Mensuration 2D

Everything else in this chapter leans on this type, so take it slowly. If every length of a figure is multiplied by k, its perimeter is multiplied by k and its area by k².

Why does this happen? A perimeter only adds lengths, so it grows exactly as a length grows. An area always multiplies two lengths, so it picks up the factor k twice. For a square, side a → ka and area a² → k²a². For a circle, radius r → kr and area πr² → πk²r². For a triangle, base and height both become k times, so 12bh becomes k² times. It does not matter which length you scale (side, diagonal, radius, diameter, height or circumference). As long as the shape stays the same, all lengths change in the same ratio.

[[fig:0]]

Similar figures. Two squares, two circles, two equilateral triangles or two regular hexagons are always similar. Two triangles are similar when their angles are equal. For similar figures:

If two circles have radii in the ratio 3 : 4, their circumferences are also 3 : 4, but their areas are 9 : 16. Going the other way, if the areas are 49 : 64, the sides (or radii, or perimeters) are 7 : 8.

Percentage change. A length increased by x% is multiplied by (1 + x100). The area gets this factor twice, so it is a successive change of x% and x%:

[[fig:1]]

If the radius goes up by 30%, the area goes up by 60 + 9 = 69%. If the side goes down by 30%, the area changes by −60 + 9 = −51%, a 51% decrease. When the two dimensions change by different amounts (length by a%, breadth by b%), use the same net-effect formula with signs: a + b + ab100.

The reverse question. Suppose the area becomes 169% of the original. The area factor is 169100, so the length factor is 1310, which means the side has gone up by 30%. Always take the square root of the area factor.

Maps and scale drawings. A map scale of 1 cm = 5 km reduces every length by the same factor, so 1 cm² on the map stands for 5 × 5 = 25 km² on the ground.

[[fig:2]]

Radhika gets a square photo enlarged so that each side becomes 1.5 times. The wooden frame (perimeter) needs 1.5 times as much wood, but the glass (area) needed is 1.5 × 1.5 = 2.25 times, which is 125% more.

A triangle has three sides and three heights. Any side can be taken as the base. The height for that base is the perpendicular drawn from the opposite vertex to that side, or to the side extended. Then

[[fig:0]]

You get the same area whichever side you choose, so 12a·h_a = 12b·h_b = 12c·h_c. This gives a very handy result: height on a side = 2 × Areathat side. It also shows that the heights are inversely proportional to the sides they stand on.

Two sides and the included angle. If two sides a, b and the angle C between them are known:

[[fig:1]]

Keep these values ready: sin 30° = 12, sin 45° = 1√2, sin 60° = sin 120° = √32, sin 90° = 1, sin 150° = 12.

Heron's formula (all three sides known). Let s = a + b + c2 be the semi-perimeter. Then

With sides 5, 5 and 6, s = 8, so the area is √8 × 3 × 3 × 2 = √144 = 12. Radhika's triangular flower bed has sides 9 m, 10 m and 17 m. Here s = 18 and the area is √18 × 9 × 8 × 1 = √1296 = 36 m².

A small check helps: (s − a) + (s − b) + (s − c) = 3s − 2s = s. So if a question gives you the three values s − a, s − b and s − c, add them and you have s.

[[fig:2]]

Sides in a ratio. If the sides are in the ratio p : q : r and the perimeter is given, first find the multiplier k = perimeterp + q + r, write the actual sides and then apply Heron. The other way is to apply Heron to p, q, r themselves and multiply that area by k².

Two more facts are used in questions. By the triangle inequality, any two sides together are longer than the third. And the largest side lies opposite the largest angle: if c² > a² + b², the angle opposite c is obtuse.

Every triangle has two special circles. The incircle sits inside the triangle and touches all three sides. Its centre, the incentre, is the point where the angle bisectors meet, and its radius is the inradius r. The circumcircle passes through all three vertices. Its centre, the circumcentre, is the point where the perpendicular bisectors of the sides meet, and its radius is the circumradius R.

Inradius. Join the incentre to the three vertices. The triangle splits into three small triangles. Their bases are the sides a, b and c, and each of them has the same height r. So

[[fig:0]]

This gives Area = r × s, or r = Areas. Whenever a question links the area, the perimeter and the inradius, this one formula is enough.

Circumradius. The standard result is

[[fig:1]]

Special triangles. Keep these as ready results.

[[fig:2]]

Try the 5-12-13 right triangle. The area is 30 and s = 15, so r = 3015 = 2, which matches 5 + 12 − 132 = 2. Also R = 132 = 6.5, which matches 5 × 12 × 134 × 30 = 780120 = 6.5. For an equilateral triangle of side 6, r = 62√3 = √3 and R = 2√3.

In a right-angled triangle, the two sides that form the 90° angle are called the legs (perpendicular p and base b). The side opposite the right angle is the hypotenuse h, and it is always the longest side. Pythagoras connects the three:

[[fig:0]]

The two legs are perpendicular to each other, so each leg is the height for the other. That is why the area is simply

Now take the hypotenuse as the base. Its height is the perpendicular from the right-angle vertex to the hypotenuse. The area is the same either way, so 12 × h × altitude = 12 × p × b, which gives

Altitude on the hypotenuse = p × bh

[[fig:1]]

Triplets save you square roots. If two sides fit a triplet, the third side is already known. Keep the base triplets in mind: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (9, 40, 41), (12, 35, 37), and any multiple of them. With legs 9 and 40, the hypotenuse is 41, the area is 12 × 9 × 40 = 180, and the altitude on the hypotenuse is 36041.

Sides in a triplet ratio. If the sides are 3k, 4k and 5k, the area is 12 × 3k × 4k = 6k² and the perimeter is 12k. With k = 7 the sides are 21, 28 and 35, the perimeter is 84 and the area is 6 × 49 = 294.

Which side is the hypotenuse? Only the side opposite the 90° angle. If the question says ∠A = 90°, then BC is the hypotenuse and AB, AC are the legs. Read this carefully before you apply Pythagoras.

[[fig:2]]

Two more results are useful. The median from the right-angle vertex to the hypotenuse is half the hypotenuse, and the circumradius is h2.

An isosceles triangle has two equal sides a and a third side, the base b. The perpendicular from the top vertex to the base is special: it falls exactly on the midpoint of the base. It therefore splits the triangle into two identical right triangles, each with hypotenuse a and base b2.

[[fig:0]]

Say the equal sides are 13 and the base is 10. Half the base is 5, so h = √169 − 25 = 12 (the 5-12-13 triplet), and the area is 12 × 10 × 12 = 60. The triplet has done the work once again.

Equal sides with the angle between them. If each equal side is a and the angle between them (the vertex angle) is θ, then Area = 12a² sin θ. With θ = 30° the area is a²4; with θ = 90° it is a²2.

[[fig:1]]

Isosceles right triangle. The two legs are equal (a and a) and the angles are 45°, 45° and 90°.

[[fig:2]]

Radhika folds a square napkin of side 6 cm along its diagonal. She gets an isosceles right triangle with legs of 6 cm, a hypotenuse of 6√2 cm and an area of 18 cm².

Sides in a ratio. If the sides are in the ratio p : p : q, take them as pk, pk and qk. Write the area in terms of k² using the height formula (or Heron), and compare it with the given area.

All three sides are equal (a) and all three angles are 60°. Drop the height from any vertex. It falls on the midpoint of the opposite side and makes a 30-60-90 triangle with hypotenuse a and base a2. So

[[fig:0]]

Perimeter = 3a

Learn the reverse forms as well, because a question often gives the height or the area instead of the side: a = 2h√3 and Area = h²√3.

A side of 8 cm gives a height of 4√3 cm, an area of 16√3 cm² and a perimeter of 24 cm. A height of 6√3 cm gives a side of 12 cm and an area of 36√3 cm².

Areas add like squares. All equilateral triangles are similar, and the area is proportional to a². So if one equilateral triangle has an area equal to the sum of the areas of two others, its side satisfies a² = a₁² + a₂², exactly as in Pythagoras. Sides 5 and 12 give a third side of 13.

[[fig:1]]

Equal area with another figure. If an equilateral triangle has the same area as some other figure, set √34x² equal to that area and solve for x.

A quadrilateral has four sides, four vertices and two diagonals, and its interior angles add up to 360°. A general quadrilateral has no ready-made area formula, but one rule applies to every one of them: a diagonal cuts the quadrilateral into two triangles.

Take a diagonal d. The other two vertices lie on opposite sides of it. Let their perpendicular distances from the diagonal be h₁ and h₂. Both triangles stand on the same base d, so

[[fig:0]]

For a diagonal of 16 m with perpendiculars of 5 m and 7 m, the area is 12 × 16 × 12 = 96 m².

If the quadrilateral is made of two triangles you already know (two right triangles, for example), add their areas.

Diagonals perpendicular to each other. This covers the kite, the rhombus, the square and any other quadrilateral whose diagonals cross at 90°:

[[fig:1]]

Radhika's kite has diagonals of 10 cm and 14 cm, so its area is 70 cm². If one diagonal is made 20% longer, the area also goes up by 20%, because the area is directly proportional to each diagonal.

Two results that come up in questions.

• Joining the midpoints of the sides of any quadrilateral gives a parallelogram whose area is exactly half the original.
• Two tangents drawn from an external point to a circle, together with the two radii to the points of contact, form a kite. It is made of two equal right triangles, because a radius is perpendicular to the tangent at the point of contact.

[[fig:2]]

A rectangle has its opposite sides equal and all four angles 90°. With length l and breadth b:

The diagonal cuts the rectangle into two right triangles with legs l and b, so every diagonal question is really a Pythagoras question. Quite often l, b and d form a triplet.

[[fig:0]]

The identity behind the harder questions. Square the sum and the difference of the sides:

[[fig:1]]

So if any two of (l + b), d and the area are given, you get the third without finding l and b. A rectangle of 20 × 21 has d = 29 (a triplet), area 420 and perimeter 82. Check the identity: (20 + 21)² = 1681, and 29² + 2 × 420 = 841 + 840 = 1681.

Sum and difference. If the perimeter and the difference l − b are given, half the perimeter gives l + b, and then l = (l + b) + (l − b)2. With a perimeter of 60 and l − b = 6, l + b = 30, so l = 18, b = 12 and the area is 216.

[[fig:2]]

Ratio. If l : b = p : q, write l = pk and b = qk. Then the area is pqk² and the perimeter is 2(p + q)k. Find k from whichever one is given.

Changes in the sides. Suppose the length and breadth are changed by fixed amounts and the area stays the same, or changes by a known amount. Write the new area as (l ± x)(b ± y), expand it and compare with lb. The lb terms cancel and a simple equation is left.

Here the rectangle formulas are used inside stories. First decide whether the story is about the boundary (fencing, a border, a frame: perimeter) or about the surface (paving, tiling, painting, carpeting, levelling: area). After that the work is straightforward.

A path outside a rectangle. A garden l × b has a path of width w running all round it on the outside. The outer rectangle is (l + 2w) × (b + 2w), because the width is added on both sides. So

A path inside, along the boundary. Now the inner rectangle is (l − 2w) × (b − 2w). So

[[fig:0]]

Radhika's garden is 30 m × 20 m. A 2 m path outside it has an area of 2 × 2 × (30 + 20 + 4) = 216 m². A 2 m path inside it has an area of 2 × 2 × (50 − 4) = 184 m².

Two roads crossing in the middle, one parallel to each side and each of width w: area = wl + wb − w². The central w × w square has been counted twice, so subtract it once. For the same garden with 3 m roads: 3 × 30 + 3 × 20 − 9 = 141 m².

[[fig:1]]

Tiles. Number of tiles = area of floorarea of one tile, after both are in the same unit. If an extra percentage is bought to cover breakage, multiply by (1 + x100) and round up. The largest square tile that fits a floor exactly has side = HCF of the length and breadth, and then the least number of tiles = lHCF × bHCF.

[[fig:2]]

Cost. For surface work, cost = area × rate per m². For fencing, cost = perimeter × rate per m. If a gate is left open, subtract its width from the perimeter. For a carpet of given width, length of carpet = area of floorwidth of carpet.

Floors and walls made of parts. An L-shaped floor is two rectangles added together. A wall with a window is the wall area minus the window area.

Walls of a room. Painting or whitewashing a room means finding the area of the four walls. Open the walls out into one long strip: its length is the perimeter of the floor and its height is h, so area of four walls = 2(l + b) × h. The floor is not a wall, and the ceiling is added only when the question says so.

From this area subtract everything that is not painted: each door and window is a small rectangle, so take its length × breadth and subtract it. The painted area then gives the cost (painted area × rate) or the number of paint cans (painted area ÷ area one can covers).

Radhika's room is 5 m long, 4 m wide and 3 m high. The walls are 2 × 9 × 3 = 54 m². One door of 2 m × 1 m and one window of 1.5 m × 1 m take away 2 + 1.5 = 3.5 m², so the painted area is 50.5 m². At ₹20 per m² the cost is ₹1,010. If the ceiling has to be whitewashed too, add 5 × 4 = 20 m² before multiplying by the rate.

[[fig:3]]

Posts and rounds of wire. A fence is made of posts at equal gaps, with wire running between them. On a closed boundary such as a rectangle, the number of posts = the number of gaps = perimeter ÷ gap. (On an open straight line the posts are one more than the gaps.) If the wire goes round n times, the length of wire is n × perimeter, and a gate that is left open reduces the length of each round by the width of the gate.

The same perimeter idea answers walking questions: time taken to go round a field = perimeter ÷ speed. Remember to bring the speed and the length to the same unit first (km/h to m/min, or metres to km).

Radhika's field is 40 m × 30 m. Posts every 5 m need 140 ÷ 5 = 28 posts. Three rounds of wire need 3 × 140 = 420 m.

[[fig:4]]

A square is a rectangle with all four sides equal. With side a:

Since d = a√2, a² = d²2. So area of a square = (diagonal)²2, and you never need the side once the diagonal is known. A diagonal of 12 cm gives an area of 1442 = 72 cm². A side of 9 cm gives a diagonal of 9√2 cm.

[[fig:0]]

From the area to the other values. Area A ⇒ side √A ⇒ perimeter 4√A. That is why the square roots of perfect squares, at least up to 30², should be memorised.

The square on the midpoints. Join the midpoints of the sides of a square and you get a smaller square with side a√2 and exactly half the area. Do it again and you get a quarter of the original, and so on.

[[fig:1]]

Square paths. A path of width w outside a square of side a has area (a + 2w)² − a² = 4w(a + w). A path inside has area a² − (a − 2w)² = 4w(a − w). A 15 m square plot with a 1 m path inside has a path area of 4 × 1 × 14 = 56 m².

[[fig:2]]

Square and rectangle together. For the same perimeter, the square has the larger area. Many questions link a square and a rectangle through equal area or equal perimeter. Write the condition as an equation in the side and solve it. When a square is cut into equal smaller squares, it is the side that gets divided: a square cut into n² equal squares has small squares of side an.

Squares in algebra. If the area is a perfect-square expression such as 9x² + 30x + 25 = (3x + 5)², the side is 3x + 5 and the perimeter is 4(3x + 5).

In a parallelogram both pairs of opposite sides are parallel and equal. The opposite angles are equal, any two adjacent angles add up to 180°, and the diagonals bisect each other. In general the diagonals are neither equal nor perpendicular.

With adjacent sides a and b:

Hold on to the word corresponding. The height must be the perpendicular distance between the base and the side opposite to it. A parallelogram has two heights, one for each pair of sides, and

[[fig:0]]

So the sides are inversely proportional to their heights, and the longer side has the shorter height. With a base of 15 cm and a height of 8 cm on it, the area is 120 cm². If the other side is 12 cm, its height is 12012 = 10 cm.

Two sides with the included angle θ. Drop the height from one end of side a onto side b: h = a sin θ. So

[[fig:1]]

With sides of 9 cm and 6 cm and an angle of 30° between them, the area is 9 × 6 × 12 = 27 cm², and the height on the 9 cm side is 6 sin 30° = 3 cm.

Diagonals. Each diagonal divides the parallelogram into two congruent triangles, each with half the area. Also, the sum of the squares of the diagonals equals the sum of the squares of the four sides: d₁² + d₂² = 2(a² + b²).

[[fig:2]]

When the parallelogram is really a rectangle or a rhombus. If a diagonal satisfies d² = a² + b², the angle between a and b is 90°, so the figure is a rectangle and its area is ab. If the diagonals are perpendicular, it is a rhombus, and the rhombus formulas of the later type apply.

A trapezium has exactly one pair of parallel sides. The parallel sides (a and b) are often called the bases, and the perpendicular distance between them is the height h.

[[fig:0]]

Why does this work? A diagonal cuts the trapezium into two triangles with bases a and b and the same height h, so the area is 12ah + 12bh. With parallel sides of 18 m and 12 m and a height of 7 m, the area is 12 × 30 × 7 = 105 m².

The midline. The segment joining the midpoints of the non-parallel sides is parallel to the bases and equal to their average, a + b2. So the area is also midline × height.

[[fig:1]]

Isosceles trapezium (equal non-parallel sides). Drop perpendiculars from the ends of the shorter base. They cut off two equal pieces from the longer base, each equal to b − a2. Each non-parallel side c is then the hypotenuse of a right triangle:

[[fig:2]]

Take parallel sides of 20 cm and 8 cm with equal slant sides of 10 cm. Each piece is 20 − 82 = 6, so h = √100 − 36 = 8 (the 6-8-10 triplet), and the area is 12 × 28 × 8 = 112 cm².

Sides in a ratio, or with a difference. If the parallel sides are pk and qk, or x and x + d, put them into the area formula and solve for k (or x).

More information than you need. Some questions give all four sides and the height as well. Once the height is known, the non-parallel sides are not needed for the area, so do not let them distract you.

A plot or banner in the shape of a trapezium. Real-life questions only add a rate to the same area formula. The area of the plot is 12(a + b)h; multiply it by the rate per m² for the price of the land, or for levelling, ploughing, tiling or printing. A fence, on the other hand, goes round the boundary, so it needs all four sides: perimeter = a + b + c + d.

Two small points save time. If one slant side stands at a right angle to the parallel sides (a right trapezium), that side is itself the height. If the height is not given but both slant sides are equal, find it from the isosceles result c² = h² + (b − a2)².

Radhika's plot has parallel sides of 45 m and 35 m, 24 m apart. Area = 12 × 80 × 24 = 960 m². At ₹1,500 per m² the land costs 960 × 1,500 = ₹14,40,000.

[[fig:3]]

A rhombus is a parallelogram with all four sides equal. Its most useful property is this: the diagonals bisect each other at right angles. So the two diagonals cut the rhombus into four identical right triangles, each with legs d₁2 and d₂2 and hypotenuse a, the side of the rhombus.

From these right triangles:

[[fig:0]]

Perimeter = 4a

A rhombus is also a parallelogram, so Area = a × h (side × height) = a² sin θ, where θ is any angle of the rhombus.

[[fig:1]]

With diagonals of 10 cm and 24 cm, the half-diagonals are 5 and 12, so the side is 13 cm (a triplet). The perimeter is 52 cm, the area is 12 × 10 × 24 = 120 cm², and the height is 12013 cm.

Side and area given, sum or difference of the diagonals asked. Square the sum:

For side 13 and area 120, (d₁ + d₂)² = 676 + 480 = 1156, so d₁ + d₂ = 34, which is indeed 10 + 24. The identity also works the other way: if the sum of the diagonals and the side are given, the area is (d₁ + d₂)² − 4a²4.

A parallelogram whose diagonals are perpendicular is a rhombus. Also, joining the midpoints of the sides of a rectangle gives a rhombus with half the rectangle's area.

[[fig:2]]

A circle is the set of all points at the same distance r (the radius) from a fixed point, the centre. The diameter d = 2r is the longest chord.

[[fig:0]]

Two results that link them save time. From C = 2πr and A = πr² we get AC = r2 and A = C²4π. So the area can be found from the circumference without working out r first.

For a radius of 10.5 cm, C = 2 × 227 × 10.5 = 66 cm and A = 227 × 10.5 × 10.5 = 346.5 cm². As a check, AC = 346.566 = 5.25 = r2.

Wheels. In one complete revolution a wheel moves forward by its circumference. So

[[fig:1]]

A wheel of radius 35 cm makes 400 revolutions, so it covers 400 × 220 cm = 88,000 cm = 880 m. If it takes 1 minute to do this, its speed is 880 m/min.

Gears and belts. When one gear turns another, the same length of teeth passes the point of contact for both, so revolutions × circumference is equal for the two gears: n₁d₁ = n₂d₂. The bigger gear turns fewer times. A 12 cm gear making 90 turns will turn an 18 cm gear 60 times.

A wire bent into a new shape. The length of the wire does not change, so the perimeter stays the same. A 66 cm wire bent into a circle has r = 10.5 and encloses 346.5 cm². Bent into a square it has side 16.5 and encloses only 272.25 cm². The circle encloses more.

[[fig:2]]

Combining circles. If a circle has an area equal to the sum of the areas of circles with radii r₁, r₂, …, then R² = r₁² + r₂² + …. If its circumference equals the sum of their circumferences, then R = r₁ + r₂ + …. For radii 5 and 12, the area-sum circle has R = 13 and the circumference-sum circle has R = 17.

Clock hands. The tip of the minute hand moves on a circle whose radius is the length of the hand. It turns through 360° in 60 minutes, so in t minutes it covers t60 of the circumference.

A semicircle is half of a circle, cut off by a diameter. Its area is half of πr², but its perimeter is not half of 2πr, because the boundary also includes the diameter.

[[fig:0]]

With π = 227, π + 2 = 367, so the perimeter of a semicircle is 367r. For r = 21 cm the perimeter is 108 cm and the area is 12 × 1386 = 693 cm².

A quadrant is one quarter of a circle, bounded by two radii at 90° and an arc.

Arc alone, or the whole perimeter? Length of the curved part or a frame along the arc means only the arc. Perimeter or boundary means the arc plus the straight edges. A semicircular window framed all round needs πr + 2r of frame. Read the wording carefully.

What is left of a circle. If a quarter of a round pizza is taken out, the remaining piece is three quarters of the circle. Its boundary is 34 × 2πr + 2r, because the two cut radii are now edges.

[[fig:1]]

Composite figures. Most questions here use a rectangle (or a trapezium or triangle) with semicircles or quadrants attached. Add the areas of the pieces. For the perimeter count only the outer boundary: the side on which a semicircle is attached is no longer part of the boundary.

[[fig:2]]

Radhika's study room is a 6 m × 5 m rectangle with a semicircular bay window of diameter 5 m on one of the shorter sides. The floor area is 30 + 12 × 3.14 × 2.5² = 30 + 9.8125 ≈ 39.81 m².

A running track (stadium shape) is a rectangle with semicircles on its two shorter sides, which is the same as the rectangle plus one full circle whose diameter equals the width.

When one figure is drawn inside another so that it just touches it, the whole solution depends on one common length. Find that length and both figures are known.

Circle inside a square (touching all four sides): diameter = side, so r = a2. Circle area : square area = πa²4 : a² = π : 4.

Square inside a circle (all four corners on the circle): the diagonal of the square = the diameter. So a√2 = 2r, and the square's area is d²2 = (2r)²2 = 2r². This is also the largest square, and the largest rectangle, that can be cut from a circle. From a circle of radius 10, the largest square has area 200.

[[fig:0]]

Rectangle inside a circle: the diagonal of the rectangle = the diameter of the circle.

Largest circle inside a rectangle: its diameter = the shorter side.

[[fig:1]]

Circle inside an equilateral triangle (the incircle): r = a2√3, and the circumcircle has R = a√3 = 2r. The ring between the two circles has area π(R² − r²) = π × a²4.

[[fig:2]]

Circle inside a right triangle: r = p + b − h2.

The same wire as a circle or a square. With the same perimeter P, the circle encloses P²4π and the square P²16. Circle : square = 4 : π, so the circle is always the bigger one.

Radhika cuts the largest possible circle from a square card of side 12 cm. Here r = 6 cm, the circle's area is 36π ≈ 113.04 cm², and the card wasted is 144 − 113.04 = 30.96 cm².

Two circles with the same centre are concentric. The region between them is a ring (annulus): a circular path around a park, the rim of a plate, the cross-section of a pipe. With outer radius R and inner radius r:

[[fig:0]]

The factorised form π(R + r)(R − r) is the quickest. It turns a subtraction of two big squares into one small multiplication.

[[fig:1]]

A path outside a circular park of radius r and width w has outer radius R = r + w. A path inside, along the boundary, has inner radius R − w. Read outside and inside carefully.

[[fig:2]]

A pond of radius 14 m has a 7 m wide path around it on the outside. Then R = 21, and the path area is 227 × (21 + 14)(21 − 14) = 227 × 35 × 7 = 770 m².

The circumferences give the width directly. Outer circumference − inner circumference = 2πR − 2πr = 2πw. So width = difference of circumferences2π. With circumferences of 440 m and 352 m, the width is 882π = 14 m.

Ratio of circumferences = ratio of radii. If the outer and inner circumferences are in the ratio p : q and the width is w, then R = pk, r = qk and (p − q)k = w. A ratio of 9 : 8 with a width of 10 gives k = 10, so the inner radius is 80.

Percentage increase in area because of an outside path: R² − r²r² × 100. This is only the scaling rule with the factor Rr.

A chord is a line segment joining two points on a circle, and the longest chord is the diameter. Two facts from Geometry do all the work in this type:

So the radius r, half the chord c2 and the distance d from the centre form a right triangle:

[[fig:0]]

With a radius of 17 cm and a chord of 30 cm, the half-chord is 15, so d = √289 − 225 = 8 cm (the 8-15-17 triplet).

Chord from its central angle θ. Chord = 2r sin θ2. The special cases are worth knowing: θ = 60° ⇒ chord = r (the triangle with the two radii is equilateral); θ = 90° ⇒ chord = r√2; θ = 120° ⇒ chord = r√3; θ = 180° ⇒ chord = 2r.

[[fig:1]]

Parallel chords. If two parallel chords lie on opposite sides of the centre, the distance between them is d₁ + d₂. If they lie on the same side, it is d₁ − d₂.

Segment. A chord cuts the circle into two segments. The smaller one is the minor segment:

[[fig:2]]

Major segment = πr² − minor segment.

Ready forms: θ = 90° ⇒ r²(π4 − 12); θ = 60° ⇒ r²(π6 − √34); θ = 120° ⇒ r²(π3 − √34). For r = 14 and θ = 90°, the sector is 154, the triangle is 98 and the segment is 56 cm².

Two equal circles, each passing through the other's centre. Their common region is made of two 120° segments, so its area is 2r²(π3 − √34) = 2πr²3 − √3r²2.

A sector is the slice of a circle between two radii, like a piece of pizza. Its curved boundary is the arc, and the angle at the centre is θ. A sector is just the fraction θ360° of the whole circle, both for the arc and for the area:

[[fig:0]]

Divide the area by the arc and you get areal = r2. So area of sector = 12 × l × r, which is very handy when the arc length is given but the angle is not.

For r = 21 cm and θ = 60°, l = 16 × 132 = 22 cm and the area is 16 × 1386 = 231 cm². The check 12 × 22 × 21 = 231 agrees. The perimeter is 22 + 42 = 64 cm.

Radian measure. The angle at the centre that cuts off an arc equal to the radius is 1 radian. So in radians θ = lr, and

[[fig:1]]

The full circle has an arc of 2πr, so the full angle is 2π radians. Hence π radians = 180°, and 1 radian = 180°π ≈ 57.3°. An arc of 15 cm in a circle of radius 6 cm makes θ = 2.5 radians. A wheel that turns n times turns through 2πn radians.

Major sector. If the minor sector has angle θ, the major sector has angle 360° − θ.

Equal arcs in different circles. Since l = rθ, for the same arc length r is inversely proportional to θ.

Grazing, and sectors at the corners of a triangle. An animal tied at a corner of a field grazes a sector whose angle is the angle at that corner. If circles of the same radius are drawn at all three vertices of a triangle, the angles of the three sectors inside the triangle add up to 180°, so together they make exactly a semicircle.

[[fig:2]]

A regular polygon has all its sides equal and all its angles equal. With n sides, each of length a:

[[fig:0]]

A regular decagon (n = 10) has an exterior angle of 36°, an interior angle of 144° and 35 diagonals.

The regular hexagon. Join the centre to all six vertices and you get six equilateral triangles of side a. So

[[fig:1]]

The distance from the centre to each vertex is also a. That is why a regular hexagon inscribed in a circle has side = radius. Its longest diagonal is 2a. A hexagon of side 8 cm has area 3√32 × 64 = 96√3 cm² and perimeter 48 cm.

So a hexagon of side a has 6 times the area of an equilateral triangle with the same side.

Regular octagon: Area = 2(1 + √2)a².

[[fig:2]]

A polygon inside a circle of radius R. Each side subtends 360°n at the centre, so side = 2R sin 180°n. For n = 6 this gives R, and for n = 4 it gives R√2.

Regular pentagon: the interior angle is 108°, and the diagonal = 1 + √52 × side ≈ 1.618 × side.

All regular polygons with the same number of sides are similar, so the scaling rule applies: area ratio = (side ratio)² = (perimeter ratio)².