Higher Tier - Year 11

Year 11 Paper 5: Higher Geometry and Measures

Covers ratio and proportion, functions and rates of change, sequences and graphs, geometry, trigonometry and Pythagoras, and mensuration and vectors.

17 questions - 80 marks - calculator allowed

Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [3 marks]

Sequences and Graphs

Find the midpoint of the line segment joining the points (2, -5) and (8, 3).

Question 2 [3 marks]

Angles, Polygons and Circle Theorems

A regular nonagon (9-sided polygon) is drawn.

Find the sum of its interior angles, and find the size of one interior angle.

Question 3 [4 marks]

Functions and Rates of Change

The functions f and g are defined by f(x) = x^2 - 4 and g(x) = 3x.

Find the values of x for which f(x) = g(x).

Question 4 [4 marks]

Mensuration and Vectors

A circular pond has radius 6.5 m.

Find its circumference to 1 decimal place.

Question 5 [4 marks]

Ratio and Proportion

y is directly proportional to x^3.

When x = 2, y = 40.

Find y when x = 5.

Question 6 [4 marks]

Mensuration and Vectors

A sector of a circle has radius 9 cm and angle 80 degrees at the centre.

Find the length of the arc, to 1 decimal place.

Question 7 [5 marks]

Ratio and Proportion

y is directly proportional to the square root of x.

When x = 16, y = 20.

Find y when x = 81.

Question 8 [5 marks]

Pythagoras and Trigonometry

A cuboid has length 8 cm, width 5 cm and height 6 cm.

Find the length of the diagonal that runs from one corner of the cuboid to the opposite corner, to 1 decimal place.

Question 9 [4 marks]

Sequences and Graphs

The curve y = x^2 + 4x + 9 is translated 2 units right and 3 units up.

Find the equation of the translated curve in the form y = x^2 + ax + b.

Question 10 [4 marks]

Pythagoras and Trigonometry

The angle of elevation of the top of a tower from a point A on level ground is 28 degrees.

From a point B, which is 40 m closer to the tower than A and on the same line, the angle of elevation is 42 degrees.

Find the height of the tower, to 1 decimal place.

Question 11 [5 marks]

Ratio and Proportion

A cyclist travels the first 24 km of a journey at 16 km/h and the remaining 20 km at 15 km/h.

Find her average speed for the whole journey, to 1 decimal place.

Question 12 [5 marks]

Angles, Polygons and Circle Theorems

ABCD is a cyclic quadrilateral. The side DC is extended to a point E, so angle BCE is an exterior angle.

Angle DAB = 112 degrees. Find angle BCD and angle BCE.

Question 13 [6 marks]

Ratio and Proportion

y is directly proportional to x and inversely proportional to z.

When x = 8 and z = 4, y = 20. Find y when x = 15 and z = 6.

Question 14 [6 marks]

Angles, Polygons and Circle Theorems

Two similar triangles, triangle ABC and triangle PQR, have area 18 cm^2 and 50 cm^2 respectively.

The length of side AB is 6 cm.

Find the length of the corresponding side PQ.

Question 15 [6 marks]

Pythagoras and Trigonometry

From the top of a vertical cliff, the angle of depression to a boat at sea is 24 degrees.

The boat is 350 m from the base of the cliff, measured along the horizontal.

Find the height of the cliff, to the nearest metre.

Question 16 [6 marks]

Functions and Rates of Change

A curve has equation y = x^3 - 4x^2 + 5.

Find the equation of the tangent to the curve at the point where x = 3.

Question 17 [6 marks]

Ratio and Proportion

Two runners start at the same point and run in the same direction along a track.

Runner A runs at 4.5 m/s. Runner B, who starts 2 seconds later than runner A, runs at 6 m/s.

Find how long after runner A starts that runner B catches up with runner A.

Model solutions

Mark scheme for Question 1 [3 marks]
Question 1[3 marks]
Answer or working	Marks
adding the x-coordinates and dividing by 2	M1
adding the y-coordinates and dividing by 2	M1
(5, -1)	A1

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
using the sum of interior angles = (n - 2) x 180	M1
(9 - 2) x 180 = 1260 degrees	M1
one interior angle = 1260 / 9 = 140 degrees	A1
Final answer: sum = 1260 degrees, one interior angle = 140 degrees

Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or working	Marks
forming the equation x^2 - 4 = 3x	M1
rearranging to x^2 - 3x - 4 = 0	M1
factorising to (x - 4)(x + 1) = 0	M1
x = 4 or x = -1	A1

Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or working	Marks
using circumference = 2 pi r	M1
2 x pi x 6.5	M1
40.840... before rounding	M1
40.8 m	A1

Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or working	Marks
y = kx^3	M1
40 = 8k, so k = 5	M1
substituting x = 5 into y = 5x^3	M1
625	A1

Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or working	Marks
using arc length = (angle/360) x 2 pi r	M1
(80/360) x 2 x pi x 9	M1
evaluating 12.566... before rounding	M1
12.6 cm	A1

Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or working	Marks
y = k sqrt(x)	M1
20 = k x 4, since sqrt(16) = 4	M1
k = 5	A1
substituting x = 81 into y = 5 sqrt(x)	M1
45	A1

Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or working	Marks
using the 3D Pythagoras formula d = sqrt(l^2 + w^2 + h^2)	M1
8^2 + 5^2 + 6^2 = 125	M1
taking the square root of 125	M1
11.18033...	A1
11.2 cm	A1

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
replacing x by x - 2	M1
expanding (x - 2)^2 + 4(x - 2) + 9	M1
simplifying to x^2 + 5, then adding 3	M1
y = x^2 + 8	A1

Mark scheme for Question 10 [4 marks]
Question 10[4 marks]
Answer or working	Marks
forming h = (x + 40) tan 28 using triangle at A, where x is the distance from B to the base	M1
forming h = x tan 42 using triangle at B	M1
setting the two expressions equal and solving for x	M1
51.9 m (1 d.p.)	A1
Final answer: 51.9 m

Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or working	Marks
time for the first leg = 24/16 = 1.5 hours	M1
time for the second leg = 20/15 = 4/3 hours	M1
total time = 1.5 + 4/3 = 17/6 hours	M1
total distance = 44 km	M1
average speed = 44 / (17/6) = 15.5 km/h (1 d.p.)	A1
Final answer: 15.5 km/h

Mark scheme for Question 12 [5 marks]
Question 12[5 marks]
Answer or working	Marks
using opposite angles in a cyclic quadrilateral sum to 180 degrees	M1
angle BCD = 180 - 112 = 68 degrees	M1
using angles BCD and BCE lie on the straight line DCE, summing to 180 degrees	M1
angle BCE = 112 degrees	A1
noting angle BCE equals angle DAB, the exterior angle of a cyclic quadrilateral equals the interior opposite angle	B1
Final answer: angle BCD = 68 degrees, angle BCE = 112 degrees

Mark scheme for Question 13 [6 marks]
Question 13[6 marks]
Answer or working	Marks
y = kx/z	M1
20 = 8k/4, so 20 = 2k	M1
k = 10	A1
substituting x = 15 and z = 6 into y = 10x/z	M1
y = 10 x 15 / 6	M1
25	A1

Mark scheme for Question 14 [6 marks]
Question 14[6 marks]
Answer or working	Marks
finding the area scale factor 50/18 = 25/9	M1
taking the square root to find the length scale factor	M1
length scale factor = 5/3	A1
PQ = AB x 5/3	M1
6 x 5/3	M1
10 cm	A1

Mark scheme for Question 15 [6 marks]
Question 15[6 marks]
Answer or working	Marks
recognising the angle of elevation from the boat equals the angle of depression, 24 degrees	M1
using tan 24 = height / 350	M1
height = 350 x tan 24	M1
evaluating 350 x tan 24 = 155.8...	M1
155.8	A1
156 m (nearest metre)	A1
Final answer: 156 m

Mark scheme for Question 16 [6 marks]
Question 16[6 marks]
Answer or working	Marks
differentiating to get dy/dx = 3x^2 - 8x	M1
substituting x = 3 into dy/dx to find the gradient	M1
gradient = 3	A1
substituting x = 3 into y = x^3 - 4x^2 + 5 to find the y-coordinate	M1
y = -4, giving the point (3, -4)	A1
y = 3x - 13	A1

Mark scheme for Question 17 [6 marks]
Question 17[6 marks]
Answer or working	Marks
runner A's distance after t seconds = 4.5t	M1
runner B's running time = (t - 2) seconds	M1
runner B's distance = 6(t - 2)	M1
forming the equation 4.5t = 6(t - 2)	M1
solving to -1.5t = -12	M1
t = 8 seconds	A1
Final answer: 8 seconds