Edexcel GCSE Maths coverage

Geometry and measures

Section G
25 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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G1

Use conventional terms/notations: points, lines, vertices, edges, planes, parallel and perpendicular lines, right angles, polygons; standard triangle labelling; draw diagrams from written description

  • Use precise geometric language: a vertex is a corner, an edge joins vertices, parallel lines never meet, and perpendicular lines meet at a right angle.
  • Label an angle by placing its vertex in the middle, so ABC\angle ABC is the angle at BB; in triangle ABCABC, side ABAB is opposite vertex CC.
  • Build a diagram one instruction at a time, marking equal lengths, parallel lines and right angles with conventional symbols rather than relying on appearance.
  • A common error is to assume a sketch is drawn to scale; only stated facts and marked relationships may be used.

Tier 1 · Easy

1 mark
ORIGINAL

In triangle PQRPQR, which vertex is opposite the side PRPR?

Tier 2 · Standard

1 mark
ORIGINAL

Lines ABAB and CDCD are parallel. A line through EE meets ABAB at FF at a right angle. State the relationship between EFEF and CDCD.

Tier 3 · Hard

3 marks
ORIGINAL

Draw quadrilateral ABCDABCD with ABCDAB\parallel CD. Draw diagonal ACAC. Mark a point EE on ACAC, then draw the line through EE perpendicular to ABAB, meeting ABAB at FF and CDCD at GG.

G2

Use standard ruler and compass constructions (perpendicular bisector, perpendicular from/at a point, angle bisector); construct figures, solve loci problems; perpendicular distance is shortest

  • A perpendicular bisector crosses a segment at its midpoint at 9090^\circ; every point on it is equidistant from the segment's endpoints.
  • Construct an angle bisector using equal-radius arcs from the vertex and then intersecting arcs from the two arms; points on it are equidistant from the arms.
  • Translate locus conditions into standard boundaries: a fixed distance from a point gives a circle, while a fixed distance from a straight line gives two parallel lines.
  • Keep compass arcs visible and do not measure to replace a construction; the perpendicular distance to a line is the shortest distance.

Tier 1 · Easy

2 marks
ORIGINAL

Describe how to construct the perpendicular bisector of a line segment ABAB using a ruler and compasses.

Tier 2 · Standard

3 marks
ORIGINAL

Two straight paths meet at OO. A lamp must be equally distant from the two paths and no more than 66 m from OO. Describe the locus of possible positions inside the angle between the paths.

Tier 3 · Hard

4 marks
ORIGINAL

Points AA and BB are 88 cm apart. A point PP must satisfy PA=PBPA=PB and PA5PA\leq5 cm. Describe and construct the complete locus of PP.

G3

Apply angles at a point, on a straight line, vertically opposite angles; use alternate and corresponding angles on parallel lines; derive and use the angle sum of a triangle and of any polygon

  • Angles at a point sum to 360360^\circ, angles on a straight line sum to 180180^\circ, and vertically opposite angles are equal.
  • When a transversal crosses parallel lines, alternate and corresponding angles are equal; co-interior angles sum to 180180^\circ.
  • Splitting an nn-sided polygon into n2n-2 triangles gives an interior-angle sum of (n2)×180(n-2)\times180^\circ.
  • Do not quote alternate or corresponding angles unless the relevant lines are stated or shown to be parallel, and name the angle fact used.

Tier 1 · Easy

1 mark
ORIGINAL

Two adjacent angles on a straight line are 6363^\circ and xx^\circ. Work out xx.

Tier 2 · Standard

3 marks
ORIGINAL

Two parallel lines are crossed by a transversal. A pair of alternate angles are (4x+7)(4x+7)^\circ and (7x38)(7x-38)^\circ. Find xx and the size of these angles.

Tier 3 · Hard

4 marks
ORIGINAL

A polygon has 1313 sides. Twelve of its interior angles are each 155155^\circ. Work out the remaining interior angle.

G4

Derive and apply properties and definitions of special quadrilaterals (square, rectangle, parallelogram, trapezium, kite, rhombus), triangles and other plane figures using appropriate language

  • Classify a shape from properties that must always hold: for example, a parallelogram has two pairs of parallel opposite sides and a trapezium has one pair of parallel sides.
  • A rectangle has four right angles, a rhombus has four equal sides, and a square satisfies both definitions as well as being a parallelogram.
  • Use diagonal properties carefully: parallelogram diagonals bisect each other, rectangle diagonals are also equal, and rhombus diagonals are also perpendicular.
  • A common error is to use a property that is merely possible rather than guaranteed; one diagram cannot establish a definition.

Tier 1 · Easy

1 mark
ORIGINAL

Name the quadrilateral that has exactly one pair of parallel sides.

Tier 2 · Standard

2 marks
ORIGINAL

One interior angle of a rhombus is 6868^\circ. Work out the other three interior angles.

Tier 3 · Hard

3 marks
ORIGINAL

The diagonals of quadrilateral WXYZWXYZ bisect each other and are equal in length. Explain why WXYZWXYZ must be a rectangle, and why the information does not prove that it is a square.

G5

Use the basic congruence criteria for triangles (SSS, SAS, ASA, RHS)

  • Congruent triangles are identical in size and shape, so all corresponding sides and angles are equal.
  • Use SSS for three side pairs, SAS for two side pairs and the included angle, ASA for two angle pairs and a side, and RHS for right triangles with equal hypotenuse and one other side.
  • Write vertices in corresponding order when stating congruence, then transfer only matching sides or angles.
  • AAA proves similarity, not congruence, and SSA is not a valid general congruence test.

Tier 1 · Easy

1 mark
ORIGINAL

Triangle ABCABC has side lengths 55 cm, 77 cm and 99 cm. Triangle PQRPQR has side lengths 55 cm, 77 cm and 99 cm. State the congruence criterion.

Tier 2 · Standard

2 marks
ORIGINAL

Triangles ABCABC and DEFDEF are right-angled at BB and EE. Also AC=DF=13AC=DF=13 cm and AB=DE=5AB=DE=5 cm. Explain why the triangles are congruent.

Tier 3 · Hard

3 marks
ORIGINAL

In quadrilateral ABCDABCD, diagonal ACAC is drawn. Given that AB=ADAB=AD and BAC=CAD\angle BAC=\angle CAD, prove that BC=CDBC=CD.

G6

Apply angle facts, congruence, similarity and quadrilateral properties to derive results about angles and sides, incl. Pythagoras' theorem and isosceles base angles, and obtain simple proofs

  • In an isosceles triangle, equal sides face equal angles; the converse also holds, so equal angles face equal sides.
  • For a right-angled triangle, use a2+b2=c2a^2+b^2=c^2 with cc opposite the right angle, and take the positive square root for a length.
  • A geometric proof should form a connected chain in which every equality follows from a stated angle fact, shape property, similarity or congruence result.
  • Do not use the fact being proved as a reason, and do not infer equal sides or angles just because they look equal in a sketch.

Tier 1 · Easy

2 marks
ORIGINAL

Triangle ABCABC is isosceles with AB=ACAB=AC. Angle BB is 4747^\circ. Work out angle AA.

Tier 2 · Standard

3 marks
ORIGINAL

A right-angled triangle has perpendicular sides 99 cm and 1212 cm. Work out the length of its hypotenuse.

Tier 3 · Hard

4 marks
ORIGINAL

In quadrilateral ABCDABCD, ABCDAB\parallel CD and AB=CDAB=CD. Diagonal ACAC is drawn. Prove that BC=ADBC=AD.

G7

Identify, describe and construct congruent and similar shapes, incl. on coordinate axes, by rotation, reflection, translation and enlargement (including fractional and negative scale factors)

  • Describe a translation by a vector, a rotation by its centre, angle and direction, a reflection by its mirror line, and an enlargement by its centre and scale factor.
  • Translations, rotations and reflections preserve lengths and angles, so the image is congruent to the object; an enlargement preserves angles and scales every length by the same factor.
  • For an enlargement about centre CC, multiply the vector from CC to each point by the scale factor; a negative factor places the image on the opposite side of CC.
  • A common error is to omit part of a full transformation description or to multiply coordinates directly when the centre of enlargement is not the origin.

Tier 1 · Easy

2 marks
ORIGINAL

Point A(3,4)A(-3,4) is translated by the vector (52)\begin{pmatrix}5\\-2\end{pmatrix}. Find the coordinates of its image.

Tier 2 · Standard

2 marks
ORIGINAL

Point P(4,1)P(4,-1) is rotated 9090^\circ anticlockwise about the origin. Find the coordinates of its image PP'.

Tier 3 · Hard

4 marks
ORIGINAL

Triangle ABCABC has vertices A(6,3)A(6,3), B(0,5)B(0,5) and C(2,1)C(-2,-1). It is enlarged by scale factor 12-\frac12 about centre (2,1)(2,-1). Find the three image coordinates.

G8

Describe the changes and invariance achieved by combinations of rotations, reflections and translations [Higher only]

  • Track a combination in the stated order: the output of the first transformation is the input to the next.
  • Two translations combine by adding their vectors, while some pairs of reflections combine to form a rotation or a translation.
  • Rotations, reflections and translations preserve lengths, angle sizes, area and parallelism; reflections reverse orientation, whereas rotations and translations preserve it.
  • Do not reverse the order of a combination: transformations need not commute, so changing the order can change the final image.

Tier 1 · Easy

2 marks
ORIGINAL

A shape is rotated and then translated. State two properties of the shape that must remain invariant.

Tier 2 · Standard

2 marks
ORIGINAL

A shape is translated first by (32)\begin{pmatrix}3\\-2\end{pmatrix} and then by (57)\begin{pmatrix}-5\\7\end{pmatrix}. Describe the single equivalent transformation.

Tier 3 · Hard

4 marks
ORIGINAL

A shape is reflected in the xx-axis and then reflected in the line y=xy=x. Describe the single equivalent transformation and state whether orientation is preserved.

G9

Identify and apply circle definitions and properties, including: centre, radius, chord, diameter, circumference, tangent, arc, sector and segment

  • A radius joins the centre to the circumference, a diameter is a chord through the centre, and every diameter is twice the radius.
  • A tangent touches a circle at one point, while a chord joins two points on the circumference; an arc is a portion of the circumference.
  • A sector is bounded by two radii and an arc, whereas a segment is bounded by a chord and an arc.
  • Do not call the curved boundary a circle: the circumference is the boundary, while the circle includes the region inside it.

Tier 1 · Easy

1 mark
ORIGINAL

A circle has radius 6.56.5 cm. Write down its diameter.

Tier 2 · Standard

2 marks
ORIGINAL

A straight segment joins two points on a circle but does not pass through its centre. Another straight line meets the circle at exactly one point. Give the geometric name of each.

Tier 3 · Hard

3 marks
ORIGINAL

A circle is centred at OO. Distinct points AA and BB are on its circumference, and ABAB is not a diameter. Describe precisely the boundaries of the minor sector AOBAOB and the minor segment cut off by ABAB.

G10

Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results [Higher only]

  • The angle at the centre is twice the angle at the circumference standing on the same arc; angles in the same segment are equal, and an angle in a semicircle is 9090^\circ.
  • Opposite angles in a cyclic quadrilateral sum to 180180^\circ, and the alternate segment theorem links the angle between a tangent and chord to the angle in the opposite segment.
  • A radius is perpendicular to a tangent at the point of contact; tangents from the same external point are equal, and the perpendicular from the centre bisects a chord.
  • Always identify the same chord or arc before applying an angle theorem, and state the theorem as the reason rather than quoting an unsupported angle.

Tier 1 · Easy

1 mark
ORIGINAL

ABAB is a diameter of a circle and CC is another point on the circumference. Find ACB\angle ACB.

Tier 2 · Standard

2 marks
ORIGINAL

In a circle with centre OO, the angle AOBAOB is 124124^\circ. Point CC lies on the major arc ABAB. Work out ACB\angle ACB.

Tier 3 · Hard

4 marks
ORIGINAL

From a point PP outside a circle with centre OO, tangents touch the circle at AA and BB. Prove that PA=PBPA=PB and that OPOP bisects APB\angle APB.

G11

Solve geometrical problems on coordinate axes

  • Use coordinate differences to find horizontal and vertical lengths, and use the midpoint formula by averaging the two xx-coordinates and the two yy-coordinates.
  • The distance between (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) is (x2x1)2+(y2y1)2\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} by Pythagoras' theorem.
  • Gradients can establish geometry: equal gradients show parallel lines, while gradients with product 1-1 show perpendicular non-vertical lines.
  • Keep signs when subtracting negative coordinates, and do not read a length directly from a sketch that is not drawn to scale.

Tier 1 · Easy

2 marks
ORIGINAL

Find the midpoint of the line segment joining (5,2)(-5,2) to (7,8)(7,8).

Tier 2 · Standard

3 marks
ORIGINAL

Points A(1,4)A(-1,4) and B(5,4)B(5,-4) are joined. Find the exact length ABAB.

Tier 3 · Hard

5 marks
ORIGINAL

Triangle ABCABC has vertices A(2,1)A(-2,1), B(4,3)B(4,3) and C(2,9)C(2,9). Show that the triangle is right-angled and isosceles, then work out its area.

G12

Identify properties of the faces, surfaces, edges and vertices of: cubes, cuboids, prisms, cylinders, pyramids, cones and spheres

  • A face is flat, a curved surface is not a face, an edge is where surfaces meet, and a vertex is a point where edges meet.
  • An nn-sided prism has two congruent polygonal ends and nn rectangular side faces, giving n+2n+2 faces, 3n3n edges and 2n2n vertices.
  • A pyramid has one polygonal base and triangular faces meeting at one apex; a cone has one circular base, one curved surface and one vertex.
  • Count each feature once, especially hidden edges, and do not give a sphere or cylinder vertices where no edges meet.

Tier 1 · Easy

2 marks
ORIGINAL

Write down the number of faces and the number of vertices of a triangular prism.

Tier 2 · Standard

2 marks
ORIGINAL

A solid has one circular plane face, one curved surface, one circular edge and one vertex. Name the solid.

Tier 3 · Hard

3 marks
ORIGINAL

A prism has a regular 99-sided polygon as each end. Work out its numbers of faces, edges and vertices.

G13

Construct and interpret plans and elevations of 3D shapes

  • A plan is the view from directly above; a front or side elevation is an orthographic view from the stated horizontal direction.
  • Project corresponding corners with parallel construction lines so widths align between the plan and front elevation and depths align between the plan and side elevation.
  • For stacks of cubes, an elevation shows the greatest height visible in each column along that viewing direction, while a numbered plan records each stack height.
  • Do not add perspective or show hidden internal edges unless asked; plans and elevations use true lengths in their viewing plane.

Tier 1 · Easy

2 marks
ORIGINAL

A cuboid is 88 cm long, 55 cm wide and 33 cm high. State the shape and dimensions of its plan.

Tier 2 · Standard

3 marks
ORIGINAL

A numbered plan of unit-cube stacks has two rows. From north to south, the rows are (2,1)(2,1) and (3,0)(3,0), with entries ordered west to east. Give the visible column heights in the front elevation viewed from the south and in the side elevation viewed from the east.

Tier 3 · Hard

4 marks
ORIGINAL

A solid uses vertical stacks of unit cubes on all four cells of a 22 by 22 plan. Its front elevation viewed from the south has heights (3,2)(3,2) from west to east. Its side elevation viewed from the east has heights (2,3)(2,3) from north to south. Find the least possible number of cubes and give one numbered plan that achieves it.

G14

Use standard units of measure and related concepts (length, area, volume/capacity, mass, time, money, etc.)

  • Standard units include millimetres, centimetres, metres and kilometres for length; square or cubic units for area and volume; litres for capacity; grams or kilograms for mass; and seconds, minutes or hours for time.
  • Convert every measurement to compatible units before calculating, remembering that area conversion factors are squared and volume conversion factors are cubed.
  • For example, because 1 m=100 cm1\text{ m}=100\text{ cm}, it follows that 1 m2=10000 cm21\text{ m}^2=10\,000\text{ cm}^2 and 1 m3=1000000 cm31\text{ m}^3=1\,000\,000\text{ cm}^3.
  • A common error is to use the linear conversion factor for an area or volume, such as multiplying square metres by 100100 instead of 1000010\,000 to obtain square centimetres.

Tier 1 · Easy

1 mark
ORIGINAL

Convert 3.75 kg3.75\text{ kg} to grams.

Tier 2 · Standard

3 marks
ORIGINAL

A rectangular water tank is 0.8 m0.8\text{ m} long, 0.5 m0.5\text{ m} wide and 0.6 m0.6\text{ m} high. It is 65%65\% full. Work out the volume of water in litres.

Tier 3 · Hard

5 marks
ORIGINAL

A floor measures 4.8 m4.8\text{ m} by 3.6 m3.6\text{ m}. Square tiles have side length 30 cm30\text{ cm} and are sold in boxes of 1212. A decorator buys at least 8%8\% more tiles than the exact number needed. Each box costs £14.7514.75. Work out the total cost.

G15

Measure line segments and angles in geometric figures, including interpreting maps and scale drawings and use of bearings

  • Lengths and angles can be measured from an accurate drawing, while a map scale links a measured map distance to the corresponding real distance.
  • For a scale 1:n1:n, multiply a map length by nn to obtain the real length in the same units; bearings are measured clockwise from north and written using three figures.
  • For example, a clockwise angle of 5252^\circ from north is written as the bearing 052052^\circ, and the reverse bearing is 232232^\circ.
  • A common error is to measure a bearing anticlockwise or from an east-west line instead of clockwise from the north line at the starting point.

Tier 1 · Easy

1 mark
ORIGINAL

A ray from point PP makes an angle of 6868^\circ clockwise from north. Write its bearing from PP.

Tier 2 · Standard

4 marks
ORIGINAL

On a map with north at the top, point BB is 6 cm6\text{ cm} east and 8 cm8\text{ cm} north of point AA. The scale is 1:200001:20\,000. Work out the real straight-line distance from AA to BB and the bearing of BB from AA. Give the bearing to the nearest degree.

Tier 3 · Hard

4 marks
ORIGINAL

On a scale drawing, a road of actual length 3.15 km3.15\text{ km} is represented by a line 8.4 cm8.4\text{ cm} long. Find the scale in the form 1:n1:n. Another road is 11.2 cm11.2\text{ cm} long on the same drawing. Work out its actual length in kilometres.

G16

Know and apply formulae to calculate: area of triangles, parallelograms, trapezia; volume of cuboids and other right prisms (including cylinders)

  • Use A=12bhA=\frac12 bh for a triangle, A=bhA=bh for a parallelogram and A=12(a+b)hA=\frac12(a+b)h for a trapezium, where each height is perpendicular to the relevant base or parallel sides.
  • A right prism has a constant cross-section, so its volume is the area of that cross-section multiplied by the prism length; for a cylinder this gives V=πr2hV=\pi r^2h.
  • For example, a trapezium with parallel sides 55 and 99 and perpendicular height 44 has area 12(5+9)×4=28\frac12(5+9)\times4=28 square units.
  • A common error is to use a sloping side as the perpendicular height or to omit the factor of 12\frac12 in a triangle or trapezium calculation.

Tier 1 · Easy

2 marks
ORIGINAL

A trapezium has parallel sides of lengths 7 cm7\text{ cm} and 13 cm13\text{ cm} and perpendicular height 6 cm6\text{ cm}. Work out its area.

Tier 2 · Standard

2 marks
ORIGINAL

A cylinder has radius 4 cm4\text{ cm} and height 9 cm9\text{ cm}. Work out its volume in terms of π\pi.

Tier 3 · Hard

4 marks
ORIGINAL

A right triangular prism is 10 cm10\text{ cm} long. Its triangular cross-section has base x cmx\text{ cm} and perpendicular height (x+2) cm(x+2)\text{ cm}. The volume is 600 cm3600\text{ cm}^3. Find xx.

G17

Know circumference of a circle = 2πr = πd and area = πr²; calculate perimeters of 2D shapes incl. circles, areas of circles and composite shapes, surface area and volume of spheres, pyramids, cones

  • For a circle, circumference is C=2πr=πdC=2\pi r=\pi d and area is A=πr2A=\pi r^2; for composite shapes, add included regions or subtract cut-out regions.
  • For a pyramid or cone, V=13×base area×perpendicular heightV=\frac13\times\text{base area}\times\text{perpendicular height}; find surface area by adding the base and every exposed triangular face, or use πrl\pi rl for a cone's curved surface.
  • For example, a sphere of radius rr has surface area 4πr24\pi r^2 and volume 43πr3\frac43\pi r^3, so doubling rr multiplies these by 44 and 88 respectively.
  • A common error is to use the diameter where a formula requires the radius, or to include an internal shared edge or face in a perimeter or surface-area total.

Tier 1 · Easy

1 mark
ORIGINAL

A circle has diameter 13 cm13\text{ cm}. Write its circumference in terms of π\pi.

Tier 2 · Standard

4 marks
ORIGINAL

A semicircle of diameter 10 cm10\text{ cm} is attached to one of the shorter sides of an 18 cm18\text{ cm} by 10 cm10\text{ cm} rectangle. The shared diameter is inside the shape. Work out the area and perimeter of the composite shape in terms of π\pi.

Tier 3 · Hard

5 marks
ORIGINAL

A solid cone has radius 6 cm6\text{ cm}, perpendicular height 8 cm8\text{ cm} and slant height 10 cm10\text{ cm}. Work out its total surface area, including the circular base, and its volume. Give both answers in terms of π\pi.

G18

Calculate arc lengths, angles and areas of sectors of circles

  • An arc or sector is the same fraction of a circle as its central angle is of 360360^\circ.
  • Use arc length =θ360×2πr=\frac{\theta}{360}\times2\pi r and sector area =θ360×πr2=\frac{\theta}{360}\times\pi r^2, rearranging when the angle or radius is unknown.
  • For example, a 9090^\circ sector is one quarter of a circle, so its arc length is 14(2πr)\frac14(2\pi r) and its area is 14πr2\frac14\pi r^2.
  • A common error is to calculate the full circumference or area without multiplying by θ360\frac{\theta}{360}, or to include radii when only arc length is requested.

Tier 1 · Easy

2 marks
ORIGINAL

Work out the arc length of a quarter-circle with radius 8 cm8\text{ cm}. Give the answer in terms of π\pi.

Tier 2 · Standard

4 marks
ORIGINAL

A sector has radius 9 cm9\text{ cm} and arc length 7π cm7\pi\text{ cm}. Work out the angle of the sector and its area in terms of π\pi.

Tier 3 · Hard

5 marks
ORIGINAL

A shape is an annular sector with angle 144144^\circ, outer radius 12 cm12\text{ cm} and inner radius 7 cm7\text{ cm}. Work out its area and perimeter in terms of π\pi.

G19

Apply the concepts of congruence and similarity, including the relationships between lengths, areas and volumes in similar figures

  • Congruent figures have the same size and shape, while similar figures have equal corresponding angles and corresponding lengths in a constant ratio.
  • Match corresponding sides in the same order, calculate one length scale factor, and multiply or divide every corresponding length by that factor.
  • At Higher tier only, a length scale factor kk gives area scale factor k2k^2 and volume scale factor k3k^3; for example, length ratio 2:52:5 gives area ratio 4:254:25 and volume ratio 8:1258:125.
  • A common error is to compare non-corresponding sides or, at Higher tier, to use the linear scale factor directly for an area or volume.

Tier 1 · Easy

1 mark
ORIGINAL

Triangle ABCABC has AB=5 cmAB=5\text{ cm}, AC=7 cmAC=7\text{ cm} and BAC=42\angle BAC=42^\circ. Triangle PQRPQR has PQ=5 cmPQ=5\text{ cm}, PR=7 cmPR=7\text{ cm} and QPR=42\angle QPR=42^\circ. State why the triangles are congruent.

Tier 2 · Standard

3 marks
ORIGINAL

Two similar shapes have corresponding lengths in the ratio smaller:larger =3:5=3:5. A side on the smaller shape is 12 cm12\text{ cm}. Another side on the larger shape is 35 cm35\text{ cm}. Work out the corresponding missing lengths.

Tier 3 · Hard

4 marks
ORIGINAL

Triangle ABCABC has side lengths 8 cm8\text{ cm}, 11 cm11\text{ cm} and 13 cm13\text{ cm}. Similar triangle DEFDEF has longest side 19.5 cm19.5\text{ cm}. Work out the other two side lengths of triangle DEFDEF and its perimeter.

G20

Know Pythagoras' theorem a² + b² = c² and the trigonometric ratios sin, cos and tan; apply them to find angles and lengths in right-angled and, where possible, general triangles in 2D and 3D figures

  • In a right-angled triangle, a2+b2=c2a^2+b^2=c^2 with cc as the hypotenuse, and sinθ=oppositehypotenuse\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, cosθ=adjacenthypotenuse\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}, tanθ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}.
  • Label the triangle relative to the required angle, choose Pythagoras or the ratio containing the known and unknown sides, and use an inverse trigonometric function when finding an angle.
  • At Foundation tier, apply Pythagoras and trigonometry to right-angled triangles in two-dimensional figures, including those found by symmetry in isosceles triangles; at Higher tier only, extend to general (non-right) triangles and identify a right-angled cross-section before applying the same method in three dimensions.
  • A common error is to treat a non-right-angled triangle as right-angled, or to use an angle and sides that do not belong to the same right-angled cross-section.

Tier 1 · Easy

2 marks
ORIGINAL

A right-angled triangle has perpendicular sides 6 cm6\text{ cm} and 8 cm8\text{ cm}. Work out the hypotenuse.

Tier 2 · Standard

4 marks
ORIGINAL

A straight ladder of length 7.2 m7.2\text{ m} rests against a vertical wall. Its foot is 2.1 m2.1\text{ m} from the wall. Work out the height reached by the ladder and the angle it makes with the ground. Give each answer to 11 decimal place.

Tier 3 · Hard

4 marks
ORIGINAL

Isosceles triangle ABCABC has AB=AC=13 cmAB=AC=13\text{ cm} and BC=10 cmBC=10\text{ cm}. Work out the perpendicular height from AA to BCBC and angle ABCABC. Give the angle to 11 decimal place.

G21

Know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90°; know the exact value of tan θ for θ = 0°, 30°, 45° and 60°

  • Know the exact sine and cosine values at 00^\circ, 3030^\circ, 4545^\circ, 6060^\circ and 9090^\circ, and the exact tangent values at 00^\circ, 3030^\circ, 4545^\circ and 6060^\circ.
  • The complete sets are sinθ:0,12,22,32,1\sin\theta:0,\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2},1 and cosθ:1,32,22,12,0\cos\theta:1,\frac{\sqrt3}{2},\frac{\sqrt2}{2},\frac12,0 for θ=0,30,45,60,90\theta=0^\circ,30^\circ,45^\circ,60^\circ,90^\circ; also tan0=0\tan0^\circ=0, tan30=13\tan30^\circ=\frac1{\sqrt3}, tan45=1\tan45^\circ=1, tan60=3\tan60^\circ=\sqrt3.
  • For example, the 1:3:21:\sqrt3:2 triangle gives sin30=12\sin30^\circ=\frac12, cos30=32\cos30^\circ=\frac{\sqrt3}{2} and tan30=13\tan30^\circ=\frac1{\sqrt3}.
  • A common error is to give a decimal approximation when an exact value is required, or to interchange the opposite and adjacent sides for 3030^\circ and 6060^\circ.

Tier 1 · Easy

1 mark
ORIGINAL

Work out the exact value of sin30+cos60\sin30^\circ+\cos60^\circ.

Tier 2 · Standard

3 marks
ORIGINAL

Work out the exact value of 2sin45cos45+cos602\sin45^\circ\cos45^\circ+\cos60^\circ.

Tier 3 · Hard

4 marks
ORIGINAL

The hypotenuse of a right-angled triangle is 12 cm12\text{ cm} and one acute angle is 3030^\circ. Work out its exact area.

G22

Know and apply the sine rule a/sin A = b/sin B = c/sin C, and cosine rule a² = b² + c² - 2bc cos A, to find unknown lengths and angles [Higher only]

  • The sine rule asinA=bsinB=csinC\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} pairs each side with its opposite angle, while the cosine rule connects three sides with one angle.
  • Use the sine rule when an opposite side-angle pair is known; use the cosine rule for two sides and their included angle or for all three sides.
  • When the sine rule gives sinB=k\sin B=k, check both B=sin1kB=\sin^{-1}k and B=180sin1kB=180^\circ-\sin^{-1}k against the triangle angle sum.
  • A common error is to pair a side with a non-opposite angle or to ignore the second possible angle in an ambiguous sine-rule problem.

Tier 1 · Easy

2 marks
ORIGINAL

In triangle ABCABC, side a=7 cma=7\text{ cm}, angle A=30A=30^\circ and angle B=90B=90^\circ. Work out side bb.

Tier 2 · Standard

3 marks
ORIGINAL

Two sides of a triangle are 7 cm7\text{ cm} and 10 cm10\text{ cm}, and the included angle is 6060^\circ. Work out the exact length of the third side.

Tier 3 · Hard

5 marks
ORIGINAL

In triangle ABCABC, A=35A=35^\circ, a=8 cma=8\text{ cm} and b=11 cmb=11\text{ cm}. Find all possible values of angles BB and CC. Give each angle to 11 decimal place.

G23

Know and apply Area = ½ ab sin C to calculate the area, sides or angles of any triangle [Higher only]

  • The area of a triangle is A=12absinCA=\frac12ab\sin C, where CC is the included angle between sides aa and bb.
  • Substitute the two sides and their included angle to find area, or rearrange before using an inverse sine to find an angle.
  • If sinC=k\sin C=k, both C=sin1kC=\sin^{-1}k and C=180sin1kC=180^\circ-\sin^{-1}k may give triangles with the same area.
  • A common error is to use an angle that is not between the two substituted sides, or to reject an obtuse solution without checking the information given.

Tier 1 · Easy

2 marks
ORIGINAL

Two sides of a triangle are 10 cm10\text{ cm} and 7 cm7\text{ cm}, and their included angle is 3030^\circ. Work out the area.

Tier 2 · Standard

3 marks
ORIGINAL

A triangle has area 48 cm248\text{ cm}^2. Two sides have lengths 12 cm12\text{ cm} and x cmx\text{ cm}, and their included angle is 3030^\circ. Work out xx.

Tier 3 · Hard

4 marks
ORIGINAL

Two sides of a triangle are 10 cm10\text{ cm} and 12 cm12\text{ cm}. Its area is 303 cm230\sqrt3\text{ cm}^2. Find both possible values of the included angle.

G24

Describe translations as 2D vectors

  • A translation moves every point through the same horizontal and vertical displacement without changing the shape's size or orientation.
  • Describe a translation by the column vector (xy)\binom{x}{y}, where positive xx is right, negative xx is left, positive yy is up and negative yy is down.
  • For example, moving 44 units left and 33 units up is the translation vector (43)\binom{-4}{3}.
  • A common error is to reverse the vector by subtracting the image coordinates from the object coordinates instead of calculating image minus object.

Tier 1 · Easy

2 marks
ORIGINAL

Point A=(3,4)A=(-3,4) is translated to A=(2,1)A'=(2,-1). Describe the translation as a vector.

Tier 2 · Standard

3 marks
ORIGINAL

Triangle ABCABC has vertices A=(1,2)A=(1,-2), B=(5,1)B=(5,1) and C=(2,4)C=(2,4). It is translated by (43)\binom{-4}{3}. Write the coordinates of the three image vertices.

Tier 3 · Hard

4 marks
ORIGINAL

A translation maps P=(6,4)P=(-6,4) to P=(1,2)P'=(1,-2). The same translation maps Q=(3,k)Q=(3,k) to Q=(10,5)Q'=(10,5). Find kk and describe the translation as a vector.

G25

Apply addition and subtraction of vectors, multiplication of vectors by a scalar, and diagrammatic and column representations of vectors; use vectors to construct geometric arguments and proofs

  • Vectors have magnitude and direction; they can be represented by directed line segments, column vectors or algebraic symbols such as a\mathbf{a} and b\mathbf{b}.
  • Add vectors component by component, subtract to reverse a displacement, and multiply by a scalar to change magnitude and, for a negative scalar, direction.
  • For a route through successive points, add vectors in travel order; at Higher tier only, compare two vector routes or show one direction vector is a scalar multiple of another to construct a geometric proof.
  • A common error is to subtract vectors in the wrong order or to forget that multiplying by a negative scalar reverses the vector's direction.

Tier 1 · Easy

1 mark
ORIGINAL

Work out (35)+(72)\binom{3}{-5}+\binom{-7}{2}.

Tier 2 · Standard

3 marks
ORIGINAL

Points AA and BB have position vectors a=(21)\mathbf{a}=\binom{2}{-1} and b=(85)\mathbf{b}=\binom{8}{5}. Point MM is the midpoint of ABAB. Find the position vector of MM and the vector AM\overrightarrow{AM}.

Tier 3 · Hard

4 marks
ORIGINAL

A walker starts at P=(3,2)P=(-3,2). The walker makes successive displacements (43)\binom{4}{3}, (15)\binom{-1}{5} and 2(12)-2\binom{1}{-2}. Work out the resultant displacement, the walker's finishing coordinates and the vector that would take the walker directly back to PP.