Maths IGCSE Y10 Flashcards

Area = 1/2 × base × height

In a right-angled triangle: a² + b² = c², where c is the hypotenuse.

x = (-b ± √(b² - 4ac)) / 2a

Simple Interest = Principal × Rate × Time

Circumference = 2πr or πd

Area = πr²

Volume = length × width × height

Mean = (Sum of values) / (Number of values)

Gradient = (Change in y) / (Change in x)

Find two numbers that multiply to give ac and add to give b, then split the middle term.

Compound Interest = Principal × (1 + Rate)^Time - Principal

Perimeter = 2 × (length + width)

Probability = (Number of favourable outcomes) / (Total number of outcomes)

1. Write the quadratic in the form ax² + bx + c. 2. Find two numbers that multiply to ac and add to b. 3. Split the middle term using these numbers. 4. Factorise in pairs.

Example: Factorise 2x² + 7x + 3. 1. a = 2, b = 7, c = 3. 2. Find two numbers that multiply to 2×3=6 and add to 7: 6 and 1. 3. Rewrite as 2x² + 6x + x + 3. 4. Factorise in pairs: 2x(x + 3) + 1(x + 3). 5. Final answer: (2x + 1)(x + 3).

1. Rearrange one equation to make x or y the subject. 2. Substitute this expression into the other equation. 3. Solve for the remaining variable. 4. Substitute back to find the other variable.

1. Multiply one or both equations to make the coefficients of x or y the same. 2. Add or subtract the equations to eliminate one variable. 3. Solve for the remaining variable. 4. Substitute back to find the other variable.

Solve: y = 2x + 1 and 3x + y = 11. 1. Substitute y = 2x + 1 into 3x + y = 11. 2. 3x + (2x + 1) = 11 → 5x + 1 = 11. 3. Solve for x: 5x = 10 → x = 2. 4. Substitute x = 2 into y = 2x + 1: y = 5. Solution: x = 2, y = 5.

Solve: 2x + y = 7 and 3x - y = 8. 1. Add the equations to eliminate y: (2x + y) + (3x - y) = 7 + 8 → 5x = 15. 2. Solve for x: x = 3. 3. Substitute x = 3 into 2x + y = 7: 2(3) + y = 7 → y = 1. Solution: x = 3, y = 1.

sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.

Use the inverse trigonometric functions: θ = sin⁻¹(opposite/hypotenuse), θ = cos⁻¹(adjacent/hypotenuse), θ = tan⁻¹(opposite/adjacent).

In 3D problems, use a² + b² + c² = d², where d is the diagonal or longest side.

For 0°, 30°, 45°, 60°, and 90°, memorise the exact values of sin, cos, and tan.

Write the surd as a product of two factors, one of which is a perfect square. Simplify the square root of the perfect square.

Multiply numerator and denominator by the conjugate of the denominator to remove the surd from the denominator.

√50 = √(25 × 2) = √25 × √2 = 5√2.

Multiply numerator and denominator by √3: (1/√3) × (√3/√3) = √3/3.

a/sin A = b/sin B = c/sin C, where a, b, c are sides and A, B, C are opposite angles.

a² = b² + c² - 2bc cos A, where A is the angle opposite side a.

cos A = (b² + c² - a²) / (2bc), where A is the angle opposite side a.

Area = 1/2 × a × b × sin C, where a and b are two sides and C is the included angle.

Sum of interior angles = (n - 2) × 180°, where n is the number of sides.

Exterior angle = 360° / n, where n is the number of sides.

Flip the shape over a given line (e.g., x-axis, y-axis, or y = x).

Turn the shape about a point (e.g., origin) by a given angle and direction (clockwise or anticlockwise).

Move the shape by a vector (x, y), where x is the horizontal movement and y is the vertical movement.

Scale the shape by a given scale factor from a centre of enlargement.

In a right-angled triangle with sides 3 cm and 4 cm, hypotenuse = √(3² + 4²) = √25 = 5 cm.