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Flashcards · 9260 · Topic 3

Geometry and measures flashcards

48 cards on Topic 3, subtopics 3.3.1 to 3.3.3. Pick a subtopic or study them all.

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3.3.1 Properties and constructions

Angles on parallel lines: corresponding, alternate and interior
Corresponding angles are equal. Alternate angles are equal. Interior (co-interior) angles add up to 180°. Use these full names as reasons.
Exterior angle of a triangle
It equals the sum of the two interior opposite angles. Example: angles of 48° and 77° give an exterior angle of 125° at the third vertex.
A regular polygon has interior angle 165°. How many sides?
Exterior angle = 180° − 165° = 15°. n = 360° ÷ 15° = 24 sides. (Sum of interior angles of any polygon = (n − 2) × 180°.)
Rhombus and kite: how do they differ?
Rhombus: four equal sides; opposite angles equal; diagonals bisect each other at 90°. Kite: two pairs of adjacent equal sides; one pair of equal opposite angles; diagonals cross at 90°.
Sector and segment
Sector: two radii and an arc. Segment: a chord and an arc.
Circle theorems: angle at the centre and angle in a semicircleExtension
The angle subtended by an arc at the centre is twice the angle at the circumference. The angle in a semicircle is 90°.
Circle theorems: same segment and cyclic quadrilateralExtension
Angles in the same segment are equal. Opposite angles of a cyclic quadrilateral sum to 180°.
Circle theorems about tangentsExtension
A tangent is perpendicular to the radius at the point of contact. Tangents from an external point are equal in length. Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.
A circle has radius 13 cm and a chord of 24 cm. How far is the chord from the centre?Extension
The perpendicular from the centre bisects the chord, so the half-chord is 12 cm. Distance = √(13² − 12²) = √25 = 5 cm.
The bearing of B from A is 287°. Find the bearing of A from B.
Back bearing: add 180° if below 180°, otherwise subtract 180°. 287° − 180° = 107°
A scale drawing has scale 1 : 1500. A line is 9 cm long. Find the real length.
9 × 1500 = 13 500 cm = 135 m. Multiply by n to go from the drawing to real life.
Loci: points equidistant from A and B, and points a fixed distance from a point O
Equidistant from A and B: the perpendicular bisector of AB. Fixed distance r from O: a circle, centre O, radius r.
How do you construct an angle of 60° at P?
Draw an arc from P cutting the line at X. With the same compass width, draw an arc from X cutting the first arc at Z. Join P to Z. Leave the arcs visible.
Conditions for congruent trianglesExtension
SSS, SAS (angle between the two sides), ASA or AAS, and RHS (right angle, hypotenuse, side). AAA and SSA are not enough.
Lines of symmetry and order of rotation symmetry of a parallelogram and of a regular n-gon
Parallelogram: 0 lines, order 2. Regular n-gon: n lines, order n.
Faces, edges and vertices of a prism with an n-sided cross-section
n + 2 faces, 3n edges, 2n vertices. Example: a triangular prism has 5 faces, 9 edges and 6 vertices.
Plan and elevation
Plan: the view from directly above. Elevation: the view from the front or a stated side.

3.3.2 Mensuration and calculation

Area and volume unit conversions
Square the length factor for area: 1 m² = 10 000 cm²; 1 cm² = 100 mm². Cube it for volume: 1 m³ = 1 000 000 cm³. 1 litre = 1000 cm³; 1 m³ = 1000 litres.
A bar of volume 250 cm³ has mass 1.75 kg. Find its density in g/cm³ and kg/m³.
Density = mass ÷ volume = 1750 g ÷ 250 cm³ = 7 g/cm³. × 1000 for kg/m³: 7000 kg/m³.
Area of a trapezium, and volume of a prism
Trapezium: ½ × (sum of the parallel sides) × perpendicular height. Prism: V = Ah, cross-sectional area × length.
Perimeter and area of a semicircle of radius r
Perimeter = πr + 2r (the arc plus the diameter). Area = ½πr². For r = 5: perimeter 5π + 10 = 25.7, area 12.5π = 39.3.
Volumes of a sphere, a cone and a pyramidExtension
Sphere: (4/3)πr³ (surface area 4πr²). Cone: (1/3)πr²h (curved surface πrl, where l is the slant height). Pyramid: (1/3) × base area × height.
How do you find the volume of a frustum?Extension
Volume of the whole cone (or pyramid) − volume of the smaller similar one removed from the top. Use the scale factor from the heights to find the small one's lengths.
Pythagoras' theorem and the three trig ratios
a² + b² = c², where c is the hypotenuse. sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent (SOH CAH TOA).
cos 40° = 15/h. Find h.
The unknown is underneath, so swap: h = 15/cos 40° = 19.6 (3 s.f.). (If the unknown is on top, multiply.)
A right-angled triangle has adjacent side 7.2 cm and hypotenuse 11.5 cm. Find the angle.
cos θ = 7.2/11.5, so θ = cos⁻¹(7.2/11.5) = 51.2° (1 d.p.).
Space diagonal of a cuboid 12 × 5 × 9 cmExtension
Base diagonal √(12² + 5²) = 13. Space diagonal = √(13² + 9²) = √250 = 15.8 cm (3 s.f.).
The sine ruleExtension
a/sin A = b/sin B = c/sin C, where side a is opposite angle A. Use it when you have a matching side and angle pair.
The cosine rule (side and angle forms)Extension
a² = b² + c² − 2bc cos A (two sides and the included angle). cos A = (b² + c² − a²)/(2bc) (three sides). A negative cosine means an obtuse angle.
A triangle has sides 6, 9 and 10. Find the angle opposite the side of length 6.Extension
cos θ = (9² + 10² − 6²)/(2 × 9 × 10) = 145/180 θ = 36.3° (1 d.p.)
Area of a triangle with two sides and the included angleExtension
Area = ½ab sin C, where C is the angle between sides a and b. Example: ½ × 12 × 9 × sin 30° = 27.
Two similar jugs have heights 9 cm and 12 cm. The small one holds 270 ml. How much does the large one hold?Extension
k = 12/9 = 4/3, so the volume factor is k³ = 64/27. 270 × 64/27 = 640 ml. (Areas scale by k², volumes by k³.)
Arc length and area of a sector of radius 15 cm and angle 48°Extension
Fraction = 48/360 = 2/15. Arc = 2/15 × 2π × 15 = 4π cm. Area = 2/15 × π × 15² = 30π cm². (Perimeter = arc + 2r.)

3.3.3 Transformations, matrices and vectors

What must you state to describe each transformation fully?
Translation: the column vector. Reflection: the equation of the mirror line. Rotation: angle, direction and centre. Enlargement: scale factor and centre.
Reflections of (x, y) in y = x, y = −x and x = k
y = x: (y, x). y = −x: (−y, −x). x = k: (2k − x, y).
Rotations of (x, y) about the origin
90° anticlockwise: (−y, x). 90° clockwise: (y, −x). 180°: (−x, −y).
Enlarge (3, 2) by scale factor 4, centre (1, 1)
Image = centre + k × (point − centre). Relative position (2, 1); × 4 gives (8, 4); add the centre: (9, 5).
What does a negative scale factor of enlargement do?Extension
The image is on the opposite side of the centre. Its size is the size of k times the object. k = −1 is the same as a rotation of 180° about the centre.
Invariant points of a reflection, a rotation and a translation
Reflection: the points on the mirror line. Rotation: the centre only. Translation: none.
Reflecting in two parallel lines, one after the other, gives which single transformation?Extension
A translation perpendicular to the lines, of twice the distance between them.
Work out (3 −2 / 1 4) × the column (5, −1)Extension
Row times column: 3 × 5 + (−2) × (−1) = 17 1 × 5 + 4 × (−1) = 1 Answer: the column (17, 1).
Identity matrix IExtension
I = (1 0 / 0 1). MI = IM = M, and it leaves any point unchanged: the "do nothing" transformation.
How do you read a transformation from its 2 × 2 matrix?Extension
The first column is the image of (1, 0); the second column is the image of (0, 1). Example: (0 −1 / 1 0) sends (1, 0) to (0, 1) and (0, 1) to (−1, 0): rotation 90° anticlockwise about the origin.
Transformation A is done first, then B. What is the single matrix?Extension
BA: the second matrix on the left, the first nearest the point. In general AB ≠ BA.
Drawing the sum of two vectors a + bExtension
Draw a, then draw b from the end of a. The sum runs from the start of a to the end of b.
Scalar multiples 3a and −aExtension
3a is the same direction as a and three times the length. −a is the same length as a but in the opposite direction.
C is (−4, 6) and D is (3, −2). Find →CD.Extension
→CD = D − C = (3 − (−4), −2 − 6) = (7, −8). Then →DC = (−7, 8).
→OP = 2a + b and →OQ = 6a + 3b. Show that O, P and Q are collinear.Extension
→OQ = 3(2a + b) = 3→OP, so OP and OQ are parallel. They share the point O, so O, P and Q are collinear.

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