A living, growing hub for GCSE Higher Maths.
GCSE exam success depends on understanding the core formulae and being able to apply them accurately in the right context. This page is your concise launchpad: each formula is stated clearly and illustrated with quick, line-by-line examples. Over time, each section will expand into lesson notes, quizzes, practice sets, and topic links across the full GCSE curriculum. For now, the focus is GCSE Higher with differentiation to follow.
The area enclosed by a circle of radius \( r \).
Given \( r = 3\,\text{cm} \)
\[ A = \pi r^2 \] \[ A = \pi \times 3^2 \] \[ A = 9\pi \approx 28.27\,\text{cm}^2 \]Reverse problem: Given \( A = 50\,\text{cm}^2 \)
\[ r = \sqrt{\dfrac{A}{\pi}} = \sqrt{\dfrac{50}{\pi}} \approx 3.99\,\text{cm} \]Perimeter (distance around) of a circle.
Given \( r = 5\,\text{cm} \)
\[ C = 2\pi r = 2\pi \times 5 = 10\pi \approx 31.42\,\text{cm} \]Given \( C = 44\,\text{cm} \)
\[ d = \dfrac{C}{\pi} \approx \dfrac{44}{\pi} \approx 14.0\,\text{cm} \] \[ r = \dfrac{d}{2} \approx 7.0\,\text{cm} \]Portion of a circle with central angle \( \theta^\circ \).
Given \( r = 6\,\text{cm} \), \( \theta = 30^\circ \)
\[ A = \dfrac{30}{360}\,\pi \times 6^2 \] \[ A = \dfrac{1}{12}\,\pi \times 36 = 3\pi \approx 9.42\,\text{cm}^2 \]Length of the curved boundary of a sector.
Given \( r = 10\,\text{cm} \), \( \theta = 45^\circ \)
\[ \ell = \dfrac{45}{360}\times 2\pi \times 10 \] \[ \ell = \dfrac{1}{8}\times 20\pi = 2.5\pi \approx 7.85\,\text{cm} \]Same cross-section all the way through (e.g. cuboid, cylinder).
Triangular prism with cross-section area \( 24\,\text{cm}^2 \), length \( 15\,\text{cm} \)
\[ V = 24 \times 15 = 360\,\text{cm}^3 \]Cylinder with \( r = 4\,\text{cm} \), length \( 12\,\text{cm} \)
\[ V = \pi r^2 \times 12 = \pi \times 4^2 \times 12 = 192\pi \approx 603.19\,\text{cm}^3 \]Base area times perpendicular height, then divide by 3.
Square pyramid with base \( 10\,\text{cm} \times 10\,\text{cm} \), height \( 18\,\text{cm} \)
\[ V = \dfrac{1}{3}\times 100 \times 18 = 600\,\text{cm}^3 \]Cone with \( r = 3\,\text{cm} \), height \( 14\,\text{cm} \)
\[ V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \times 9 \times 14 = 42\pi \approx 131.95\,\text{cm}^3 \]Use the triangle method for remembering rearrangements.
Distance \( 150\,\text{km} \) in \( 2\,\text{h} \)
\[ v = \dfrac{150}{2} = 75\,\text{km/h} \]Mass \( 500\,\text{g} \), volume \( 250\,\text{cm}^3 \)
\[ \rho = \dfrac{500}{250} = 2\,\text{g/cm}^3 \]Force \( 200\,\text{N} \) over area \( 0.5\,\text{m}^2 \)
\[ P = \dfrac{200}{0.5} = 400\,\text{Pa} \]In a right triangle: the square on the hypotenuse equals the sum of squares on the legs.
Given legs \( a=6 \), \( b=8 \)
\[ c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]Given hypotenuse \( c=13 \) and leg \( a=5 \)
\[ b = \sqrt{c^2 - a^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \]Use with right triangles to find sides or angles.
Opposite \( 7 \), hypotenuse \( 25 \)
\[ \theta = \sin^{-1}\!\left(\dfrac{7}{25}\right) \approx 16.26^\circ \]Adjacent \( 9 \), hypotenuse \( 15 \)
\[ \theta = \cos^{-1}\!\left(\dfrac{9}{15}\right) \approx 53.13^\circ \]Opposite \( 5 \), adjacent \( 4 \)
\[ \theta = \tan^{-1}\!\left(\dfrac{5}{4}\right) \approx 51.34^\circ \]Use in non-right triangles with a matching side–angle pair.
Given \( A=40^\circ \), \( a=8 \), \( B=72^\circ \)
\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} \] \[ b = a \cdot \dfrac{\sin B}{\sin A} = 8\cdot \dfrac{\sin 72^\circ}{\sin 40^\circ} \approx 11.9 \]Use when two sides and the included angle are known, or all three sides.
Given \( b=7 \), \( c=9 \), \( A=50^\circ \)
\[ a^2 = 7^2 + 9^2 - 2\cdot 7\cdot 9\cos 50^\circ \] \[ a^2 \approx 130 - 126\cdot 0.6428 \approx 48.8 \] \[ a \approx 6.99 \]Find an angle from three sides.
Given \( a=10 \), \( b=13 \), \( c=7 \)
\[ \cos A = \dfrac{169 + 49 - 100}{2\cdot 13\cdot 7} = \dfrac{118}{182} \approx 0.6484 \] \[ A \approx 49.4^\circ \]Use two sides and the included angle.
Given \( a=12 \), \( b=9 \), \( C=35^\circ \)
\[ A = \dfrac{1}{2}\cdot 12 \cdot 9 \cdot \sin 35^\circ \approx 31.0 \]Solves \( ax^2 + bx + c = 0 \) for real/complex roots.
\( x^2 - 5x + 6 = 0 \)
\[ a=1,\; b=-5,\; c=6 \] \[ x = \dfrac{5 \pm \sqrt{25 - 24}}{2} = \dfrac{5 \pm 1}{2} \] \[ x = 3 \quad \text{or} \quad x = 2 \]Memorise key values: \(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\).
Using a \(30\text{–}60\text{–}90\) triangle with sides \(1:\sqrt{3}:2\)
\[ \sin 30^\circ = \dfrac{1}{2},\quad \cos 30^\circ = \dfrac{\sqrt{3}}{2},\quad \tan 30^\circ = \dfrac{1}{\sqrt{3}} \]\( m \) is the gradient; \( c \) is the y-intercept.
Through \( (0,-3) \) with gradient \( 2 \)
\[ y = 2x - 3 \]Parallel to \( y = -\tfrac{1}{2}x + 4 \) through \( (2,1) \)
\[ m = -\tfrac{1}{2} \quad \Rightarrow \quad y = -\tfrac{1}{2}x + 2 \]Rate of change of \( y \) with respect to \( x \) between two points.
Between \( (2,5) \) and \( (8,17) \)
\[ m = \dfrac{17 - 5}{8 - 2} = \dfrac{12}{6} = 2 \]Coordinates of the midpoint of a segment.
Between \( (-3,4) \) and \( (7,-2) \)
\[ \left(\dfrac{-3+7}{2},\,\dfrac{4-2}{2}\right) = (2,1) \]For quick reference, you can view or download the formulae sheet directly from MathsGenie. It includes all the key equations provided in the exam materials.