 = 25,000 m². If he keeps a 40 m by 25 m corner for a shed and a threshing floor, that corner is 1000 m², and the land left for the crop is 25,000 - 1000 = 24,000 m², or 2.4 hectares.

Fencing against covering. Suppose that same 40 m by 25 m rectangle is to be both fenced and turfed. The fence follows the boundary, so it needs the perimeter: 2 × (40 + 25) = 130 m of wire. The turf covers the surface, so it needs the area: 1000 m². If wire costs ₹90 a metre and turf costs ₹35 a square metre, the fence comes to 130 × 90 = ₹11,700 and the turf to 1000 × 35 = ₹35,000. One rectangle, two completely different calculations, and the only thing telling you which to use is the verb in the question.

Here is the checklist that catches nearly everything in this chapter.

• Same units first. Never multiply metres by centimetres. Convert, then calculate.
• Square the conversion factor. One square metre is 10,000 cm², not 100 cm².
• Perpendicular height, not the slanting side. This applies to parallelograms, triangles and trapeziums alike.
• Do not lose the half. Triangles, trapeziums and rhombuses all carry a 1/2 that rectangles and parallelograms do not.
• Radius, not diameter. Read the question twice, because using d in place of r makes the area four times too big.
• For a sector, keep the fraction. Multiply by the angle divided by 360, and add two radii if the question asks for the perimeter.
• Write the unit on the answer. An area ends in cm², m², hectares or km².
• Estimate before, check after. Round the numbers, get a rough answer, and see whether the exact one lands in the same range.

That last item deserves a demonstration. A triangle has base 9.8 cm and height 6.1 cm. Rounding to 10 and 6 gives an estimate of 1/2 × 10 × 6 = 30 cm². The exact working is 1/2 × 9.8 × 6.1 = 1/2 × 59.78 = 29.89 cm². The estimate and the answer sit right next to each other, which is what you want. Had the exact working produced 2.989 or 298.9, the estimate would have caught it in a second.

One closing thought about why all of this hangs together. Every formula in this chapter grew from just two ideas: a rectangle's area is length × breadth, and rearranging a shape does not change how much space it covers. Cut and slide, and a parallelogram becomes a rectangle. Double it, and a triangle becomes a parallelogram. Turn a copy around, and a trapezium becomes a parallelogram. Slice finely and interleave, and even a circle becomes a rectangle. So if you ever forget a formula from this chapter, you do not have to guess. You can rebuild it from a rectangle and a pair of scissors.