Exploring Some Geometric Themes

Angles, symmetry, constructions, transformations and circles all come from a few simple ideas that you can prove for yourself. This chapter shows you the rules and, more importantly, the reason each one is true.

Lines, angles and the rules they force on each other

Quick answer When two lines cross, or when a line cuts a pair of parallel lines, the angles are locked together. A handful of rules covers almost every angle question you will meet.

Geometry begins with two very plain objects. A point marks a position and has no size. A line is perfectly straight and carries on for ever in both directions. A piece of a line with two end points is a line segment, and a piece with one end point that runs on for ever in one direction is a ray. Two rays that start from the same point make an angle, and that shared starting point is the vertex.

Angles are named by size. An acute angle is less than 90 degrees, a right angle is exactly 90 degrees, an obtuse angle is between 90 and 180 degrees, a straight angle is exactly 180 degrees, and a reflex angle is more than 180 degrees but less than a full turn of 360 degrees. Two angles that add up to 90 degrees are complementary, and two that add up to 180 degrees are supplementary. The complement of 37 degrees is 90 - 37 = 53 degrees, and the supplement of the same angle is 180 - 37 = 143 degrees. Only the total matters, so the two angles do not have to be drawn next to each other.

The first rule that does real work for you is the linear pair. When two adjacent angles sit on a straight line, together they sweep out a straight angle, so they must add to 180 degrees.

Example 1. Two angles form a linear pair and one of them is four times the other. Write them as x and 4x. Then x + 4x = 180, so 5x = 180 and x = 36. The two angles are 36 degrees and 4 × 36 = 144 degrees. Check: 36 + 144 = 180. Correct.

Next come vertically opposite angles. When two straight lines cross, the two angles facing each other across the crossing point are equal, and here is the reason. Call the four angles a, b, c and d as you go round the point. Angles a and b form a linear pair, so a + b = 180. Angles b and c also form a linear pair, so b + c = 180. Both totals are 180, so a + b = b + c, and taking b away from both sides leaves a = c. The same argument gives b = d. The four angles also fill up all the space round the crossing point, so a + b + c + d = 360 degrees. In fact the angles around any point add to 360 degrees, however many of them there are.

Example 2. Three angles at a point measure 90 degrees, 85 degrees and 100 degrees. The fourth is 360 - (90 + 85 + 100) = 360 - 275 = 85 degrees. Check: 90 + 85 + 100 + 85 = 360. Correct.

Two lines drawn on the same flat surface that never meet, however far you extend them, are parallel. A line that cuts across two other lines is a transversal. When the two lines being cut are parallel, the transversal locks their angles together in three ways. Corresponding angles are equal: these are the pairs sitting in matching positions at the two crossings, like the top left angle at each crossing. Alternate interior angles are equal: these lie between the two parallel lines but on opposite sides of the transversal. Co-interior angles are supplementary: these lie between the two parallel lines on the same side of the transversal, and they add to 180 degrees. Every other pair you can spot follows from these three together with linear pairs and vertically opposite angles.

Example 3. A transversal cuts two parallel lines and one co-interior angle is 108 degrees. Its partner is 180 - 108 = 72 degrees. The angle corresponding to the 108 degree angle is also 108 degrees, and the angle alternate to it is 108 degrees as well.

These transversal rules explain why the three angles of a triangle add to 180 degrees. Take triangle ABC and draw a line through A that is parallel to BC. The angle at B equals the alternate angle formed on one side of A, and the angle at C equals the alternate angle formed on the other side of A. Those two angles, together with angle A squeezed between them, make a straight angle along the drawn line. So angle A + angle B + angle C = 180 degrees. Notice how a rule about parallel lines has quietly produced a rule about triangles; that is how geometry grows.

One useful consequence is the exterior angle rule. Extend one side of a triangle past a vertex. The angle formed outside the triangle equals the sum of the two interior angles that are not next to it. The proof is one line: the exterior angle and the interior angle beside it form a linear pair adding to 180 degrees, and the three interior angles also add to 180 degrees, so the exterior angle must equal what is left, which is the other two interior angles.

Example 4. A triangle has two interior angles of 48 degrees and 67 degrees. The exterior angle at the third vertex is 48 + 67 = 115 degrees, and the third interior angle is 180 - 115 = 65 degrees. Check: 48 + 67 + 65 = 180. Correct.

Linear pair: a + b = 180 degrees a and b are adjacent angles whose outer arms form a straight line.
Angles at a point: a + b + c + ... = 360 degrees Use whenever several angles fill up the space round a single point with no gap or overlap.
Co-interior angles on parallel lines: x + y = 180 degrees x and y lie between the parallel lines on the same side of the transversal.
Angle sum of a triangle: angle A + angle B + angle C = 180 degrees True for every triangle, whatever its shape or size.
Exterior angle = sum of the two opposite interior angles The exterior angle is formed by extending one side; the two opposite angles are the ones not touching it.
Remember
  • Angles on a straight line (a linear pair) add to 180 degrees, and all the angles around a point add to 360 degrees.
  • Vertically opposite angles are equal, and the reason is two linear pairs that share one angle.
  • For a transversal cutting parallel lines: corresponding angles are equal, alternate interior angles are equal, co-interior angles are supplementary.
  • The three angles of a triangle add to 180 degrees, proved by drawing a parallel line through one vertex.
  • An exterior angle of a triangle equals the sum of the two interior angles that are not next to it.
  • The parallel-line rules only work when the lines really are parallel, so look for the arrow marks before using them.

Line symmetry: folding a shape onto itself

Quick answer A line of symmetry is a fold that makes the two halves land exactly on top of each other. The mirror line is always the perpendicular bisector of the join from a point to its image.

A shape has line symmetry, also called reflection symmetry, if you can fold it along some straight line so that the two halves land exactly on top of each other, with nothing sticking out anywhere. That fold line is a line of symmetry, and you can also think of it as a mirror line: the half you can see and the half hidden behind the mirror are the same.

The mirror line does something very precise, and this is the part worth remembering. Take any point P of the shape that is not on the mirror line, and find the matching point P dash on the other side. Join them with a straight segment. The mirror line always cuts that segment exactly in half and meets it at a right angle. In short, the mirror line is the perpendicular bisector of the segment joining any point to its image. This one fact turns symmetry from something you judge by eye into something you can construct and test, and you will meet the very same idea again when you build a perpendicular bisector with a compass.

Counting lines of symmetry is a matter of trying every fold that could possibly work and rejecting the ones that do not. A scalene triangle, with all three sides different, has none. An isosceles triangle with exactly two equal sides has one: the fold that runs from the apex to the midpoint of the base. An equilateral triangle has three, one from each vertex.

A square has four lines of symmetry: two that pass through the midpoints of opposite sides, and two along the diagonals. A rectangle that is not a square has only two, both through midpoints of opposite sides. Its diagonals are not lines of symmetry, and there is a clean reason why. Take a rectangle 6 cm long and 3 cm wide. Folding along a diagonal would try to lay the 6 cm side on top of the 3 cm side, and a 6 cm segment cannot possibly sit exactly on a 3 cm segment. The two halves are congruent triangles, yes, but congruent is not the same as landing on each other in that particular fold.

A rhombus that is not a square has two lines of symmetry, and this time they are the diagonals. All four sides of a rhombus are equal, so a fold along a diagonal lays one equal side onto another equal side, and it works. A parallelogram that is neither a rectangle nor a rhombus has no line of symmetry at all: its opposite sides are equal, but no fold brings the slanting sides together. A kite that is not a rhombus has exactly one line of symmetry, the diagonal joining the two vertices where the equal sides meet. An isosceles trapezium also has one, the line through the midpoints of its two parallel sides.

A regular polygon with n sides has exactly n lines of symmetry. When n is even, half of them join opposite vertices and half join midpoints of opposite sides. A regular hexagon, for instance, has 3 vertex-to-vertex lines and 3 midpoint-to-midpoint lines, giving 3 + 3 = 6. When n is odd, each line runs from a vertex to the midpoint of the side opposite it, so a regular pentagon has 5 such lines. A circle beats them all: every single line drawn through its centre is a line of symmetry, so it has infinitely many.

Capital letters are a handy testing ground. In ordinary block capitals, A, H, I, M, O, T, U, V, W, X and Y have a vertical line of symmetry. B, C, D, E, H, I, K, O and X have a horizontal one. Four letters, H, I, O and X, appear in both lists, so each of them has a vertical line of symmetry and a horizontal one. Letters such as F, G, J, L, N, P, Q, R, S and Z have none.

Example 1. How many of the five letters in the word MATHS, written in block capitals, have at least one line of symmetry? M has a vertical one, A has a vertical one, T has a vertical one, and H has both a vertical and a horizontal one. S has none. So 4 of the 5 letters are symmetric.

Example 2. A shape is drawn on squared paper and one of its corners sits 3 squares to the left of a vertical mirror line. Where is the image of that corner? Three squares to the right of the mirror line, at the same height. The join between the two corners is 3 + 3 = 6 squares long, and the mirror line cuts it exactly in the middle, 3 squares from each end, at a right angle. That agrees with the perpendicular bisector rule.

Regular polygon with n sides: number of lines of symmetry = n Equilateral triangle 3, square 4, regular pentagon 5, regular hexagon 6.
Mirror line = perpendicular bisector of P P-dash P is any point of the shape that is not on the mirror line and P-dash is its image; points on the mirror line do not move at all.
Rectangle 2 lines, rhombus 2 lines, square 4 lines, kite 1 line, parallelogram 0 lines The rectangle and rhombus counts are for shapes that are not squares, and the kite count is for a kite that is not a rhombus.
Circle: every line through the centre is a line of symmetry This is why a circle is the most symmetric flat shape there is.
Remember
  • A line of symmetry is a fold along which the two halves of the shape land exactly on top of each other.
  • The mirror line is the perpendicular bisector of the segment joining any point to its mirror image.
  • A regular polygon with n sides has n lines of symmetry, and a circle has infinitely many.
  • A rectangle that is not a square has 2 lines of symmetry and its diagonals are not among them; a rhombus that is not a square has 2, and they are its diagonals.
  • A parallelogram that is neither a rectangle nor a rhombus has no line of symmetry at all.
  • Congruent halves are not enough: the fold must actually carry one half onto the other.

Rotational symmetry, centre and order

Quick answer If turning a shape about a fixed point leaves it looking unchanged, it has rotational symmetry. The order counts how many times that happens in one full turn.

Folding is not the only way a shape can repeat itself. Pin a shape to the page at one point and turn it. If, before you have gone all the way round, the shape sits exactly where it started and looks unchanged, it has rotational symmetry. The pin point is the centre of rotation, and the number of times the shape looks unchanged during one complete turn of 360 degrees is its order of rotational symmetry.

Every shape returns to itself after a full turn, so every shape has order at least 1. Order 1 is simply the polite way of saying that the shape has no rotational symmetry worth talking about.

If a shape has order n, then the n matching positions are spread evenly around the full turn. That gives the rule you will use constantly: the smallest angle of rotation is 360 divided by the order, and every whole multiple of that angle works too. So a shape of order 5 matches itself after 72 degrees, 144 degrees, 216 degrees, 288 degrees and 360 degrees, since 360 ÷ 5 = 72 and 72 × 5 = 360.

Example 1. A square. Turn it a quarter turn about the point where its diagonals cross and each vertex moves to where the next one was; the square looks identical. The order is 4 and the smallest angle is 360 ÷ 4 = 90 degrees. The turns that work are 90, 180, 270 and 360 degrees.

Example 2. An equilateral triangle has order 3, with a smallest angle of 360 ÷ 3 = 120 degrees. A regular hexagon has order 6, with a smallest angle of 360 ÷ 6 = 60 degrees. A regular polygon with n sides always has order n, and the reason is worth stating: all its vertices sit at the same distance from the centre and are spaced evenly, so a turn of 360 ÷ n degrees carries each vertex exactly onto the next one and every side onto the next side.

Example 3. A fan has 5 identical blades set evenly around its hub. Its order is 5 and the smallest turn that leaves it looking the same is 360 ÷ 5 = 72 degrees. Turning it through two blade positions is 2 × 72 = 144 degrees, which also works. A cycle wheel with 8 identical spokes has order 8 and a smallest angle of 360 ÷ 8 = 45 degrees; check that 45 × 8 = 360.

Now for the quadrilaterals. A parallelogram has rotational symmetry of order 2. Turn it a half turn about the point where its diagonals cross and it lands back on itself, because the diagonals of a parallelogram bisect each other, so that crossing point is exactly halfway along each diagonal and a half turn swaps each vertex with the opposite one. A rectangle and a rhombus also have order 2, and a square has order 4. A kite that is not a rhombus has order 1, and so does an isosceles trapezium: neither of them can be turned onto itself except by a full turn. A circle matches itself after a turn of any angle at all, so its rotational symmetry has no finite order.

Here is the point students most often miss: line symmetry and rotational symmetry are separate properties, and a shape can have one without the other. A parallelogram has no line of symmetry but has order 2. The block capital letters S, N and Z have no line of symmetry at all, yet each has rotational symmetry of order 2, because a half turn brings each of them back to itself. An isosceles triangle is the other way round: it has 1 line of symmetry but only order 1. The letters H, I, O and X manage both. Where the two counts do line up is easy to state: whenever a shape has at least one line of symmetry, the number of lines equals the order. A regular polygon with n sides has n lines and order n, a rectangle has 2 lines and order 2, and a kite has 1 line and order 1. The counts can only differ when a shape has no line of symmetry at all, and then the line count is 0 while the order is still at least 1, as with the parallelogram and the letter S.

Example 4. A design looks unchanged when turned through 40 degrees. What is its order? The order is 360 ÷ 40 = 9. Check: 9 × 40 = 360, so nine equally spaced positions fit into the full turn.

Smallest angle of rotation = 360 / order Use when the order is known; the answer is in degrees.
Order = 360 / smallest angle of rotation Use when the turning angle is given, for example 40 degrees gives order 9.
Regular polygon with n sides: order = n and smallest angle = 360 / n Square gives 4 and 90 degrees, regular hexagon gives 6 and 60 degrees.
Turns that work for a shape of order n: k × (360 / n) degrees, for k = 1, 2, 3, ... , n For order 5 these are 72, 144, 216, 288 and 360 degrees.
Remember
  • The order of rotational symmetry counts how many times a shape looks unchanged during one full turn of 360 degrees.
  • Smallest angle of rotation = 360 divided by the order, and every multiple of that angle also works.
  • A regular polygon with n sides has order n; a circle looks unchanged after a turn of any angle.
  • A half turn about the crossing point of the diagonals maps any parallelogram onto itself, so a plain parallelogram, a rectangle and a rhombus each have order 2, while a square has order 4.
  • Order 1 means the shape has no useful rotational symmetry, since every shape returns to itself after a full turn.
  • Line symmetry and rotational symmetry are independent: S has order 2 with no line of symmetry, while an isosceles triangle has a line of symmetry but no rotational symmetry, since its order is only 1.

Reflections, rotations and translations

Quick answer Three moves shift a figure without changing its size or shape. Knowing what each one keeps fixed lets you predict the single move that two reflections add up to.

A transformation is a rule that sends every point of a figure to a new position. Three of them keep the figure exactly the same size and shape, so the image is congruent to the object. They are called rigid motions, because you could carry out any of them with a stiff cardboard cut-out.

A reflection needs a mirror line. Every point moves straight across the line to the point at the same perpendicular distance on the other side, so the mirror line is the perpendicular bisector of every join from a point to its image. Points that already lie on the mirror line do not move at all.

A rotation needs three things: a centre, an angle and a direction, either clockwise or anticlockwise. Every point keeps its distance from the centre and swings round by the given angle. The centre itself is the only point that stays put.

A translation needs a distance and a direction, nothing more. Every point slides by the same amount in the same direction, so the joins from each point to its image are all equal in length and all parallel. No point stays put.

What survives all three? Every length, every angle, the area, and the fact that parallel lines stay parallel. What changes is the position, and for a reflection one extra thing changes as well. Label the corners of a triangle A, B, C so that they read clockwise. After a translation or a rotation they still read clockwise. After a reflection they read anticlockwise. This flip in the sense of the labelling is the give-away that tells a reflection apart from a rotation, and it is why the writing on an ambulance bonnet looks correct only in a mirror.

On squared paper the three moves are easy to work with. Take the point (2, 3).

  • Reflected in the y-axis it becomes (-2, 3), since it is 2 units to the right of the axis, so the image is 2 units to the left.
  • Reflected in the x-axis it becomes (2, -3).
  • Reflected in the vertical line through 5 on the x-axis it becomes (8, 3), because the point is 5 - 2 = 3 units to the left of that line, so the image is 3 units to the right, at 5 + 3 = 8.
  • Translated 4 units right and 2 units up it becomes (2 + 4, 3 + 2) = (6, 5).
  • Rotated through 180 degrees about the origin it becomes (-2, -3).

Now the interesting part. What happens if you reflect a figure twice? Suppose the two mirrors are parallel and a distance d apart. The result is not a reflection at all but a single translation through 2d, at right angles to the mirrors, in the direction that takes you from the first mirror to the second. Here is a check on a number line. Put the first mirror at 0 and the second at 4, so d = 4. A point at 3 is reflected in the first mirror to -3. Reflecting -3 in the mirror at 4 gives 2 × 4 - (-3) = 8 + 3 = 11. The point has moved from 3 to 11, a shift of 11 - 3 = 8, which is 2 × 4. Try another point to be sure: -1 reflects to 1, and 1 reflects in the second mirror to 2 × 4 - 1 = 7, a shift of 7 - (-1) = 8 again. Every point moves by the same 8, which is exactly what a translation does.

Suppose instead the two mirrors meet at a point, at an angle A between them. Then two reflections give a single rotation about that meeting point, through an angle of 2A, in the direction from the first mirror to the second. So mirrors set at 35 degrees produce a rotation of 2 × 35 = 70 degrees. If the two mirrors are at right angles, the result is a rotation of 2 × 90 = 180 degrees, a half turn, which is why a corner made of two mirrors sends a ray of light straight back the way it came. In both cases the pattern is the same: two reflections give you twice, twice the gap or twice the angle.

Transformations also give a cleaner way to say what symmetry means. A shape has line symmetry when there is a reflection that maps the shape onto itself, and rotational symmetry when there is a rotation of less than a full turn that maps the shape onto itself. Symmetry, in other words, is just a transformation that leaves no visible trace.

Two reflections in parallel mirrors d apart = one translation of 2d The translation is perpendicular to the mirrors, running from the first mirror towards the second.
Two reflections in mirrors meeting at angle A = one rotation of 2A The rotation is about the point where the two mirrors meet.
Reflection in the x-axis: (x, y) goes to (x, -y) The x-coordinate is untouched because the point moves straight up or down.
Reflection in the y-axis: (x, y) goes to (-x, y) The y-coordinate is untouched because the point moves straight left or right.
Translation of a units right and b units up: (x, y) goes to (x + a, y + b) Use a negative a for a shift left and a negative b for a shift down.
Remember
  • Reflection, rotation and translation are rigid motions: lengths, angles and area are unchanged, so object and image are congruent.
  • Fixed points differ: a reflection fixes the whole mirror line, a rotation fixes only its centre, a translation fixes nothing.
  • A reflection reverses the sense, so corners labelled clockwise come out anticlockwise; rotations and translations keep the sense.
  • Two reflections in parallel mirrors a distance d apart give one translation of 2d, at right angles to the mirrors.
  • Two reflections in mirrors meeting at angle A give one rotation of 2A about their meeting point.
  • Symmetry restated: line symmetry means a reflection maps the shape onto itself, rotational symmetry means a rotation does.

Constructions with compass and straight-edge, and why they work

Quick answer Two tools only: a straight-edge whose markings you ignore, and a compass. Every step is justified by equal radii and congruent triangles, so the result is exact rather than nearly right.

In a proper construction you are allowed exactly two tools. A straight-edge draws straight lines, and you must ignore its markings. A compass draws circles and arcs, and it also carries a length from one place on the paper to another. The restriction sounds annoying until you notice what it buys you: every point you produce is forced to be in exactly the right place, and you can say precisely why. A measured drawing is only as good as your eyesight; a construction is exact.

Copying a segment. To copy segment AB, open the compass so that its point is at A and its pencil at B. Draw a ray from a new point P and, without changing the opening, put the compass point at P and cut the ray at Q. Then PQ = AB, because the opening never changed. That is the whole justification, and it is the idea behind everything else here.

The perpendicular bisector of AB. Open the compass to more than half the length of AB. With the point at A, draw arcs above and below the segment. Keeping the same opening, put the point at B and draw arcs that cross the first two at X and Y. Draw the line XY. It cuts AB exactly in half and meets it at a right angle.

Why does that work? Because all four arcs had the same opening r, we know AX = BX = r and AY = BY = r. So X is the same distance from A as from B, and so is Y. Now compare triangles AXY and BXY. They have AX = BX, AY = BY, and the side XY in common, so they are congruent by SSS, which gives angle AXY = angle BXY. Next compare triangles AXM and BXM, where M is the point where XY crosses AB. They have AX = BX, angle AXM = angle BXM from the last step, and the side XM in common, so they are congruent by SAS. Therefore AM = MB, which is the bisecting part, and angle AMX = angle BMX. Those last two angles form a linear pair, so they add to 180 degrees; being equal as well, each must be 180 ÷ 2 = 90 degrees, which is the perpendicular part. A shorter way to see the same thing: AXBY has four equal sides, so it is a rhombus, and the diagonals of a rhombus bisect each other at right angles.

Why must the opening be more than half of AB? If it is less, the arcs around A stop short of the arcs around B and they never cross, so there is no X to find. If it is exactly half, the arcs merely touch at the midpoint and you get one point instead of two, which is not enough to draw a line.

The bisector of an angle. With the compass point at the vertex O and any convenient radius, draw an arc cutting the two arms at A and B. Then, with the same opening in both cases, draw arcs from A and from B that cross at C. The ray OC bisects the angle. The reason is one congruence: OA = OB because they are radii of the same arc, AC = BC because those arcs had the same opening, and OC is shared. So triangles OAC and OBC are congruent by SSS, which forces angle AOC = angle BOC.

A perpendicular from a point to a line. With the compass point at P, draw an arc that cuts the line at two points. Now construct the perpendicular bisector of those two points in the usual way; it passes through P and meets the line at a right angle. The construction is not new, only reused, and that is typical of this subject.

An angle of 60 degrees. Draw a ray OA. With centre O and any radius r, draw an arc cutting OA at P. Without changing the opening, put the compass point at P and draw an arc cutting the first one at Q. Join OQ. Then OP = r and OQ = r, since both are radii of the first arc, and PQ = r because the second arc had the same opening. Triangle OPQ therefore has three equal sides, so it is equilateral, and the three equal angles of an equilateral triangle add to 180 degrees, making each one 180 ÷ 3 = 60 degrees. Angle AOQ is exactly 60 degrees, and not one thing was measured.

From 60 degrees you can reach a whole family of angles. Step the same radius round the arc a second time to get 120 degrees, since 60 + 60 = 120. Bisecting 60 gives 30, and bisecting 30 gives 15. Bisect the angle between the 60 mark and the 120 mark and you get (60 + 120) ÷ 2 = 90 degrees; halve that for 45 and halve again for 22.5. Bisect between the 60 ray and the 90 ray for (60 + 90) ÷ 2 = 75 degrees, between the 90 ray and the 120 ray for (90 + 120) ÷ 2 = 105 degrees, and between the 90 ray and the straight line for (90 + 180) ÷ 2 = 135 degrees. Interestingly, 20 degrees is out of reach: mathematicians proved long ago that cutting a 60 degree angle into three equal parts cannot be done with these two tools alone, even though halving it is easy.

Two habits worth building. Keep the pencil sharp, because a fat arc is a fat guess. And never rub out your arcs when the figure is finished; they are the record of how the point was found, and without them a correct construction looks like a lucky drawing.

Perpendicular bisector: compass opening must be more than half of AB A smaller opening leaves the arcs from A and B too short to meet.
SSS congruence justifies both the perpendicular bisector and the angle bisector Equal radii supply the equal sides; the shared side completes the third pair.
60 degrees = each angle of an equilateral triangle = 180 / 3 Three equal compass radii force the triangle to be equilateral.
Bisecting halves an angle: 60 gives 30 gives 15, and 90 gives 45 gives 22.5 Each bisection is the same construction repeated on the new angle.
Angle midway between two constructed rays = (first + second) / 2 Bisecting between the 60 and 120 rays gives 90 degrees; between 60 and 90 gives 75 degrees.
Remember
  • A construction uses only a straight-edge (markings ignored) and a compass, so every point is exactly placed rather than measured.
  • The perpendicular bisector works because the four arcs share one opening, making X and Y equidistant from A and from B.
  • The angle bisector works because triangles OAC and OBC are congruent by SSS, forcing the two half angles to be equal.
  • The 60 degree construction works because three equal compass openings build an equilateral triangle, whose angles are each 60 degrees.
  • Repeated bisection gives 30, 15, 90, 45, 75, 105 and 135 degrees from the basic 60 and 120 degree marks.
  • Set the compass to more than half the segment, or the arcs will never cross.

Circles: parts, symmetry and simple measures

Quick answer The circle is the most symmetric shape there is. Its parts have names worth learning, and two formulas plus Pythagoras handle most of the questions.

A circle is the set of all points on a flat surface that are the same distance from one fixed point. The fixed point is the centre, and the fixed distance is the radius. That single sentence is the source of every circle property in this chapter.

The names of the parts are worth learning properly. A chord is a segment joining two points on the circle. A diameter is a chord that passes through the centre, and it is the longest chord of all, with length d = 2r. An arc is a piece of the curved boundary; the shorter piece is the minor arc and the longer piece the major arc, and a diameter splits the circle into two equal arcs called semicircles. A sector is the slice bounded by two radii and an arc, shaped like a piece of cake. A segment is the region cut off by a chord together with its arc. The circumference is the boundary itself, and also the word for its length. A tangent is a straight line that touches the circle at exactly one point without crossing it.

Measure the circumference of any round object with a thread and divide by its diameter, and you always get the same number, a little more than 3. That number is called pi. It never stops and never repeats, so in schoolwork we use 22/7 or 3.14, whichever the question suggests. This gives the two formulas you need: circumference C = pi × d = 2 × pi × r, and area A = pi × r².

Example 1. A circle has radius 7 cm; take pi = 22/7. Then C = 2 × (22/7) × 7 = 2 × 22 = 44 cm, and A = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm². Radii that are multiples of 7 are chosen exactly so the 7 cancels.

Example 2. A circle has diameter 28 cm, so r = 28 ÷ 2 = 14 cm. Then C = 2 × (22/7) × 14 = 2 × 22 × 2 = 88 cm, and A = (22/7) × 196 = 22 × 28 = 616 cm².

Example 3. Working backwards from a circumference of 132 cm. From C = 2 × (22/7) × r we get 44r ÷ 7 = 132, so r = 132 × 7 ÷ 44 = 21 cm. Check by putting it back: 2 × (22/7) × 21 = 2 × 22 × 3 = 132. Correct. The diameter is 42 cm and the area is (22/7) × 441 = 22 × 63 = 1386 cm².

Example 4. The curved edge of a semicircular protractor of radius 7 cm has length pi × r = (22/7) × 7 = 22 cm. Its complete boundary also includes the straight edge, which is the diameter, so the whole perimeter is 22 + 14 = 36 cm.

Now the symmetry, which is where the circle connects back to the rest of this chapter. Every line drawn through the centre is a line of symmetry, so a circle has infinitely many. It also looks unchanged after a turn about its centre through any angle whatsoever, so its rotational symmetry has no finite order. No other flat shape is anywhere near as symmetric, and that is exactly why wheels are round: whatever angle the wheel has turned through, the centre stays the same height above the road.

One chord fact does a lot of work: the perpendicular drawn from the centre to a chord bisects that chord. The proof uses congruence. Let the perpendicular from the centre O meet the chord AB at M, and join OA and OB. Both are radii, so OA = OB. Triangles OMA and OMB each have a right angle at M, share the side OM, and have equal hypotenuses OA and OB, so they are congruent by RHS. Therefore AM = MB.

Because triangle OMA has a right angle at M, Pythagoras applies and gives a very handy relation: r² = (half the chord)² + (distance of the chord from the centre)².

Example 5. A circle has radius 13 cm and a chord of length 24 cm. Half the chord is 12 cm, so the distance from the centre is √(13² - 12²) = √(169 - 144) = √25 = 5 cm.

Example 6. A circle has radius 10 cm and a chord lying 6 cm from the centre. Half the chord is √(10² - 6²) = √(100 - 36) = √64 = 8 cm, so the chord is 2 × 8 = 16 cm long. Notice the pattern: in the same circle, at a distance of 0 the chord is the diameter, 20 cm; at 6 cm it is 16 cm; at 8 cm it is 2 × √(100 - 64) = 2 × 6 = 12 cm. The nearer a chord is to the centre, the longer it is, and chords of equal length are always the same distance from the centre.

Finally, a construction that ties the last two sections together. Trace a bangle on paper and you have a circle with no marked centre. Draw any two chords that are not parallel and construct the perpendicular bisector of each. They cross at the centre. The reason is exactly the perpendicular bisector property: the centre is one radius from both ends of any chord, so it must lie on the perpendicular bisector of that chord, and lying on two of them pins it down to a single point.

C = 2 × pi × r = pi × d r is the radius and d the diameter; use pi = 22/7 when the radius is a multiple of 7.
A = pi × r² Area is in square units, so write the answer as cm² or m².
d = 2r, so r = d / 2 Read the question carefully to see which of the two is given.
r² = (half the chord)² + (distance from the centre)² Comes from the right-angled triangle made by the radius, half the chord and the perpendicular from the centre.
Perimeter of a semicircle = pi × r + 2r The curved edge plus the straight diameter; forgetting the diameter is a common slip.
Remember
  • Circumference C = 2 × pi × r = pi × d, and area A = pi × r², with pi taken as 22/7 or 3.14.
  • The diameter is twice the radius and is the longest chord of the circle.
  • The perpendicular from the centre to a chord bisects the chord, proved by RHS congruence.
  • Pythagoras then gives r² = (half the chord)² + (distance of the chord from the centre)².
  • A circle has infinitely many lines of symmetry and looks unchanged after a turn through any angle about its centre.
  • The centre of a traced circle is found where the perpendicular bisectors of two chords cross.

Quick checks and the slips to avoid

Quick answer Most lost marks in geometry come from a small set of avoidable habits. Here they are, each with the check that catches it.

Do not mix up the two symmetry counts. The number of lines of symmetry and the order of rotational symmetry answer different questions, and they do not have to be the same number. A parallelogram has 0 lines but order 2, and the block capital S has 0 lines but order 2 as well. The two counts agree whenever a shape has at least one line of symmetry, so an equilateral triangle gives 3 and 3, a rectangle gives 2 and 2, and a kite gives 1 and 1; they can only disagree when the shape has no line of symmetry at all. Read the question, decide which count is wanted, and answer that one.

The diagonals of a rectangle are not lines of symmetry. Test it with numbers rather than by squinting at the page. In a rectangle 6 cm by 3 cm, a fold along a diagonal would have to lay the 6 cm side onto the 3 cm side, and those lengths are different, so the fold cannot work. In a rhombus the diagonals are lines of symmetry, because all four sides are equal there.

Parallel-line rules need parallel lines. Corresponding, alternate and co-interior relationships are only true when the two lines being cut are parallel. If the question does not say so and the diagram carries no arrow marks, you cannot use them. And keep the two families apart: alternate interior angles are equal, while co-interior angles are supplementary. Confusing them turns an answer of 70 degrees into 110 degrees. The quick test: if the two angles lie between the parallel lines and on the same side of the transversal, they add to 180 degrees; if they are on opposite sides, they are equal.

Do not measure inside a construction. If a question says construct, the arcs must do the work; taking out a protractor and marking 60 degrees is not a construction, whatever the finished figure looks like. Use the ruler as a straight-edge only, keep every arc on the page, and if you want to check your work with a protractor afterwards, do it as a test and never as the method.

Treat a diagram as a guide, not as data. A line that looks like a bisector may not be one. Two segments that look equal may not be. Use only what is given in words or shown by proper marks such as ticks for equal sides, arcs for equal angles and small squares for right angles.

Check radius against diameter. In circle questions this single confusion causes more wrong answers than any other. The circumference is 2 × pi × r but pi × d, so putting a diameter where a radius belongs doubles your answer, and in an area calculation it multiplies it by four. Read the question twice, write down r explicitly, and only then substitute.

Reflection reverses the sense. If you have drawn an image and its corners still read round in the same direction as the object, you have performed a rotation or a translation, not a reflection. Check the direction of the labelling before you commit.

Two reflections give you twice. Twice the gap when the mirrors are parallel, twice the angle when they meet. If you find yourself writing the gap or the angle unchanged, you have missed the doubling.

Finish with an arithmetic check. The three angles of a triangle must add to exactly 180 degrees. The four angles of a quadrilateral must add to 360 degrees. The angles round a point must add to 360 degrees. If you solved for an unknown x, put your value back into every expression in the question and add them up; if the total is not what it should be, you have found the mistake yourself. Finally, get the units right: lengths in cm or m, areas in cm² or m², and angles in degrees. An area written without the square unit is an incomplete answer.

One last habit that pays off across the whole of geometry. Whenever you use a rule, ask yourself in one line why it is true. Vertically opposite angles are equal because of two linear pairs. The angle bisector construction works because of an SSS congruence. The perpendicular from the centre bisects a chord because of an RHS congruence. Rules remembered with their reasons stay put; rules memorised alone tend to swap places with each other exactly when you need them.

Triangle 180 degrees, quadrilateral 360 degrees, angles at a point 360 degrees The three totals to check any angle answer against before writing it down.
Alternate interior angles equal, co-interior angles add to 180 degrees Same side of the transversal means supplementary; opposite sides means equal.
C = 2 × pi × r but C = pi × d Use one form or the other, never both, or the answer comes out doubled.
Two reflections: 2 × the gap for parallel mirrors, 2 × the angle for meeting mirrors The doubling is easy to forget and is worth writing down before you calculate.
Remember
  • Lines of symmetry and order of rotational symmetry are different counts: they agree whenever the shape has at least one line of symmetry, and differ only when it has none, as with the parallelogram at 0 lines and order 2.
  • Corresponding, alternate and co-interior rules apply only when the two lines cut by the transversal are parallel.
  • In a construction, never measure: the arcs are the method and the record, so leave them on the page.
  • Check whether a circle question gives the radius or the diameter before substituting into C = 2 × pi × r or A = pi × r².
  • Verify every answer by adding up: 180 degrees in a triangle, 360 degrees in a quadrilateral or round a point.
  • Learn each rule together with its one-line reason, so the rules do not get swapped under pressure.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

Linear pair: a + b = 180 degrees
Angles at a point: a + b + c + ... = 360 degrees
Co-interior angles on parallel lines: x + y = 180 degrees
Angle sum of a triangle: angle A + angle B + angle C = 180 degrees
Exterior angle = sum of the two opposite interior angles
Regular polygon with n sides: number of lines of symmetry = n
Mirror line = perpendicular bisector of P P-dash
Rectangle 2 lines, rhombus 2 lines, square 4 lines, kite 1 line, parallelogram 0 lines
Circle: every line through the centre is a line of symmetry
Smallest angle of rotation = 360 / order
Order = 360 / smallest angle of rotation
Regular polygon with n sides: order = n and smallest angle = 360 / n
Turns that work for a shape of order n: k × (360 / n) degrees, for k = 1, 2, 3, ... , n
Two reflections in parallel mirrors d apart = one translation of 2d
Two reflections in mirrors meeting at angle A = one rotation of 2A
Reflection in the x-axis: (x, y) goes to (x, -y)
Reflection in the y-axis: (x, y) goes to (-x, y)
Translation of a units right and b units up: (x, y) goes to (x + a, y + b)
Perpendicular bisector: compass opening must be more than half of AB
SSS congruence justifies both the perpendicular bisector and the angle bisector
60 degrees = each angle of an equilateral triangle = 180 / 3
Bisecting halves an angle: 60 gives 30 gives 15, and 90 gives 45 gives 22.5
Angle midway between two constructed rays = (first + second) / 2
C = 2 × pi × r = pi × d
A = pi × r²
d = 2r, so r = d / 2
r² = (half the chord)² + (distance from the centre)²
Perimeter of a semicircle = pi × r + 2r
Triangle 180 degrees, quadrilateral 360 degrees, angles at a point 360 degrees
Alternate interior angles equal, co-interior angles add to 180 degrees
C = 2 × pi × r but C = pi × d
Two reflections: 2 × the gap for parallel mirrors, 2 × the angle for meeting mirrors

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1

Two angles form a linear pair and one of them is four times the other. What is the smaller angle?

Q2

A transversal cuts two parallel lines. One co-interior angle measures 108 degrees. What does its partner measure?

Q3

How many lines of symmetry does a rectangle that is not a square have?

Q4

Two interior angles of a triangle are 48 degrees and 67 degrees. What is the exterior angle at the third vertex?

Q5

What is the smallest angle of rotation that maps a regular hexagon onto itself?

Q6

Which of these transformations reverses the sense, so that corners labelled clockwise come out anticlockwise?

Q7

A figure is reflected in one mirror and then in a second mirror parallel to the first, 5 cm away. What single move has the same effect?

Q8

In the compass construction of a 60 degree angle, why is the angle exactly 60 degrees?

Q9

A chord of length 16 cm is drawn in a circle of radius 10 cm. How far is the chord from the centre?

Q10

Taking pi as 22/7, what is the circumference of a circle of radius 21 cm?

Q11

Which block capital letter has rotational symmetry of order 2 but no line of symmetry at all?

Q12

How many lines of symmetry does a circle have?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 8

1 Two straight lines AB and CD cross at O. If angle AOC = (3x + 10) degrees and angle BOD = (5x - 30) degrees, find x and all four angles at O.

Angles AOC and BOD face each other across the crossing point, so they are vertically opposite and therefore equal.

3x + 10 = 5x - 30

Take 3x from both sides: 10 = 2x - 30. Add 30 to both sides: 40 = 2x, so x = 40 ÷ 2 = 20.

Now substitute:

  • angle AOC = 3(20) + 10 = 60 + 10 = 70 degrees
  • angle BOD = 5(20) - 30 = 100 - 30 = 70 degrees (the two agree, as they must)
  • angle AOD forms a linear pair with angle AOC, so it is 180 - 70 = 110 degrees
  • angle BOC is vertically opposite angle AOD, so it is 110 degrees

Check: the four angles fill the space round O, and 70 + 110 + 70 + 110 = 360. Correct.

2 Lines l and m are parallel and are cut by a transversal. One of a pair of co-interior angles is five times the other. Find both angles.

Co-interior angles between parallel lines are supplementary, so they add to 180 degrees.

Let the smaller angle be x degrees, so the larger is 5x degrees.

x + 5x = 180

6x = 180, so x = 180 ÷ 6 = 30.

The angles are 30 degrees and 5 × 30 = 150 degrees.

Check: 30 + 150 = 180, and 150 is indeed five times 30. Correct.

3 For each shape, write the number of lines of symmetry and the order of rotational symmetry: (a) equilateral triangle, (b) square, (c) rectangle that is not a square, (d) rhombus that is not a square, (e) parallelogram that is neither a rectangle nor a rhombus, (f) regular hexagon, (g) circle.

For each shape, count the folds that work, then count the positions in one full turn of 360 degrees.

  • (a) Equilateral triangle: 3 lines of symmetry, order 3. Smallest turn = 360 ÷ 3 = 120 degrees.
  • (b) Square: 4 lines of symmetry (2 through midpoints of opposite sides, 2 along the diagonals), order 4. Smallest turn = 360 ÷ 4 = 90 degrees.
  • (c) Rectangle, not a square: 2 lines of symmetry, order 2. Smallest turn = 180 degrees. Its diagonals are not lines of symmetry.
  • (d) Rhombus, not a square: 2 lines of symmetry, and here they are the diagonals; order 2, smallest turn 180 degrees.
  • (e) Parallelogram, neither: 0 lines of symmetry, order 2. A half turn about the point where the diagonals cross maps it onto itself.
  • (f) Regular hexagon: 6 lines of symmetry (3 vertex to vertex, 3 midpoint to midpoint), order 6. Smallest turn = 360 ÷ 6 = 60 degrees.
  • (g) Circle: infinitely many lines of symmetry, one through the centre in every direction, and it matches itself after a turn through any angle.

Note: every shape in this list that has at least one line of symmetry has an order equal to its number of lines, and that includes the rectangle and the rhombus, which are not regular. The parallelogram is the exception, with 0 lines but order 2, so the two counts must always be worked out separately.

4 Construct the perpendicular bisector of a line segment AB of length 7.4 cm. Write the steps and explain why the construction gives a line that is both a bisector and a perpendicular.

Steps.

  1. Draw AB = 7.4 cm with a ruler.
  2. Open the compass to more than half of 7.4 cm, that is more than 3.7 cm. An opening of about 5 cm is convenient.
  3. With the point at A, draw arcs above and below AB.
  4. Keeping exactly the same opening, put the point at B and draw arcs cutting the first two at X (above) and Y (below).
  5. Draw the line XY. It meets AB at a point M.

Why it works. All four arcs were drawn with one opening r, so AX = BX = r and AY = BY = r. In triangles AXY and BXY we have AX = BX, AY = BY and the common side XY, so the two triangles are congruent by SSS, which gives angle AXY = angle BXY.

Now in triangles AXM and BXM we have AX = BX, angle AXM = angle BXM (just proved) and the common side XM, so they are congruent by SAS. Hence AM = MB, so the line bisects AB, and angle AMX = angle BMX. These two form a linear pair, so they add to 180 degrees; being equal as well, each is 180 ÷ 2 = 90 degrees.

Result. AM = MB = 7.4 ÷ 2 = 3.7 cm, and XY meets AB at 90 degrees. Check with a ruler: 3.7 + 3.7 = 7.4. Correct.

5 Using only a compass and a straight-edge, construct an angle of 75 degrees. Explain the plan and justify each angle you use.

Plan. 75 is the number exactly midway between 60 and 90, since (60 + 90) ÷ 2 = 75. So build 60, build 90, then bisect the angle between them.

  1. Draw a ray OA. With centre O and any radius r, draw an arc cutting OA at P.
  2. With the same opening and centre P, cut the arc at Q. Then OP = OQ = PQ = r, so triangle OPQ is equilateral and angle AOQ = 180 ÷ 3 = 60 degrees.
  3. With the same opening and centre Q, cut the arc again at R. By the same reasoning angle QOR = 60 degrees, so angle AOR = 60 + 60 = 120 degrees.
  4. Bisect angle QOR: with centres Q and R and a suitable opening, draw arcs meeting at S. Then angle AOS = (60 + 120) ÷ 2 = 90 degrees.
  5. Now bisect angle QOS, which lies between the 60 degree ray and the 90 degree ray. Call the bisector OT. Then angle AOT = (60 + 90) ÷ 2 = 75 degrees.

Why each bisection is exact. In every bisection the two arcs are drawn with one compass opening from two points that are the same distance from O, so the two triangles formed are congruent by SSS and the two half angles must be equal.

Check: 60 + 15 = 75, and 15 is half of the 30 degree gap between the 60 and 90 rays. Correct.

6 A chord of a circle of radius 25 cm has length 48 cm. Find the distance of the chord from the centre.

Drop a perpendicular from the centre O to the chord AB, meeting it at M. The perpendicular from the centre bisects the chord, so AM = 48 ÷ 2 = 24 cm.

Join OA. It is a radius, so OA = 25 cm, and triangle OMA has a right angle at M.

By Pythagoras: OA² = AM² + OM²

25² = 24² + OM²

625 = 576 + OM²

OM² = 625 - 576 = 49, so OM = √49 = 7 cm.

Check: 7² + 24² = 49 + 576 = 625 = 25². Correct.

7 The circumference of a circular garden is 176 m. Taking pi = 22/7, find its radius, its diameter and its area. If fencing costs 35 rupees per metre, find the cost of fencing the garden once.

Radius. C = 2 × pi × r, so 176 = 2 × (22/7) × r = 44r ÷ 7.

Multiply both sides by 7: 1232 = 44r, so r = 1232 ÷ 44 = 28 m.

Check: 2 × (22/7) × 28 = 2 × 22 × 4 = 176. Correct.

Diameter. d = 2r = 2 × 28 = 56 m.

Area. A = pi × r² = (22/7) × 28² = (22/7) × 784. Since 784 ÷ 7 = 112, A = 22 × 112 = 2464 m².

Cost of fencing. The fence runs once round the boundary, a length of 176 m, at 35 rupees per metre.

Cost = 176 × 35 = 176 × 30 + 176 × 5 = 5280 + 880 = ₹6160.

8 A triangle has vertices A(1, 2), B(4, 2) and C(1, 6) on squared paper. Write the coordinates of the image of each vertex after (a) a reflection in the y-axis, (b) a translation of 3 units right and 2 units down, (c) a rotation of 180 degrees about the origin. State one thing that is the same in all three images.

(a) Reflection in the y-axis. The rule is (x, y) goes to (-x, y), so only the sign of the x-coordinate changes.

  • A(1, 2) goes to (-1, 2)
  • B(4, 2) goes to (-4, 2)
  • C(1, 6) goes to (-1, 6)

(b) Translation 3 right and 2 down. The rule is (x, y) goes to (x + 3, y - 2).

  • A(1, 2) goes to (1 + 3, 2 - 2) = (4, 0)
  • B(4, 2) goes to (4 + 3, 2 - 2) = (7, 0)
  • C(1, 6) goes to (1 + 3, 6 - 2) = (4, 4)

(c) Rotation of 180 degrees about the origin. The rule is (x, y) goes to (-x, -y).

  • A(1, 2) goes to (-1, -2)
  • B(4, 2) goes to (-4, -2)
  • C(1, 6) goes to (-1, -6)

What stays the same. All three are rigid motions, so every side length is unchanged and each image is congruent to the original triangle. In the original, AB = 4 - 1 = 3 units, AC = 6 - 2 = 4 units, and by Pythagoras BC = √(3² + 4²) = √(9 + 16) = √25 = 5 units. Every image is therefore a 3, 4, 5 right-angled triangle as well, with the right angle at the image of A.

Previous-year board questions 6

Q1 AB is parallel to CD and a transversal cuts them. One angle measures (2x + 15) degrees and the angle corresponding to it measures (3x - 25) degrees. Find x and the size of the angle. 3 marks mark

Corresponding angles on parallel lines are equal, so set the two expressions equal.

2x + 15 = 3x - 25

Take 2x from both sides: 15 = x - 25. Add 25 to both sides: x = 40.

Substitute back into each expression:

  • 2x + 15 = 2(40) + 15 = 80 + 15 = 95 degrees
  • 3x - 25 = 3(40) - 25 = 120 - 25 = 95 degrees

The two agree, which confirms x = 40. The angle is 95 degrees, and the co-interior angle beside it is 180 - 95 = 85 degrees.

Q2 State the number of lines of symmetry and the order of rotational symmetry for a regular pentagon and for the block capital letter Z. Also find the smallest angle of rotation for the pentagon. 3 marks mark

Regular pentagon. It has 5 equal sides and 5 equal angles. Each line of symmetry runs from a vertex to the midpoint of the opposite side, and there are 5 vertices, so there are 5 lines of symmetry.

Its order of rotational symmetry is 5, because a turn about the centre carries each vertex onto the next one.

Smallest angle of rotation = 360 ÷ 5 = 72 degrees. The turns that work are 72, 144, 216, 288 and 360 degrees.

Letter Z. No fold maps Z onto itself, so it has 0 lines of symmetry. A half turn about its centre does map it onto itself, so its order of rotational symmetry is 2, with a smallest angle of 360 ÷ 2 = 180 degrees.

Z is a neat reminder that a shape can have rotational symmetry with no line symmetry at all.

Q3 Construct an angle of 45 degrees using only a compass and a straight-edge, and justify why the angle produced is exactly 45 degrees. 4 marks mark

Steps.

  1. Draw a ray OA. With centre O and any radius r, draw an arc cutting OA at P.
  2. With the same opening and centre P, cut the arc at Q, then with centre Q cut it again at R.
  3. Join OQ and OR. Bisect angle QOR, giving a ray OS.
  4. Bisect angle AOS, giving a ray OT. Then angle AOT = 45 degrees.

Justification. In step 2, OP = OQ = PQ = r, so triangle OPQ is equilateral and angle AOQ = 180 ÷ 3 = 60 degrees. The same argument for triangle OQR gives angle QOR = 60 degrees, so angle AOR = 60 + 60 = 120 degrees.

Bisecting angle QOR gives angle AOS = (60 + 120) ÷ 2 = 90 degrees.

Bisecting angle AOS gives angle AOT = 90 ÷ 2 = 45 degrees.

Each bisection is exact because the two arcs used are drawn with a single compass opening from two points equally far from O, making the two triangles congruent by SSS and the two half angles equal.

Q4 A circular table cover has a diameter of 70 cm and a lace border runs all round its edge. Taking pi = 22/7, find the length of lace needed and the area of the cover. If lace costs 12 rupees per metre, find the cost to the nearest rupee. 4 marks mark

Radius. r = d ÷ 2 = 70 ÷ 2 = 35 cm.

Length of lace = circumference. C = 2 × (22/7) × 35 = 2 × 22 × 5 = 220 cm, which is 220 ÷ 100 = 2.2 m.

Area. A = pi × r² = (22/7) × 35² = (22/7) × 1225. Since 1225 ÷ 7 = 175, A = 22 × 175 = 3850 cm².

Cost. 2.2 m of lace at 12 rupees per metre costs 2.2 × 12 = 26.40 rupees.

To the nearest rupee the cost is ₹26.

Check on the area: 175 × 22 = 175 × 20 + 175 × 2 = 3500 + 350 = 3850. Correct.

Q5 Two plane mirrors are placed so that they meet at an angle of 40 degrees. A figure is reflected first in one mirror and then in the other. Describe the single transformation that has the same effect, and say what would happen if the mirrors were at right angles instead. 3 marks mark

Two reflections in mirrors that meet are equivalent to a single rotation about the point where the mirrors meet.

The angle of that rotation is twice the angle between the mirrors, and the direction runs from the first mirror towards the second.

Angle of rotation = 2 × 40 = 80 degrees, about the point where the two mirrors meet.

If the mirrors were at right angles: angle of rotation = 2 × 90 = 180 degrees, a half turn. This is why a corner made of two mirrors set at 90 degrees sends a ray of light back along the direction it arrived from.

Note on sense: each reflection reverses the labelling of the corners, so two reflections reverse it twice and the figure ends up the right way round, which is exactly what a rotation does.

Q6 In a circle, a chord of length 30 cm lies 8 cm from the centre. Find the radius of the circle, and then find the length of a chord of the same circle that lies 15 cm from the centre. 5 marks mark

Finding the radius. The perpendicular from the centre bisects the chord, so half the chord is 30 ÷ 2 = 15 cm, and the distance from the centre is 8 cm.

By Pythagoras: r² = 15² + 8² = 225 + 64 = 289

r = √289 = 17 cm.

Finding the second chord. The new chord lies 15 cm from the centre, and the radius is still 17 cm.

(half the chord)² = r² - 15² = 289 - 225 = 64

Half the chord = √64 = 8 cm, so the chord is 2 × 8 = 16 cm long.

Check: 8² + 15² = 64 + 225 = 289 = 17². Correct. Notice also that the second chord (16 cm) is shorter than the first (30 cm) because it lies further from the centre, which fits the rule that chords closer to the centre are longer.

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