Geometry & Measures

A right-angled triangle has legs 3 cm and 4 cm. What is the hypotenuse?

• A7 cm
• B12 cm
• C25 cm
• D5 cm✓

Answer: By Pythagoras, h² = 3² + 4² = 9 + 16 = 25, so h = √25 = 5 cm.

In a right-angled triangle, which ratio is sin θ?

• Aopposite/hypotenuse✓
• Bhypotenuse/opposite
• Cadjacent/hypotenuse
• Dopposite/adjacent

Answer: SOH: sin θ = opposite ÷ hypotenuse.

What is the area of a circle with radius 6 cm? (Use π ≈ 3.14)

• A113.0 cm²✓
• B18.8 cm²
• C37.7 cm²
• D36 cm²

Answer: Area = πr² = 3.14 × 6² = 3.14 × 36 ≈ 113.0 cm².

The angle in a semicircle (from the diameter to a point on the circle) is:

• A90°✓
• B60°
• C45°
• D180°

Answer: A circle theorem: the angle in a semicircle is always a right angle, 90°.

What is the sum of the interior angles of a hexagon?

• A540°
• B720°✓
• C900°
• D360°

Answer: Interior angle sum = (n − 2) × 180° = (6 − 2) × 180° = 720°.

Two shapes are similar with a length scale factor of 3. What is the area scale factor?

• A3
• B6
• C9✓
• D27

Answer: Area scale factor = (length scale factor)² = 3² = 9.

What is the volume of a cylinder with radius 2 cm and height 5 cm? (Use π ≈ 3.14)

• A31.4 cm³
• B125.6 cm³
• C20 cm³
• D62.8 cm³✓

Answer: Volume = πr²h = 3.14 × 2² × 5 = 3.14 × 20 = 62.8 cm³.

What is the value of cos 60°?

• A√3/2
• B0
• C1
• D0.5✓

Answer: cos 60° = 0.5 (a standard exact trig value).

A bearing is always measured:

• AAnticlockwise from south
• BFrom the nearest object
• CIn radians
• DClockwise from north, as 3 figures✓

Answer: Bearings are measured clockwise from north and written with three figures (e.g. 075°).

In triangle ABC, angle A = 40°, angle B = 75° and side BC = 9 cm. Use the sine rule to work out the length of AC, to 1 decimal place.

• A3.0 cm
• B6.0 cm
• C12.7 cm
• D13.5 cm✓

Answer: BC is opposite angle A and AC is opposite angle B, so AC/sin 75° = 9/sin 40°. That gives AC = 9 × sin 75° ÷ sin 40° = 9 × 0.9659 ÷ 0.6428 = 13.5 cm.

In triangle PQR, angle P = 32°, QR = 5 cm and PR = 8 cm. Use the sine rule to work out angle Q, to 1 decimal place, giving every possible answer.

• A19.3° or 160.7°
• B58.0° only
• C58.0° or 122.0°✓
• D58.0° or 148.0°

Answer: QR is opposite P and PR is opposite Q, so sin Q = 8 × sin 32° ÷ 5 = 0.8479. The calculator gives 58.0°, but sine is also positive in the second quadrant, so 180° − 58.0° = 122.0° is a second valid triangle because 32° + 122.0° is still less than 180°.

A triangle has two sides of length 7 cm and 10 cm with an angle of 55° between them. Use the cosine rule to work out the length of the third side, to 1 decimal place.

• A5.9 cm
• B8.3 cm✓
• C12.2 cm
• D15.1 cm

Answer: a² = 7² + 10² − 2 × 7 × 10 × cos 55° = 149 − 140 × 0.5736 = 68.70, so a = 8.3 cm. Leaving out the cosine term would be treating a non-right-angled triangle as right-angled.

A triangle has sides of length 6 cm, 9 cm and 11 cm. Work out the size of its largest angle, to 1 decimal place.

• A33.0°
• B54.8°
• C87.9°
• D92.1°✓

Answer: The largest angle is opposite the longest side, 11 cm. cos C = (6² + 9² − 11²) ÷ (2 × 6 × 9) = (117 − 121) ÷ 108 = −0.03704. A negative cosine gives an obtuse angle, C = 92.1°.

A triangle has two sides of length 12 cm and 15 cm with an angle of 38° between them. Work out its area, to 1 decimal place.

• A55.4 cm²✓
• B70.9 cm²
• C90.0 cm²
• D110.8 cm²

Answer: Area = ½ab sin C = ½ × 12 × 15 × sin 38° = 90 × 0.6157 = 55.4 cm². Omitting the sine would treat the two sides as perpendicular.

A, B and C are points on the circumference of a circle with centre O. B lies on the major arc AC, and the angle AOC at the centre standing on the minor arc AC is 118°. Work out the size of angle ABC.

• A236°
• B118°
• C62°
• D59°✓

Answer: The angle at the centre is twice the angle at the circumference when both stand on the same arc, so angle ABC = 118° ÷ 2 = 59°.

ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle ABC = 104° and angle BCD = 71°. Work out angles ADC and DAB.

• Aangle ADC = 76°, angle DAB = 109°✓
• Bangle ADC = 104°, angle DAB = 71°
• Cangle ADC = 109°, angle DAB = 76°
• Dangle ADC = 76°, angle DAB = 71°

Answer: Opposite angles of a cyclic quadrilateral add to 180°. ADC is opposite ABC, so ADC = 180° − 104° = 76°, and DAB is opposite BCD, so DAB = 180° − 71° = 109°.

AT is a tangent to a circle at the point A, and B and C are points on the circle. The angle between the tangent AT and the chord AB is 63°. Work out the size of angle ACB, the angle in the alternate segment.

• A27°
• B63°✓
• C117°
• D126°

Answer: By the alternate segment theorem the angle between a tangent and a chord equals the angle subtended by that chord in the alternate segment, so angle ACB = 63°.

A circle has centre O and radius 8 cm. A tangent touches the circle at P, and Q is a point on that tangent with OQ = 17 cm. Work out the length PQ.

• A25 cm
• B18.8 cm
• C15 cm✓
• D9 cm

Answer: A tangent meets a radius at 90°, so OPQ is right-angled at P with hypotenuse OQ. PQ = √(17² − 8²) = √225 = 15 cm.

Column vectors are written as (x, y). Given a = (5, −2) and b = (−3, 8), work out 2a − b.

• A(7, 4)
• B(7, −12)
• C(13, 4)
• D(13, −12)✓

Answer: 2a = (10, −4). Subtracting b component by component gives (10 − (−3), −4 − 8) = (13, −12); subtracting a negative x-component increases it.

A vector is written as (x, y). Work out the magnitude of the vector v = (−7, 24).

• A17
• B23.0
• C25✓
• D31

Answer: Magnitude uses Pythagoras on the components: |v| = √((−7)² + 24²) = √(49 + 576) = √625 = 25. Squaring removes the minus sign, so the components are added, not subtracted.

OABC is a parallelogram in which OA is parallel to CB and AB is parallel to OC. Vector OA = a and vector OC = c. M is the midpoint of AB. Write vector OM in terms of a and c.

• Aa + ½c✓
• B½a + c
• C½(a + c)
• Da + c

Answer: AB is parallel and equal to OC, so vector AB = c. Travelling from O to M means going along OA then half of AB, giving OM = a + ½c.

Two similar triangles have corresponding sides of length 6 cm and 15 cm. The smaller triangle has an area of 24 cm². Work out the area of the larger triangle.

• A9.6 cm²
• B60 cm²
• C150 cm²✓
• D375 cm²

Answer: The length scale factor is 15 ÷ 6 = 2.5, so the area scale factor is 2.5² = 6.25. The larger area is 24 × 6.25 = 150 cm²; using 2.5 alone scales lengths, not areas.

Two mathematically similar solid cylinders have total surface areas of 50 cm² and 200 cm². The smaller cylinder has a volume of 75 cm³. Work out the volume of the larger cylinder.

• A4,800 cm³
• B600 cm³✓
• C300 cm³
• D150 cm³

Answer: The area scale factor is 200 ÷ 50 = 4, so the length scale factor is √4 = 2 and the volume scale factor is 2³ = 8. The larger volume is 75 × 8 = 600 cm³.

A cuboid measures 8 cm by 6 cm by 5 cm. Work out the length of the longest straight rod that will fit inside it, to 1 decimal place.

• A19.0 cm
• B11.2 cm✓
• C10.0 cm
• D7.8 cm

Answer: The longest rod lies along the space diagonal, so all three dimensions are squared: √(8² + 6² + 5²) = √125 = 11.2 cm. Using only two dimensions gives a face diagonal, which is shorter.

A cuboid ABCDEFGH has a horizontal rectangular base ABCD measuring 12 cm by 9 cm and a vertical height of 8 cm, with G vertically above C. Work out the angle between the diagonal AG and the base, to 1 decimal place.

• A28.1°✓
• B32.2°
• C33.7°
• D61.9°

Answer: First find the base diagonal AC = √(12² + 9²) = 15 cm. The angle at A in the right-angled triangle ACG has opposite 8 cm and adjacent 15 cm, so the angle is tan⁻¹(8 ÷ 15) = 28.1°.