A general parallelogram has 0 lines of symmetry and rotational symmetry of order 2: a half-turn maps it onto itself.

A right regular prism has one plane through its middle parallel to the bases, plus one vertical plane for each line of symmetry of its regular cross-section. The polygon therefore has 5 lines of symmetry and is a regular pentagon.

With centre A, draw an arc of radius \(10\text{ cm}\). With centre B, draw an arc of radius \(8.5\text{ cm}\). Label their intersection C, then join AC and BC with a ruler. Either valid intersection is acceptable. Keep both arcs visible.

A kite has these properties.

Counting rows from the top and columns from the left, shade row 3, column 1 and row 4, column 4. The completed pattern has both diagonals as lines of symmetry.

(a) On the original drawing, \(AB=4.2\text{ cm}\). Thus the actual distance is \(4.2(0.5)=\boxed{2.1\text{ km}}\).

(b) Convert the distances to drawing lengths: \(4.4/0.5=8.8\text{ cm}\) and \(3.3/0.5=6.6\text{ cm}\). Draw arcs centred on A and B with these radii. Label their intersection east of both boats L and retain the construction arcs.

(a) Counting rows from the top and columns from the left, shade row 2, column 1. Reflection in the top-left to bottom-right diagonal then pairs every shaded square. Alternatively, shade row 3, column 4 to make the other diagonal the symmetry line.

(b) Shade row 2, column 4. A half-turn pairs this with row 3, column 1, and pairs the other two shaded squares.

(a) There are 4 vertical planes: each passes through E and a line of symmetry of the square base.

(b) The axis passes through E and M.

A rhombus has both diagonals as symmetry lines and half-turn symmetry.

The two triangular ends and three rectangular faces form a triangular prism.

The required lengths on the drawing are \(AC=21/3=7\text{ cm}\) and \(BC=15.9/3=5.3\text{ cm}\). Draw an arc centred at A with radius 7 cm and an arc centred at B with radius 5.3 cm. Label an intersection C, then join AC and BC with a ruler. Leave both arcs visible.

The square above 1 is opposite 5, so it is 2. The square to the right of 5 is opposite 1, so it is 6. The bottom square is opposite 4, so it is 3.

A regular decagon has ten equally spaced vertices, giving rotational symmetry of order 10.

(a) Draw an arc of radius 7 cm from A and one of radius 5 cm from B. Label their intersection C and join AC and BC, leaving the arcs visible.

(b) A measurement gives about \(\boxed{82^\circ}\). The cosine-rule check is \(\cos C=(7^2+5^2-8^2)/(2\times7\times5)=1/7\), giving 81.8°.

(c)(i) \(8(10000)=80000\text{ cm}=\boxed{0.8\text{ km}}\). (ii) From B, A is due west, so \(\boxed{270^\circ}\).

(a) Order 2, since a half turn maps the pattern onto itself. (b) Both diagonals of the outer square are lines of symmetry.

(a) \(\triangle DEF\cong\triangle BCD\) and \(\triangle ADF\cong\triangle ADB\). Regular-hexagon equal sides and diagonals give the matching side lengths.

(b) \(OP=OQ\) (radii). \(\angle OPT=\angle OQT=90^\circ\) because a tangent is perpendicular to the radius. With common hypotenuse \(OT\) and equal radii, triangles are congruent by RHS. Corresponding sides \(TP=TQ\), so tangents are equal.

From the left end of the 12 cm base draw an arc of radius 9.5 cm; from the right end draw an arc of radius 8 cm. Join an intersection of the arcs to both ends. Leave the arcs visible.

\(\angle BAX=\angle PQX\) (alternate angles); \(\angle AXB=\angle PXQ\) (vertically opposite angles). Together with \(AX=XQ\), congruence is ASA. Corresponding sides give \(PX=XB\).

\(\angle ADC=\angle ADB=90^\circ\). Side \(AD\) is common. Equal hypotenuses \(AC=AB\) and this common side give RHS congruence.

(a) A and C. (b) ASA: their 6 cm side lies between equal 60° and 25° angles.

(a) Order 2: a half-turn reproduces the shape. (b) Two lines: horizontal and vertical through the central intersection.

(a) Order 4. (b) Four symmetry axes: horizontal, vertical and both diagonals through the centre.

Draw arcs of radius 6.5 cm centred at each endpoint of the 8 cm diagonal. The intersections above and below it are the other vertices. Join each intersection to both diagonal endpoints, retaining the arcs.

Order 2: the pattern is unchanged after a 180° rotation, but not after 90°.

Set the compasses to5 cm and draw an arc centred at A. Set them to6.5 cm and draw an arc centred at B. Their intersection above AB is C. Join AC and BC with a ruler, retaining the arcs. The supplied AB is8 cm on the original scale drawing.

Draw an arc of radius6.2 cm centred at A and an arc of radius7.6 cm centred at B. Their intersection above AB is C. Join AC and BC using a ruler, keeping both arcs.

(a) Kite: AB=DB and AC=DC after reflection. (b) Angle ACB=180°−58°−82°=40°. Reflection gives angle BCD=40°, so angle ACD=80°.

(a) Draw the75° ray from AB at A to intersect CX at D. (b) BC measures9.5 cm, representing \(9.5(5)=47.5\) m.

(c) Construct an arc centred at A with radius7 cm; use the part outside it. Construct the perpendicular bisector of BC using equal compass radii from B and C; use C’s side. Construct the internal angle bisector of BCD using equal arcs from points on BC and CD; use the side nearer CD. Shade only the intersection of these three regions inside ABCD, retaining the construction arcs. (d)5 m=500 cm, so1 :500.

(a) 4 vertical planes: two through opposite base vertices and two through midpoints of opposite base edges. (b) Draw a 4 cm base. Set the compass to 6 cm and draw intersecting arcs from each endpoint. Join their intersection to both endpoints; keep the construction arcs.

Construct the internal bisector of angle PQR, using an arc from Q meeting both sides and equal arcs from these two intersections. Draw an arc centred at R with radius \(13/2=6.5\) cm. The intersection of this arc with the bisector inside the garden is the bird-table position.

(a) With the same compass radius greater than half PQ, draw intersecting arcs from P and Q above and below the line. Join the two intersections. (b) Draw an arc centred at B meeting BA and BC. From these two points, draw equal-radius arcs meeting inside the angle; join B to their intersection.

(a) 5. (b) 1: it maps to itself only after a full turn.

Draw equal-radius arcs centred at A and B, with radius greater than half AB. Join the two intersections: the perpendicular bisector of AB.

(a)(i) 72/8=9 cm. (ii) Draw arcs of radius 9 cm centred at A and C. Their intersection on the upper side of AB is D; join AD and CD. (b) Construct the perpendicular bisector of AB and the internal bisector of angle ABC. Their intersection is the tree’s position.

(a) Construct the internal angle bisector at B, retaining both pairs of compass arcs. (b) Shade the portion on the BA side of that bisector and outside the given circle centred at C.

(a) Construct the internal bisector at C with compass arcs. (b) Draw the portion of a circle centred at E of radius 3 cm lying inside the pentagon. (c) Shade inside this circle, on the DC side of the angle bisector, within the pentagon.

Set the compass radius to AB. Draw arcs centred at A and B. Label an intersection C and join AC and BC, keeping the arcs visible. All three sides equal AB.

(a) Draw an arc centred at A meeting AB and AC. Using equal radii, draw two intersecting arcs centred on these intersection points. Join A to their intersection. (b) Shade the part between AC and the bisector; points on the bisector are equidistant from AB and AC.

A kite.

Draw the perpendicular bisector of the segment joining the trees using equal-radius arcs from each tree. Draw a circle centred on the bird bath of radius 5 cm on the original scale (10 m). The intersection inside the garden is S. Use the 1 cm reference when displaying the drawing at a different size.

Nearer to BC than to AB. The angle bisector separates points equally distant from the two lines; the shaded side borders BC.

(a) Draw equal-radius arcs from A and B, meeting above and below AB. Join the intersections to form the perpendicular bisector of AB. (b) Draw an arc centred at A that cuts AB and AC. From these two intersections, draw equal-radius arcs and join their intersection to A to bisect \(\angle BAC\). (c) Their intersection is E. Draw the circle centred at E with radius \(250/50=5\) cm. (d) Shade the part inside the field, outside this circle, on the AC side of the angle bisector.

A parallelogram (with neither equal adjacent sides nor right angles).

The region is more than 20 m from D, since it lies outside the circle of radius 20 m centred at D. It is nearer to CD than to CB, since it lies on the CD side of the angle bisector CE.

(a) Draw an arc centred at B cutting BA and BC at D and E. With equal compass radius, draw arcs centred at D and E that intersect at F inside the angle. Draw BF and retain the construction arcs. (b) Draw the part of the line parallel to AC that is 3 cm from AC on the inside of the triangle.

Choose a compass radius greater than half AB. Draw arcs above and below AB with centre A, then arcs with the same radius with centre B. Join the two intersections. This line passes through the midpoint of AB at right angles; retain both pairs of construction arcs.

(a) Draw the part of a circle with centre B and radius 5 cm inside the triangle. (b) Bisect angle ABC with compass arcs. (c) Shade the part outside the 5 cm circle on the AB side of this bisector.

(a) Draw an arc centred at A cutting AB and AC. From these intersections draw two equal-radius arcs that intersect inside the angle. Join A to their intersection.

(b) Equidistant from the lines AB and AC.

(a) Draw the part of the circle with centre B and radius 5.7 cm inside the triangle. Use the 1 cm reference if viewing a resized diagram. (b) Shade the part outside that circle on the BC side of CN. CN is the boundary where distances to BC and AC are equal.

(a) Circle with centre C and radius 4 cm. Use the 1 cm reference when viewing a resized figure. (b) Draw intersecting equal-radius arcs from A and B above and below AB; join their intersections to form the perpendicular bisector. (c) Shade inside the circle on the B side of this bisector.

(a) Measure BC approximately 9.5 cm on the drawing. Since 1 cm represents 50 m, \(9.5(50)=475\) m. (b) Construct the perpendicular bisector of AB with equal-radius intersecting arcs from A and B. Its intersection with AB is M; its intersection with CD is X. (c) Construct a circle with centre B and drawing radius \(290/50=5.8\) cm, and the internal angle bisector of DCB. Within polygon MBCX, label P the region inside the circle on the CD side of the angle bisector.

Parallelogram. The pictured rhombus has line symmetry; the trapezium lacks rotational symmetry of order 2.

(a) Measure AB approximately 12.5 cm, giving approximately 125 km. (b) Bearing approximately 290°, clockwise from north at B. (c) 10 km = 1000000 cm, so 1 : 1000000. (d) Construct the perpendicular bisector of CB using equal-radius intersecting arcs; construct the internal bisector of angle ACB using equal-distance marks on its arms and intersecting arcs. Draw a line parallel to CB, 1.5 cm from it inside the triangle. Shade the quadrilateral bounded by these three lines and AB. (e) Area factor \(10^2=100\) km² per map cm². Thus \(0.4(100)=40\) km².

(a) Matching the 46° angle and the 9.65 cm side gives x = 74°. (b) The corresponding side is y = 8.69 cm.

(a) Approximately 046°. (b) SL measures approximately 8.6 cm, so \(SL=8.6(3/2)=12.9\) km. (c) Reverse bearing 340°. (d) 3 km = 300000 cm, giving \(2:300000=1:150000\). (e) Construct the perpendicular bisector of SL using equal-radius arcs; construct the internal bisector of angle PSL using arcs. Their intersection is B. (f) The area scale is \((1.5)^2=2.25\) km² per cm², so \(1.5(2.25)=3.375\) km².

(a)(i) Draw equal-radius arcs centred at A and C and join their intersections. (ii) Draw an arc centred at C crossing CA and CB, then equal-radius arcs from those crossing points; join C to their intersection. (b) Shade the intersection on A’s side of the perpendicular bisector and on AC’s side of the angle bisector, inside ABC.

(a) E, B, A. (b) Z.

(a) BC measures approximately 5.2 cm, giving 104 m. (b) Draw a circle centred at C with radius 8 cm and construct the internal bisector of angle BAD using equal compass arcs. F is their intersection inside the park.

8: a turn of 45° maps the shape onto itself.

(a) Draw a circle centred at E with radius 3 cm. (b) Construct the internal angle bisector at C, using an arc crossing CD and CB and equal arcs centred at those crossing points. (c) Shade inside the circle on the CB side of this bisector.

(a) Equidistant from A and B (also from AD and BC). (b) Use DB as the boundary for equal distances to AB and BC, and MN for equal distances to A and B. Shade the intersection on the AB side of DB and the A side of MN.

(a) Draw the arc centred at A, radius \(40/8=5\) cm. (b) With equal compass radii greater than half DB, make arcs from D and B; join their intersections. (c) Shade outside A’s 5 cm circle on the D side of the perpendicular bisector.

(a)(i) Construct the perpendicular bisector of QR, the internal bisector of SPQ, and a circle arc centred at R with radius \(60/10=6\) cm. The shaded region below is bounded by these three loci and QR. (ii) The clockwise bearing from north is approximately 218°. (b)(i) \(\pi(45)^2=6361.725\ldots\approx6360\) m². (ii) \((210/360)2\pi(45)=164.9336\ldots\approx165\) m.

(a)(i) Draw an arc centred at B crossing BA and BC; from those crossings use equal-radius arcs and join their intersection to B. (ii) Construct the perpendicular bisector of BC with equal-radius arcs from B and C. (b) Shade the intersection on the BC side of the angle bisector and the C side of the perpendicular bisector, as shown.

(a) Join the two top endpoints, completing a rectangle. Its horizontal and vertical centre lines are symmetry axes. (b) Join the top endpoints and add one diagonal. A half-turn maps the resulting rectangle and diagonal onto themselves.

(a) Draw the part of a radius-3 cm circle centred at A lying outside the rectangle; omit the arc inside. (b) Construct the perpendicular bisector of a horizontal side with equal-radius compass arcs. It is the vertical centre line of the rectangle.

(a) At B measure 67° from BA towards AD; the ray meets AD at C. (b) With the same compass radius, greater than half AB, draw arcs from A and B above and below AB. Join the two arc intersections. The result is perpendicular to AB through its midpoint.

(a) I. (b) I and N. The circular O has more than two reflection axes and more than order 2 rotational symmetry; it is excluded by “exactly”.

Construct the perpendicular bisector of GH and keep the G side. Construct the internal angle bisector of HGF and keep the side nearer HG. Since 1 cm represents 100 m, draw an arc of radius 5.5 cm centred at F; keep the exterior. Shade the intersection of these three regions inside the park. The scale bar allows the diagram to be resized accurately.

(a)(i) Draw equal-radius arcs from A and B above and below AB, and join the intersections to form its perpendicular bisector. (ii) Draw a circle with centre C and radius AC. (b) Shade the part inside both circles on the B side of the bisector.

(a) 3: one plane through the centre parallel to each pair of opposite faces. (b) 4, since a quarter turn maps the shape onto itself.

Draw equal-radius arcs from R and T, using a radius greater than half RT. Join their two intersections to form the perpendicular bisector of RT.

(a)(i) Draw an arc centred at B meeting BA and BC. Use equal-radius arcs from these intersections; join their intersection to B. (ii) Construct the perpendicular bisector of AB with equal-radius arcs centred at A and B. (b) Shade the intersection of the A side of the perpendicular bisector and the AB side of the angle bisector, inside ABC.

(a) Draw the part of the circle with centre D and radius 6 cm lying inside the quadrilateral. (b)(i) Draw intersecting equal-radius arcs from A and B above and below AB, then join their intersections. (ii) Bisect angle ABC: use an arc centred on B meeting BA and BC, then equal-radius arcs centred at these two points; join B to their intersection. (c)(i) Q is the intersection of the two bisectors. (ii) Its perpendicular distance to AB is approximately 2.8 cm. (d) Shade the region inside the 6 cm arc, to the A side of the perpendicular bisector, and between AB and the angle bisector. Compass arcs are shown below. Use the 1 cm reference if the diagram is resized.

(a) Draw equal-radius arcs from O and A on both sides of OA. Join their intersections. This line is perpendicular to OA through its midpoint. (b) If M is this midpoint, \(OM=2.5,\ OD=5\) cm, so \(\cos\angle AOD=2.5/5=1/2\). Therefore 60°. The compass arcs and the two intersections are shown below.

(a) \(((1+5)/2,(2+5)/2)=(3,3.5)\). (b) \(\binom{5-1}{5-2}=\binom43\). (c) Draw equal-radius arcs from A and B on both sides of AB, and join the arc intersections. This is the perpendicular bisector, through (3,3.5), with gradient −4/3. Its equation is \(y=-4x/3+7.5\).

(a) Draw an arc centred at B meeting BA and BC; from the two intersections draw equal-radius arcs. Join B to their intersection. The bisector passes through (3.5,3.5), with equation \(y=-5x+21\). (b) Gradient \((4-1)/(6-4)=3/2\). Using B gives \(y=\tfrac32x-5\). (c) The two perpendicular sides are equal, so \(AB^2/2=6.5\Rightarrow AB=\sqrt{13}\approx3.61\) cm.

(a)(i) Shade the top-right square. Reflection in the vertical centre line pairs the top corner squares and the two shaded middle-row squares. (ii) Shade the bottom-right square: it pairs with the top-left under a 180° rotation. (b) 2 planes, each through the apex and the midpoints of opposite base sides.

(a) 1: only a full turn maps the entire diagram onto itself. (b) 1, the vertical centre line.

(a) 0. (b) 2. (c) One example is the horizontal plane halfway between top and bottom, parallel to these faces. The shaded section below shows this plane through the cuboid’s centre.

Construct the perpendicular bisector of AC, using equal-radius arcs centred at A and C. Construct the bisector of angle BAC, using an arc centred at A and equal-radius arcs from its intersections with AB and AC. Inside ABC, shade the region on A’s side of the perpendicular bisector and on AB’s side of the angle bisector. The compass arcs and the correct region are shown below.

Take A = (0,0), B = (10,0), C = (10,6), D = (0,6). The three conditions are \(y<x,\ y>3,\ (x-5)^2+y^2<25\). Put R in their intersection, for example (7,3.5).

(a) 5: one plane perpendicular to its length and four extending along the square’s symmetry lines. (b) 2, since a half-turn restores it.

(a) 5, with rotations of 72°. (b) 0; a reflection reverses the blade curvature.

Draw parallel segments 3 cm outside each side. Join adjacent offset segments by circular arcs of radius 3 cm centred on the corresponding vertices. Use only the external arcs; the resulting boundary has rounded corners.

(a)(i) Draw equal-radius intersecting arcs centred on C and D; join their intersections to make the perpendicular bisector of CD. (ii) Construct the internal angle bisector of DAB with an arc from A and equal-radius arcs from the two marked side points. (b) SHOP is the intersection of the D-side of the perpendicular bisector and the AD-side of the angle bisector. (c)(i) Draw the arc centred on B with radius \(100/20=5\) cm. (ii) Shade the part of ABCD outside this arc and outside the shop region.