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GCSE / IGCSE Maths – Geometry

Angles, circle theorems, Pythagoras, area and volume, transformations.

13 questions  ·  83 marks  ·  answers included  ·  GCSE/IGCSE (Higher)
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This worksheet covers:

  • Angle facts and polygons
  • Circle theorems
  • Pythagoras in 2D and 3D
  • Area and volume (including sectors, cones, spheres and frustums)
  • Similar shapes
  • Transformations

Section A is non-calculator; use a calculator for Sections B and C. Allow about 75 minutes.

Questions

Section A – Skills

Non-calculator
1

(a) Work out the size of each interior angle of a regular decagon.(b) A regular polygon has interior angles of $156^\circ$. Work out the number of sides.(c) Two parallel lines are crossed by a transversal. One of the angles formed is $68^\circ$. Write down the size of the co-interior (allied) angle to it, giving a reason.

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(a) Exterior $= 360 \div 10 = 36^\circ$, interior $= 144^\circ$(b) Exterior $= 24^\circ$, $n = 360 \div 24 = 15$(c) $112^\circ$; co-interior angles between parallel lines add to $180^\circ$
[6]
2

(a) A right-angled triangle has shorter sides of length $7\,\text{cm}$ and $24\,\text{cm}$. Work out the length of the hypotenuse.(b) A cuboid measures $3\,\text{cm}$ by $4\,\text{cm}$ by $12\,\text{cm}$. Work out the length of the longest diagonal of the cuboid.

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(a) $\sqrt{49 + 576} = \sqrt{625} = 25\,\text{cm}$(b) $\sqrt{9 + 16 + 144} = \sqrt{169} = 13\,\text{cm}$
[5]
3

(a) A trapezium has parallel sides of length $8\,\text{cm}$ and $14\,\text{cm}$ and a perpendicular height of $5\,\text{cm}$. Work out its area.(b) A sector of a circle has radius $6\,\text{cm}$ and angle $60^\circ$. Work out the exact area of the sector and the exact arc length, leaving your answers in terms of $\pi$.

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(a) $\dfrac{1}{2}(8 + 14)(5) = 55\,\text{cm}^2$(b) Area $= \dfrac{60}{360}\pi(6^2) = 6\pi\,\text{cm}^2$; arc $= \dfrac{60}{360}(2\pi)(6) = 2\pi\,\text{cm}$
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4

A cone has base radius $3\,\text{cm}$ and perpendicular height $4\,\text{cm}$.(a) Show that the slant height is $5\,\text{cm}$.(b) Work out the volume of the cone, giving your answer in terms of $\pi$.(c) Work out the curved surface area, giving your answer in terms of $\pi$. (Curved surface area of a cone $= \pi r l$.)

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(a) $l = \sqrt{3^2 + 4^2} = 5\,\text{cm}$(b) $\dfrac{1}{3}\pi(3^2)(4) = 12\pi\,\text{cm}^3$(c) $\pi(3)(5) = 15\pi\,\text{cm}^2$
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5

The point $P$ has coordinates $(2, 3)$.(a) Write down the coordinates of the image of $P$ after a rotation of $90^\circ$ clockwise about the origin.(b) Write down the coordinates of the image of $P$ after a reflection in the line $y = x$.(c) The point $Q(3, 2)$ is enlarged by scale factor $-2$ with centre $(1, 1)$. Find the coordinates of the image of $Q$.(d) Describe fully the single transformation that maps triangle $A$ to triangle $B$, where $B$ is the same shape and size as $A$, is the same way round (not a mirror image), and every point has moved $5$ units right and $2$ units down.

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(a) $(3, -2)$(b) $(3, 2)$(c) $(1 – 2(2),\ 1 – 2(1)) = (-3, -1)$(d) Translation by the vector with components $5$ (right) and $-2$ (down), i.e. $\left(5, -2\right)$ as a column vector
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6

Points $A$, $B$ and $C$ lie on a circle with centre $O$. Angle $AOB = 110^\circ$ and $C$ is on the major arc.(a) Work out angle $ACB$, giving a reason.(b) $D$ is a point on the minor arc $AB$. Work out angle $ADB$, giving a reason.(c) A tangent to the circle at $A$ meets the line $OA$. Write down the angle between the tangent and $OA$.

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(a) $55^\circ$; the angle at the centre is twice the angle at the circumference(b) $125^\circ$; opposite angles of a cyclic quadrilateral $ACBD$ add to $180^\circ$(c) $90^\circ$ (tangent is perpendicular to the radius)
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Section B – Exam-style

Calculator allowed
7

A cuboid $ABCDEFGH$ has base $ABCD$ with $AB = 6\,\text{cm}$ and $BC = 8\,\text{cm}$, and vertical height $CG = 5\,\text{cm}$. $G$ is vertically above $C$.(a) Work out the length $AC$.(b) Work out the length $AG$, giving your answer to 3 significant figures.(c) Work out the angle that $AG$ makes with the base $ABCD$, giving your answer to 1 decimal place.

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(a) $\sqrt{36 + 64} = 10\,\text{cm}$(b) $\sqrt{100 + 25} = \sqrt{125} = 11.2\,\text{cm}$(c) $\tan\theta = \dfrac{5}{10}$, $\theta = 26.6^\circ$
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8

A solid is made from a cylinder of radius $3\,\text{cm}$ and height $8\,\text{cm}$ with a hemisphere of radius $3\,\text{cm}$ fixed on top.(a) Work out the volume of the solid, giving your answer in terms of $\pi$.(b) The solid is made of metal with density $8.9\,\text{g/cm}^3$. Work out the mass of the solid to the nearest gram. (Volume of a sphere $= \dfrac{4}{3}\pi r^3$.)

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(a) $\pi(3^2)(8) + \dfrac{1}{2}\cdot\dfrac{4}{3}\pi(3^3) = 72\pi + 18\pi = 90\pi\,\text{cm}^3$(b) $90\pi \times 8.9 = 2516.4… \approx 2516\,\text{g}$
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9

Two mathematically similar bottles have heights $12\,\text{cm}$ and $18\,\text{cm}$.(a) The label on the smaller bottle has area $30\,\text{cm}^2$. Work out the area of the corresponding label on the larger bottle.(b) The larger bottle holds $540\,\text{ml}$. Work out how much the smaller bottle holds.

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(a) Length scale factor $\dfrac{3}{2}$, area factor $\dfrac{9}{4}$: $30 \times \dfrac{9}{4} = 67.5\,\text{cm}^2$(b) Volume factor $\dfrac{27}{8}$: $540 \times \dfrac{8}{27} = 160\,\text{ml}$
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10

$A$, $B$ and $C$ are points on a circle with centre $O$. The line $TA$ is a tangent to the circle at $A$. Angle $OAB = 35^\circ$.(a) Work out angle $AOB$, giving a reason.(b) Work out angle $ACB$, where $C$ is on the major arc, giving a reason.(c) Work out angle $TAB$ (the angle between the tangent and the chord $AB$) and state the circle theorem that connects your answers to (b) and (c).

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(a) Triangle $OAB$ is isosceles ($OA = OB$, radii), so $AOB = 180 – 2(35) = 110^\circ$(b) $ACB = 55^\circ$; angle at the centre is twice the angle at the circumference(c) $TAB = 90 – 35 = 55^\circ$; alternate segment theorem: the angle between tangent and chord equals the angle in the alternate segment
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11

A shape is made from a square of side $10\,\text{cm}$ with a quarter circle of radius $5\,\text{cm}$ removed from each of two adjacent corners (the quarter circles are centred at the two corners and do not overlap).(a) Work out the area of the shape to 3 significant figures.(b) Work out the perimeter of the shape to 3 significant figures.

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(a) $100 – 2 \times \dfrac{1}{4}\pi(5^2) = 100 – 12.5\pi = 60.7\,\text{cm}^2$(b) Straight edges: $10 + 10 + 5 + 5 = 30$; arcs: $2 \times \dfrac{1}{4}(2\pi)(5) = 5\pi$; perimeter $= 30 + 5\pi = 45.7\,\text{cm}$
[6]

Section C – Challenge

Calculator allowed
12

A solid cone has base radius $6\,\text{cm}$ and height $12\,\text{cm}$. The top of the cone is removed by a cut parallel to the base, $4\,\text{cm}$ from the vertex, leaving a frustum.(a) Show that the radius of the circular cut is $2\,\text{cm}$.(b) Work out the volume of the frustum, giving your answer in terms of $\pi$.(c) The frustum is melted down and recast as a sphere. Work out the radius of the sphere to 3 significant figures.

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(a) Similar triangles: $r = 6 \times \dfrac{4}{12} = 2\,\text{cm}$(b) $\dfrac{1}{3}\pi(36)(12) – \dfrac{1}{3}\pi(4)(4) = 144\pi – \dfrac{16\pi}{3} = \dfrac{416\pi}{3}\,\text{cm}^3$ ($\approx 436\,\text{cm}^3$)(c) $\dfrac{4}{3}\pi r^3 = \dfrac{416\pi}{3} \Rightarrow r^3 = 104$, $r = 4.70\,\text{cm}$
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13

Triangle $A$ has vertices $(1, 1)$, $(3, 1)$ and $(1, 4)$.(a) Triangle $A$ is reflected in the $x$-axis to give triangle $B$. Write down the vertices of $B$.(b) Triangle $B$ is then rotated $180^\circ$ about the origin to give triangle $C$. Write down the vertices of $C$.(c) Describe fully the single transformation that maps triangle $A$ directly to triangle $C$.(d) Triangle $A$ is enlarged by scale factor $3$ about the origin. Work out the area of the enlarged triangle.

Show answer
(a) $(1, -1)$, $(3, -1)$, $(1, -4)$(b) $(-1, 1)$, $(-3, 1)$, $(-1, 4)$(c) Reflection in the $y$-axis (the line $x = 0$)(d) Area of $A = \dfrac{1}{2}(2)(3) = 3$; enlarged area $= 3 \times 9 = 27$ square units
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