For Class 8, 9 & 10
Master the basics - and everything after gets easier
Concept-first questions with clear model answers in Physics, Chemistry, Maths and Biology, all NCERT-aligned. Start early, build the habit, and walk into your boards, NEET and JEE already ahead.
Interactive lessons
learn by playingDrag, slide and build - watch each concept come alive, then reveal the answer.
200 interactive lessons
Ohm's law
Class 10 Physics
Slide V & R, watch the bulb glow
Open →pH scale
Class 10 Chemistry
Slide across acids and bases
Open →Atomic number and mass number
Class 9 Chemistry
Add protons & neutrons, build shells
Open →Laws of reflection
Class 8 Physics
Change the angle, watch it bounce
Open →Volume of a sphere
Class 9 Maths
Grow the radius, see the volume
Open →Area of a trapezium
Class 8 Maths
Drag the sides, read the area
Open →Power of a lens
Class 10 Physics
Move the object, trace the rays
Open →Food chain and energy flow
Class 10 Biology
Follow the energy as it flows
Open →Speed
Class 8 Physics
Slide distance & time, watch the speed
Open →Density
Class 9 Physics
Pack mass into volume, float or sink
Open →Work done
Class 9 Physics
Push harder or farther, watch work grow
Open →Kinetic energy
Class 9 Physics
Speed it up - energy grows with the square
Open →Power of a lens
Class 10 Physics
Shorten the focal length, boost the power
Open →Mole concept
Class 9 Chemistry
Weigh out grams, count the moles
Open →Avogadro's number
Class 9 Chemistry
Add moles, count the particles
Open →Microscope magnification
Class 8 Biology
Grow the image, read the magnification
Open →Population density
Class 10 Biology
Add individuals, shrink the land, see crowding
Open →Simple interest
Class 8 Maths
Slide money, rate & time, watch interest
Open →Pythagoras theorem
Class 9 Maths
Stretch the two sides, get the hypotenuse
Open →Probability of an event
Class 10 Maths
Change the outcomes, watch the odds
Open →Newton's second law
Class 9 Physics
Push a mass, pick an acceleration
Open →Momentum
Class 9 Physics
Slide mass & velocity, build momentum
Open →Pressure
Class 8 Physics
Shrink the area, feel the pressure rise
Open →Weight
Class 9 Physics
Change the planet's gravity, watch your weight
Open →Refractive index
Class 10 Physics
Slow light in the medium, raise the index
Open →Resistors in series
Class 10 Physics
Add two resistors in a line
Open →Mass percentage of a solution
Class 9 Chemistry
Dissolve solute, read the strength
Open →Concentration of a solution
Class 9 Chemistry
Pack solute into less liquid
Open →Population change
Class 10 Biology
Balance births against deaths
Open →Compound microscope
Class 8 Biology
Combine eyepiece & objective lenses
Open →Area of a circle
Class 8 Maths
Grow the radius, watch the area square
Open →Volume of a cuboid
Class 8 Maths
Stretch length, breadth & height
Open →Electronic configuration and valency
Class 9 Chemistry
Slide the atomic number, build the atom
Open →Homologous series (alkanes)
Class 10 Chemistry
Add carbons, name the compound
Open →Mass number
Class 9 Chemistry
Add protons & neutrons, get the mass number
Open →Power
Class 9 Physics
More work in less time = more power
Open →Potential energy
Class 9 Physics
Lift a mass higher, store energy
Open →Wave speed
Class 9 Physics
Tune frequency & wavelength, set the speed
Open →Electric current
Class 10 Physics
Push charge per second, get the current
Open →Percentage
Class 8 Maths
Compare part to whole as a %
Open →Electron dot structure
Class 9 Chemistry
Draw valence electrons as dots
Open →Acceleration
Class 9 Physics
Speed up over time, find acceleration
Open →Distance, speed and time
Class 8 Physics
Set speed & time, cover the distance
Open →Frequency and time period
Class 9 Physics
Shorten the period, raise the frequency
Open →Heating effect of current
Class 10 Physics
Raise the current, watch heating soar
Open →Area of a triangle
Class 8 Maths
Set base & height, halve the rectangle
Open →Area of a rectangle
Class 8 Maths
Set length & breadth, fill the area
Open →Mean (average)
Class 9 Maths
Share the total equally across items
Open →Discount
Class 8 Maths
Slide price & % off, see the saving
Open →Heart rate
Class 10 Biology
Set heart rate & time, count the beats
Open →Resistors in parallel
Class 10 Physics
Wire two resistors side by side
Open →Electric charge
Class 10 Physics
Flow current over time, collect charge
Open →Electrical energy and units
Class 10 Physics
Run appliances, add up the units
Open →Kelvin temperature scale
Class 9 Chemistry
Slide Celsius, read the Kelvin
Open →Moles from number of particles
Class 9 Chemistry
Divide particles by Avogadro's number
Open →Ten percent law
Class 10 Biology
See 10% of energy reach the next level
Open →Area of a square
Class 8 Maths
Grow the side, square the area
Open →Volume of a cube
Class 8 Maths
Grow the edge, cube the volume
Open →Circumference of a circle
Class 8 Maths
Grow the radius, roll out the rim
Open →Surface area of a cube
Class 9 Maths
Grow the edge, cover six faces
Open →Potential difference
Class 10 Physics
Share work across charge, get volts
Open →Resistance from Ohm's law
Class 10 Physics
Divide voltage by current, get resistance
Open →Echo and SONAR
Class 9 Physics
Time the echo, find the distance
Open →Mass from moles
Class 9 Chemistry
Multiply moles by molar mass
Open →Breathing rate
Class 10 Biology
Set breathing rate & time
Open →Volume of a cylinder
Class 10 Maths
Set radius & height, fill the can
Open →Compound interest
Class 8 Maths
Compound money over years
Open →Profit and loss percentage
Class 8 Maths
Set cost & selling price, see profit %
Open →Perimeter of a rectangle
Class 8 Maths
Set length & breadth, walk the border
Open →Surface area of a sphere
Class 10 Maths
Grow the radius, wrap the ball
Open →Time period
Class 9 Physics
Raise the frequency, shrink the period
Open →Relative velocity
Class 9 Physics
Two objects approach - add their speeds
Open →Average velocity
Class 9 Physics
Average the start and end speeds
Open →Equations of motion (v = u + at)
Class 9 Physics
Accelerate from u for a time t
Open →Kelvin to Celsius
Class 9 Chemistry
Slide Kelvin, read the Celsius
Open →Population growth rate
Class 10 Biology
Balance births vs deaths per population
Open →Perimeter of a square
Class 8 Maths
Grow the side, walk four edges
Open →Perimeter of a triangle
Class 8 Maths
Add the three sides
Open →Area of a parallelogram
Class 8 Maths
Set base & height, slide the shape
Open →Area of a rhombus
Class 8 Maths
Set the two diagonals
Open →Equations of motion (distance)
Class 9 Physics
Start, accelerate, cover ground
Open →Joule's law of heating
Class 10 Physics
Raise current, resistance or time
Open →Electric power (P = VI)
Class 10 Physics
Multiply voltage by current
Open →Average atomic mass of isotopes
Class 9 Chemistry
Mix two isotopes by abundance
Open →Seed germination percentage
Class 9 Biology
Count sprouted seeds out of the total
Open →Volume of a cone
Class 9 Maths
Set radius & height, fill the cone
Open →Surface area of a cylinder
Class 9 Maths
Wrap the side and both ends
Open →Surface area of a cuboid
Class 9 Maths
Cover all six rectangular faces
Open →nth term of an AP
Class 10 Maths
Step from the first term by d
Open →Sum of an AP
Class 10 Maths
Add up the first n terms
Open →Focal length of a mirror
Class 10 Physics
Halve the radius to find the focus
Open →Speed of light in a medium
Class 10 Physics
Raise the index, slow the light
Open →Percentage purity
Class 9 Chemistry
Weigh the pure part of a sample
Open →Slope of a line
Class 10 Maths
Rise over run gives the steepness
Open →Percentage change
Class 8 Maths
Compare a new value to the old
Open →Volume of a hemisphere
Class 9 Maths
Grow the radius of half a ball
Open →Area of a sector
Class 10 Maths
Cut a slice of angle from a circle
Open →Length of an arc
Class 10 Maths
Measure the curved edge of a slice
Open →Slant height of a cone
Class 9 Maths
Combine radius & height for the slant
Open →Unit conversion (km/h to m/s)
Class 9 Physics
Convert km/h into m/s
Open →Equations of motion (v^2 = u^2 + 2as)
Class 9 Physics
Accelerate over a distance, find v
Open →Impulse
Class 9 Physics
Hit harder or longer, change momentum
Open →Wavelength
Class 9 Physics
Speed over frequency gives wavelength
Open →Oscillations
Class 9 Physics
Vibrate at a frequency for a time
Open →Cost of electricity
Class 10 Physics
Units times rate gives the bill
Open →Number of neutrons
Class 9 Chemistry
Take protons away from the mass number
Open →Curved surface area of a cone
Class 9 Maths
Wrap the slanted side of a cone
Open →Total surface area of a cone
Class 9 Maths
Add the base circle to the cone's side
Open →Curved surface area of a hemisphere
Class 9 Maths
Cover the dome of a hemisphere
Open →Diagonal of a square
Class 9 Maths
Cross a square corner to corner
Open →Unit conversion (m/s to km/h)
Class 9 Physics
Convert m/s into km/h
Open →Distance from velocities
Class 9 Physics
From two speeds, find the distance
Open →Diagonal of a rectangle
Class 9 Maths
Cross a rectangle corner to corner
Open →Diagonal of a cuboid
Class 9 Maths
The longest rod that fits in a box
Open →Area by Heron's formula
Class 9 Maths
Area from just the three sides
Open →Interior angle sum of a polygon
Class 8 Maths
Add up a polygon's inside angles
Open →Exterior angle of a regular polygon
Class 8 Maths
Share 360 among a polygon's corners
Open →Number of diagonals of a polygon
Class 8 Maths
Count the diagonals of a polygon
Open →Discriminant
Class 10 Maths
Test how many roots a quadratic has
Open →Sum of roots
Class 10 Maths
Sum of a quadratic's roots
Open →Punnett square (monohybrid cross)
Class 10 Biology
Cross two parents, predict the offspring
Open →Balancing chemical equations
Class 10 Chemistry
Slide coefficients until atoms balance
Open →Writing chemical formulae (valency)
Class 9 Chemistry
Criss-cross valencies into a formula
Open →Current from power
Class 10 Physics
Divide power by voltage for current
Open →Power (P = V^2 / R)
Class 10 Physics
Voltage squared over resistance
Open →Sine ratio
Class 10 Maths
Opposite over hypotenuse
Open →Cosine ratio
Class 10 Maths
Adjacent over hypotenuse
Open →Tangent ratio
Class 10 Maths
Opposite over adjacent
Open →Area of an equilateral triangle
Class 9 Maths
Area of an equilateral triangle
Open →Curved surface area of a cylinder
Class 9 Maths
Wrap only the curved side
Open →Loss percentage
Class 8 Maths
Sell below cost, find the loss %
Open →Amount with simple interest
Class 8 Maths
Principal plus its simple interest
Open →Distance formula
Class 10 Maths
Straight distance between two points
Open →States of matter
Class 9 Chemistry
Heat particles solid → liquid → gas
Open →Parts of a plant cell
Class 8 Biology
Tap a cell part to see its job
Open →Diagonal of a cube
Class 9 Maths
Longest diagonal through a cube
Open →Total surface area of a hemisphere
Class 9 Maths
Dome plus its flat circle
Open →Sum of first n natural numbers
Class 10 Maths
Add 1 + 2 + ... + n instantly
Open →Range of data
Class 9 Maths
Spread from smallest to largest
Open →Class mark
Class 9 Maths
Midpoint of a class interval
Open →Selling price from profit percent
Class 8 Maths
Mark up cost by a profit %
Open →Perimeter of a sector
Class 10 Maths
Two radii plus the curved arc
Open →Circumference from diameter
Class 8 Maths
Circumference straight from diameter
Open →Power (P = F x v)
Class 9 Physics
Force times velocity gives power
Open →Percentage of a number
Class 8 Maths
Find a percentage of a number
Open →Series and parallel circuits
Class 10 Physics
Break a bulb in series vs parallel
Open →Symbols of elements
Class 9 Chemistry
Match each element to its symbol
Open →Free fall (velocity)
Class 9 Physics
Drop from a height, hit this speed
Open →Free fall (time)
Class 9 Physics
How long a drop takes
Open →Free fall (distance)
Class 9 Physics
Distance fallen in a given time
Open →Complement of an event
Class 10 Maths
Chance an event does NOT happen
Open →Product of roots
Class 10 Maths
Product of a quadratic's roots
Open →Exterior angle theorem
Class 9 Maths
Exterior angle = sum of remote interiors
Open →Complementary angles
Class 10 Maths
What adds to 90 degrees
Open →Supplementary angles
Class 9 Maths
What adds to 180 degrees
Open →Perimeter of a semicircle
Class 10 Maths
Curved half plus the diameter
Open →Area of a semicircle
Class 10 Maths
Half the area of a circle
Open →Turning effect (moments)
Class 9 Physics
Balance the see-saw with moments
Open →Reflex arc
Class 10 Biology
Step through a reflex, stimulus to action
Open →Buoyant force (upthrust)
Class 9 Physics
Displace liquid, feel the upthrust
Open →Relative density
Class 9 Physics
Compare a density to water's
Open →Power in lifting a load
Class 9 Physics
Lift a load, faster needs more power
Open →Cosecant ratio
Class 10 Maths
Hypotenuse over opposite
Open →Secant ratio
Class 10 Maths
Hypotenuse over adjacent
Open →Cotangent ratio
Class 10 Maths
Adjacent over opposite
Open →Height from angle of elevation
Class 10 Maths
Height from an angle of elevation
Open →Area of a quadrant
Class 10 Maths
A quarter of a circle's area
Open →Interior angle of a regular polygon
Class 8 Maths
One inside angle of a regular polygon
Open →Sum of first n odd numbers
Class 10 Maths
Add the first n odd numbers
Open →Sum of first n even numbers
Class 10 Maths
Add the first n even numbers
Open →Quadratic formula (a root)
Class 10 Maths
Larger root of a quadratic
Open →LCM from HCF
Class 10 Maths
LCM from the product and HCF
Open →Depreciation
Class 8 Maths
Value drops by a % each year
Open →Cost price from selling price
Class 8 Maths
Work back to the cost price
Open →Downstream speed
Class 8 Maths
Row with the current
Open →Upstream speed
Class 8 Maths
Row against the current
Open →Average speed for a round trip
Class 8 Maths
Average speed there and back
Open →Sales tax / GST
Class 8 Maths
Tax added on a price
Open →Area of a ring (annulus)
Class 10 Maths
Area of a ring between two circles
Open →Edge of a cube from volume
Class 9 Maths
Edge back from the volume
Open →Radius from area
Class 10 Maths
Radius back from a circle's area
Open →Side from area of a square
Class 8 Maths
Side back from a square's area
Open →Height of a triangle from area
Class 9 Maths
Height back from area and base
Open →Rate from simple interest
Class 8 Maths
Rate back from the interest
Open →Time from simple interest
Class 8 Maths
Time back from the interest
Open →Principal from simple interest
Class 8 Maths
Principal back from the interest
Open →Mean proportional
Class 10 Maths
Geometric mean of two numbers
Open →Fourth proportional
Class 8 Maths
Complete the proportion a : b = c : ?
Open →Marked price from selling price
Class 8 Maths
Marked price back from the sale price
Open →Chambers of the human heart
Class 10 Biology
Tap a heart chamber to see its job
Open →Equation of a line (y = mx + c)
Class 9 Maths
Read y off a straight line
Open →Average term of an AP
Class 10 Maths
Average of first and last term
Open →Number of terms in an AP
Class 10 Maths
How many terms in an AP
Open →Midpoint of two points
Class 10 Maths
x-coordinate of a midpoint
Open →Empirical mode
Class 10 Maths
Estimate the mode from mean & median
Open →Length of a shadow
Class 10 Maths
Shadow from height and sun angle
Open →Train crossing a pole
Class 8 Maths
Speed to cross a pole
Open →Time and work
Class 8 Maths
More workers, fewer days
Open →Dividing in a ratio
Class 8 Maths
Split a total in a ratio
Open →Unitary method
Class 8 Maths
Cost of a single item
Open →Your progress — Foundation - Class 8 to 10
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Showing 78 questions in Mathematics for Class 10. Tap a card to reveal the answer.
MathematicsReal NumbersmediumFind the HCF of 96 and 404 using Euclid's division algorithm.
Reveal answer ↓
What it is
Euclid's algorithm finds the HCF by repeated division until the remainder becomes zero.
Answer
Apply repeated division: 404 = 96 x 4 + 20; 96 = 20 x 4 + 16; 20 = 16 x 1 + 4; 16 = 4 x 4 + 0. The last non-zero remainder is 4, so HCF(96, 404) = 4.
a = bq + r, then HCF(a,b) = HCF(b,r)
- •404 = 96 x 4 + 20
- •96 = 20 x 4 + 16
- •20 = 16 x 1 + 4
- •16 = 4 x 4 + 0 -> HCF = 4
Why learn this
HCF simplifies fractions, shares things equally and even underpins computer cryptography (RSA).
💡 Memory trick
Divide, take the remainder, divide again. Last non-zero remainder = HCF.
MathematicsPolynomialseasyFind the zeroes of the polynomial x^2 - 7x + 10 and verify the relationship between the zeroes and the coefficients.
Reveal answer ↓
What it is
The zeroes of a quadratic are where it equals zero; their sum is -b/a and product is c/a.
Answer
Factorise: x^2 - 7x + 10 = (x - 2)(x - 5), so the zeroes are 2 and 5. Sum of zeroes = 2 + 5 = 7 = -(-7)/1 = -b/a. Product of zeroes = 2 x 5 = 10 = 10/1 = c/a. Both relations hold.
Sum = -b/a ; Product = c/a
- •(x - 2)(x - 5) -> zeroes 2, 5
- •Sum = 7 = -b/a
- •Product = 10 = c/a
Why learn this
Finding roots is how we solve projectile paths, profit models and design problems in engineering.
💡 Memory trick
Sum = -b/a (has the minus), Product = c/a (plain).
MathematicsPair of Linear Equations in Two VariablesmediumSolve the pair of equations: 2x + 3y = 13 and x - y = -1.
Reveal answer ↓
What it is
Two linear equations meet at one point - the (x, y) that satisfies both at once.
Answer
From the second equation, x = y - 1. Substitute into the first: 2(y - 1) + 3y = 13, so 5y - 2 = 13, giving 5y = 15 and y = 3. Then x = y - 1 = 2. Solution: x = 2, y = 3.
Substitution: express one variable, substitute into the other
- •x = y - 1 from second equation
- •2(y-1) + 3y = 13 -> 5y = 15
- •y = 3, x = 2
Why learn this
It solves 'two unknowns' problems: prices, mixtures, speed-time and business break-even.
💡 Memory trick
Substitution: make one variable the subject, then plug it into the other equation.
MathematicsQuadratic EquationsmediumFind the roots of x^2 - 5x + 6 = 0 and state the nature of the roots using the discriminant.
Reveal answer ↓
What it is
The discriminant D = b^2 - 4ac reveals the nature of a quadratic's roots before you solve it.
Answer
Factorise: (x - 2)(x - 3) = 0, so the roots are x = 2 and x = 3. The discriminant D = b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1. Since D > 0 and is a perfect square, the roots are real, distinct and rational.
D = b^2 - 4ac ; x = (-b +/- sqrt(D)) / 2a
- •(x-2)(x-3) = 0 -> roots 2, 3
- •D = b^2 - 4ac = 1
- •D > 0 -> two distinct real roots
Why learn this
Engineers use it to know at a glance if a design has real solutions (bridges, projectiles).
💡 Memory trick
D > 0 real and distinct; D = 0 real and equal; D < 0 no real roots.
MathematicsArithmetic ProgressionseasyThe first term of an AP is 3 and the common difference is 5. Find its 10th term.
Reveal answer ↓
What it is
In an AP each term rises by a fixed step d; the nth term is a + (n - 1)d.
Answer
The nth term of an AP is a_n = a + (n - 1)d. Here a = 3, d = 5, n = 10, so a_10 = 3 + (10 - 1) x 5 = 3 + 45 = 48.
a_n = a + (n - 1)d
- •a_n = a + (n - 1)d
- •a = 3, d = 5, n = 10
- •a_10 = 48
Why learn this
APs model salaries with fixed raises, seating rows, EMIs and steady savings.
💡 Memory trick
nth term = start + (steps taken) x step = a + (n - 1)d.
MathematicsArithmetic ProgressionsmediumFind the sum of the first 20 terms of the AP 2, 5, 8, 11, ...
Reveal answer ↓
What it is
The sum of an AP pairs the first and last terms: S = n/2 [2a + (n - 1)d].
Answer
Here a = 2 and d = 3. The sum of n terms is S_n = n/2 [2a + (n - 1)d]. So S_20 = 20/2 [2(2) + 19(3)] = 10 [4 + 57] = 10 x 61 = 610.
S_n = n/2 [2a + (n - 1)d]
- •a = 2, d = 3, n = 20
- •S_n = n/2 [2a + (n-1)d]
- •S_20 = 10 x 61 = 610
Why learn this
It quickly totals things that grow steadily - savings, stacked logs, salaries over years.
💡 Memory trick
Sum = (number of terms / 2) x (2a + (n - 1)d): half the count times the bracket.
MathematicsTrianglesmediumState the Basic Proportionality Theorem (Thales' theorem).
Reveal answer ↓
What it is
A line parallel to one side of a triangle cuts the other two sides in the same ratio.
Answer
The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides those two sides in the same ratio. In triangle ABC, if DE is parallel to BC and meets AB at D and AC at E, then AD/DB = AE/EC.
If DE || BC then AD/DB = AE/EC
- •Line parallel to one side of a triangle
- •Divides the other two sides in the same ratio
- •AD/DB = AE/EC
- •Basis of similarity of triangles
Why learn this
It's the basis of similarity, scale drawings, maps and finding heights from shadows.
💡 Memory trick
Parallel line inside a triangle -> equal ratios: AD/DB = AE/EC.
MathematicsCoordinate GeometryeasyFind the distance between the points (2, 3) and (5, 7).
Reveal answer ↓
What it is
The distance between two points is the hypotenuse of the right triangle of their coordinate gaps.
Answer
The distance formula is d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Here d = sqrt[(5 - 2)^2 + (7 - 3)^2] = sqrt[9 + 16] = sqrt[25] = 5 units.
d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
- •d = sqrt[(x2-x1)^2 + (y2-y1)^2]
- •= sqrt[9 + 16] = sqrt[25]
- •Distance = 5 units
Why learn this
GPS, video games, robotics and maps all compute 'how far' with this exact formula.
💡 Memory trick
It's just Pythagoras: sqrt of (x-gap)^2 + (y-gap)^2.
MathematicsIntroduction to TrigonometrymediumIf sin(theta) = 3/5, find cos(theta) and tan(theta), where theta is acute.
Reveal answer ↓
What it is
The identity sin^2 + cos^2 = 1 lets you find one trig ratio from another for the same angle.
Answer
Using the identity sin^2(theta) + cos^2(theta) = 1, cos^2(theta) = 1 - (3/5)^2 = 1 - 9/25 = 16/25, so cos(theta) = 4/5 (positive since theta is acute). Then tan(theta) = sin/cos = (3/5) / (4/5) = 3/4.
sin^2(theta) + cos^2(theta) = 1 ; tan = sin/cos
- •sin^2 + cos^2 = 1
- •cos(theta) = 4/5
- •tan(theta) = sin/cos = 3/4
Why learn this
Trig ratios drive navigation, physics waves, construction and computer graphics.
💡 Memory trick
Think of the 3-4-5 triangle: sin 3/5 -> cos 4/5 -> tan 3/4.
MathematicsSome Applications of TrigonometrymediumThe angle of elevation of the top of a tower from a point 30 m from its base is 30 degrees. Find the height of the tower.
Reveal answer ↓
What it is
An angle of elevation turns a triangle into a height problem using tan = opposite / adjacent.
Answer
Let the height be h. tan(30 degrees) = h / 30, and tan(30 degrees) = 1/sqrt(3). So h = 30 x (1/sqrt(3)) = 30/sqrt(3) = 10 sqrt(3) m, which is about 17.3 m.
tan(theta) = height / base distance
- •tan(angle) = opposite/adjacent = h/30
- •tan(30 deg) = 1/sqrt(3)
- •h = 10 sqrt(3) m (about 17.3 m)
Why learn this
Surveyors, pilots and builders measure unreachable heights (towers, hills) exactly this way.
💡 Memory trick
Draw the right triangle; tan(angle) = height / base distance.
MathematicsCirclesmediumA point P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent drawn from P to the circle.
Reveal answer ↓
What it is
A tangent touches a circle at one point and is perpendicular to the radius at that point.
Answer
The tangent is perpendicular to the radius at the point of contact, so the radius, tangent and line OP form a right-angled triangle with OP as the hypotenuse. Length of tangent = sqrt[OP^2 - r^2] = sqrt[13^2 - 5^2] = sqrt[169 - 25] = sqrt[144] = 12 cm.
tangent length = sqrt[OP^2 - r^2]
- •Tangent is perpendicular to radius at contact
- •Right triangle: OP hypotenuse
- •Length = sqrt[13^2 - 5^2] = 12 cm
Why learn this
It's used in gear design, road curves and belt-and-pulley engineering.
💡 Memory trick
Radius, tangent and line to the point form a right triangle: tangent = sqrt(OP^2 - r^2).
MathematicsSurface Areas and VolumeseasyFind the volume of a cone whose base radius is 3 cm and height is 4 cm. (Take pi = 22/7.)
Reveal answer ↓
What it is
A cone's volume is exactly one-third of a cylinder with the same base and height.
Answer
Volume of a cone = (1/3) x pi x r^2 x h = (1/3) x (22/7) x 3^2 x 4 = (1/3) x (22/7) x 36 = (22 x 12) / 7 = 264/7, which is about 37.7 cubic cm.
V(cone) = (1/3) pi r^2 h
- •V = (1/3) pi r^2 h
- •= (1/3)(22/7)(9)(4)
- •= 264/7 = about 37.7 cm^3
Why learn this
It's how we measure ice-cream cones, funnels, heaps of grain and tent capacity.
💡 Memory trick
Cone = 1/3 x pi r^2 h - a cone is a third of its cylinder.
MathematicsStatisticsmediumThe mean of a distribution is 27 and its median is 30. Estimate the mode using the empirical relationship.
Reveal answer ↓
What it is
For a moderately skewed data set, the three averages link as Mode = 3 Median - 2 Mean.
Answer
The empirical relationship between the three measures of central tendency is Mode = 3 Median - 2 Mean. Substituting, Mode = 3(30) - 2(27) = 90 - 54 = 36.
Mode = 3 Median - 2 Mean
- •Mode = 3 Median - 2 Mean
- •= 3(30) - 2(27)
- •= 90 - 54 = 36
Why learn this
It estimates the most common value from the other two - handy in surveys and economics.
💡 Memory trick
Mode = 3 Median - 2 Mean ('3M minus 2-Mean gives the Mode').
MathematicsProbabilityeasyOne card is drawn at random from a well-shuffled deck of 52 playing cards. What is the probability that it is a king?
Reveal answer ↓
What it is
Probability = favourable outcomes / total outcomes, always a number between 0 and 1.
Answer
There are 4 kings in a deck of 52 cards. Probability = number of favourable outcomes / total outcomes = 4/52 = 1/13.
P(event) = favourable outcomes / total outcomes
- •4 kings in 52 cards
- •P = favourable/total = 4/52
- •= 1/13
Why learn this
It powers weather forecasts, insurance, games, medical trials and AI.
💡 Memory trick
Favourable over Total: 4 kings out of 52 -> 4/52 = 1/13.
MathematicsIntroduction to TrigonometrymediumEvaluate 2 tan^2(45 degrees) + cos^2(30 degrees) - sin^2(60 degrees).
Reveal answer ↓
What it is
The standard angles 0, 30, 45, 60 and 90 degrees have fixed trig values worth memorising.
Answer
Use standard values: tan(45 deg) = 1, cos(30 deg) = sqrt(3)/2, sin(60 deg) = sqrt(3)/2. So the expression = 2(1)^2 + (sqrt(3)/2)^2 - (sqrt(3)/2)^2 = 2 + 3/4 - 3/4 = 2.
tan45 = 1 ; cos30 = sqrt(3)/2 ; sin60 = sqrt(3)/2
- •tan45 = 1, cos30 = sin60 = sqrt(3)/2
- •2(1) + 3/4 - 3/4
- •= 2
Why learn this
They appear in nearly every physics and maths problem - knowing them saves huge time.
💡 Memory trick
For sin, use sqrt(0..4)/2 across 0,30,45,60,90; cos is the same list reversed.
MathematicsProbabilityeasyWhat is the probability of rolling a 4 on a fair die? Change the outcomes to explore.
Reveal answer ↓
What it is
Probability measures how likely an event is - favourable outcomes out of all equally likely outcomes.
Answer
Probability = favourable outcomes / total outcomes = 1 / 6 = 0.167 (about 17%). A fair die has 6 equally likely faces and only one of them is a 4. Every probability lies between 0 (impossible) and 1 (certain).
P(event) = favourable outcomes / total outcomes
- •P = favourable outcomes / total outcomes
- •Always between 0 and 1
- •One face of a die: 1/6
Why learn this
It's the maths behind games, weather forecasts and risk.
💡 Memory trick
P = favourable / total, always between 0 and 1.
MathematicsSurface Areas and VolumesmediumFind the volume of a cylinder of radius 3 and height 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The volume of a cylinder is the area of its circular base times its height.
Answer
Volume = pi x r^2 x h = 3.14159 x 3^2 x 7 = 3.14159 x 9 x 7 = 197.9 cubic units. The base is a circle of area pi x r^2, and stacking it to a height h gives the volume.
V = pi x r^2 x h
- •Volume = pi x r^2 x h
- •Base area (pi x r^2) times height
- •Unit: cubic units
Why learn this
It measures the capacity of cans, pipes, tanks and drums.
💡 Memory trick
Volume = pi x r^2 x h - base area times height.
MathematicsSurface Areas and VolumesmediumFind the surface area of a sphere of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The surface area of a sphere is four times pi times the square of its radius.
Answer
Surface area = 4 x pi x r^2 = 4 x 3.14159 x 7^2 = 4 x 3.14159 x 49 = 615.75 square units. Remarkably, a sphere's surface is exactly four times the area of its flat circular cross-section (pi x r^2).
S = 4 x pi x r^2
- •Surface area = 4 x pi x r^2
- •Four times the flat circle pi x r^2
- •Unit: square units
Why learn this
It measures the skin of balls, bubbles and planets.
💡 Memory trick
Surface area = 4 x pi x r^2 - four times the flat circle.
MathematicsArithmetic ProgressionsmediumFind the 5th term of the AP starting at 2 with common difference 3. Slide to explore.
Reveal answer ↓
What it is
In an arithmetic progression, each term is found by adding the common difference to the first term repeatedly.
Answer
nth term = a + (n - 1) d = 2 + (5 - 1) x 3 = 2 + 12 = 14. We add the common difference (n - 1) times because the first term already counts as term 1.
a_n = a + (n - 1) d
- •nth term = a + (n - 1) d
- •a is the first term, d the common difference
- •Add d one time fewer than n
Why learn this
It's how we jump straight to any term without listing them all.
💡 Memory trick
nth term = a + (n - 1) d.
MathematicsArithmetic ProgressionsmediumFind the sum of the first 5 terms of the AP starting at 2 with common difference 3. Slide to explore.
Reveal answer ↓
What it is
The sum of the first n terms of an arithmetic progression is n/2 times the sum of twice the first term and (n - 1) times the common difference.
Answer
Sum = n/2 x (2a + (n - 1) d) = 5/2 x (2x2 + (5 - 1) x 3) = 5/2 x (4 + 12) = 5/2 x 16 = 40. This equals the number of terms times the average of the first and last term.
S_n = n/2 x (2a + (n - 1) d)
- •Sum = n/2 x (2a + (n - 1) d)
- •Equals n x average of first and last term
- •For evenly spaced terms
Why learn this
It adds up long evenly-spaced lists, like seats or savings, in one step.
💡 Memory trick
Sum = n/2 x (2a + (n - 1) d).
MathematicsCoordinate GeometrymediumA line rises 6 units over a run of 3 units. Find its slope. Slide to explore.
Reveal answer ↓
What it is
The slope of a line is how much it rises for each unit it runs across.
Answer
Slope = rise / run = 6 / 3 = 2. A slope of 2 means the line climbs 2 units for every 1 unit across. A negative slope would mean the line falls as you move right.
slope = rise / run
- •Slope = rise / run
- •Positive climbs, negative falls
- •Zero slope is a flat line
Why learn this
It measures steepness for graphs, roads, ramps and roofs.
💡 Memory trick
Slope = rise / run. Up is positive, down is negative.
MathematicsAreas Related to CirclesmediumFind the area of a 90 degree sector of a circle of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The area of a sector is the fraction of the circle its angle covers, times the circle's area.
Answer
Sector area = (angle / 360) x pi x r^2 = (90 / 360) x 3.14159 x 7^2 = 0.25 x 3.14159 x 49 = 38.48 square units. A 90 degree sector is one quarter of the circle, so it has a quarter of the circle's area.
sector area = (angle / 360) x pi x r^2
- •Sector area = (angle / 360) x pi x r^2
- •It's a fraction of the whole circle
- •At 360 degrees it is the full circle
Why learn this
It measures pie-chart slices, fan sweeps and pizza pieces.
💡 Memory trick
Sector area = (angle / 360) x pi x r^2.
MathematicsAreas Related to CirclesmediumFind the length of a 90 degree arc of a circle of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The length of an arc is the fraction of the circumference that its angle subtends.
Answer
Arc length = (angle / 360) x 2 x pi x r = (90 / 360) x 2 x 3.14159 x 7 = 0.25 x 43.98 = 11.0 units. A 90 degree arc is one quarter of the full circumference of the circle.
arc length = (angle / 360) x 2 x pi x r
- •Arc length = (angle / 360) x 2 x pi x r
- •A fraction of the circumference
- •Unit: units of length
Why learn this
It measures curved paths, tracks and the edges of sectors.
💡 Memory trick
Arc length = (angle / 360) x 2 x pi x r.
MathematicsQuadratic EquationsmediumFind the discriminant of x^2 + 5x + 6 = 0. Slide a, b and c to explore.
Reveal answer ↓
What it is
The discriminant of a quadratic equation tells how many real roots it has.
Answer
Discriminant D = b^2 - 4ac = 5^2 - 4 x 1 x 6 = 25 - 24 = 1. Since D is positive, the equation has two distinct real roots. If D were 0 there would be one repeated root, and if D were negative there would be no real roots.
D = b^2 - 4ac
- •D = b^2 - 4ac
- •D > 0: two real roots; D = 0: one; D < 0: none
- •Found without solving
Why learn this
It reveals the nature of the roots without actually solving the equation.
💡 Memory trick
D = b^2 - 4ac. Positive -> 2 roots, zero -> 1, negative -> none.
MathematicsQuadratic EquationsmediumFind the sum of the roots of x^2 - 5x + 6 = 0. Slide a and b to explore.
Reveal answer ↓
What it is
The sum of the roots of a quadratic equation is minus b divided by a.
Answer
Sum of roots = -b / a = -(-5) / 1 = 5. Indeed the roots of x^2 - 5x + 6 = 0 are 2 and 3, which add to 5. The product of the roots is c / a = 6, matching 2 x 3.
sum of roots = -b / a
- •Sum of roots = -b / a
- •Product of roots = c / a
- •Relates roots to coefficients
Why learn this
It relates the roots directly to the coefficients, without solving.
💡 Memory trick
Sum of roots = -b / a; product = c / a.
MathematicsIntroduction to TrigonometrymediumIn a right triangle the opposite side is 3 and the hypotenuse is 5. Find sin of the angle. Slide to explore.
Reveal answer ↓
What it is
In a right triangle, the sine of an angle is the opposite side divided by the hypotenuse.
Answer
sin(theta) = opposite / hypotenuse = 3 / 5 = 0.6. The sine of an acute angle is always between 0 and 1, because the opposite side is never longer than the hypotenuse.
sin(theta) = opposite / hypotenuse
- •sin = opposite / hypotenuse (SOH)
- •Always between 0 and 1
- •3-4-5 triangle gives sin = 0.6
Why learn this
Sine links an angle to side lengths - the heart of trigonometry.
💡 Memory trick
SOH: Sine = Opposite / Hypotenuse.
MathematicsIntroduction to TrigonometrymediumIn a right triangle the adjacent side is 4 and the hypotenuse is 5. Find cos of the angle. Slide to explore.
Reveal answer ↓
What it is
In a right triangle, the cosine of an angle is the adjacent side divided by the hypotenuse.
Answer
cos(theta) = adjacent / hypotenuse = 4 / 5 = 0.8. Like sine, it lies between 0 and 1, and the two obey the identity sin^2(theta) + cos^2(theta) = 1.
cos(theta) = adjacent / hypotenuse
- •cos = adjacent / hypotenuse (CAH)
- •Always between 0 and 1
- •sin^2 + cos^2 = 1
Why learn this
Cosine pairs with sine to resolve forces, waves and vectors.
💡 Memory trick
CAH: Cosine = Adjacent / Hypotenuse.
MathematicsIntroduction to TrigonometrymediumIn a right triangle the opposite side is 3 and the adjacent side is 4. Find tan of the angle. Slide to explore.
Reveal answer ↓
What it is
In a right triangle, the tangent of an angle is the opposite side divided by the adjacent side.
Answer
tan(theta) = opposite / adjacent = 3 / 4 = 0.75. It also equals sin(theta) / cos(theta), and it grows without limit as the angle approaches 90 degrees.
tan(theta) = opposite / adjacent
- •tan = opposite / adjacent (TOA)
- •tan = sin / cos
- •Grows large near 90 degrees
Why learn this
Tangent gives the slope of a line and heights from angles of elevation.
💡 Memory trick
TOA: Tangent = Opposite / Adjacent.
MathematicsCoordinate GeometrymediumFind the distance between (0, 0) and (3, 4). Slide the coordinates to explore.
Reveal answer ↓
What it is
The distance between two points is the square root of the sum of the squares of the differences in their coordinates.
Answer
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2) = sqrt((3 - 0)^2 + (4 - 0)^2) = sqrt(9 + 16) = sqrt(25) = 5 units. It is simply the Pythagoras theorem applied to the horizontal and vertical gaps between the points.
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
- •Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
- •Pythagoras on coordinates
- •(0,0) to (3,4) is 5
Why learn this
It measures straight-line gaps on a graph or map.
💡 Memory trick
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2).
MathematicsArithmetic ProgressionseasyFind the sum 1 + 2 + 3 + ... + 10. Slide n to explore.
Reveal answer ↓
What it is
The sum of the first n natural numbers is n times n plus one, divided by two.
Answer
Sum = n(n + 1) / 2 = 10 x 11 / 2 = 55. This is an arithmetic progression with first term 1 and common difference 1; pairing the first and last terms gives the neat n(n + 1) / 2 formula.
sum = n(n + 1) / 2
- •Sum = n(n + 1) / 2
- •1 to 100 sums to 5050
- •An AP with a = 1, d = 1
Why learn this
It totals a long list of consecutive numbers in one step.
💡 Memory trick
Sum = n(n + 1) / 2.
MathematicsAreas Related to CirclesmediumFind the perimeter of a 90 degree sector of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The perimeter of a sector is its two straight radii plus the curved arc.
Answer
Perimeter = 2r + arc = 2 x 7 + (90 / 360) x 2 x 3.14159 x 7 = 14 + 11.0 = 25.0 units. It is the two radii bounding the sector plus the curved arc along the circle.
perimeter = 2r + (angle / 360) x 2 pi r
- •Perimeter = 2r + arc length
- •Arc = (angle / 360) x 2 pi r
- •Two radii plus the curved edge
Why learn this
It measures the boundary of a pie slice or a fan.
💡 Memory trick
Perimeter = 2r + arc = 2r + (angle / 360) x 2 pi r.
MathematicsProbabilityeasyWhat is the probability of NOT rolling a 4 on a die? Change the outcomes to explore.
Reveal answer ↓
What it is
The probability that an event does not happen is one minus the probability that it does.
Answer
P(not E) = 1 - P(E) = 1 - 1/6 = 5/6 (about 0.833). The event and its complement together cover every outcome, so their probabilities always add up to 1.
P(not E) = 1 - P(E)
- •P(not E) = 1 - P(E)
- •An event and its complement add to 1
- •Not rolling a 4 has probability 5/6
Why learn this
It's often far easier to find the chance of the opposite event.
💡 Memory trick
P(not E) = 1 - P(E).
MathematicsQuadratic EquationsmediumFind the product of the roots of x^2 - 5x + 6 = 0. Slide a and c to explore.
Reveal answer ↓
What it is
The product of the roots of a quadratic equation is c divided by a.
Answer
Product of roots = c / a = 6 / 1 = 6. The roots of x^2 - 5x + 6 = 0 are 2 and 3, and indeed 2 x 3 = 6. Their sum is -b / a = 5, matching 2 + 3.
product of roots = c / a
- •Product of roots = c / a
- •Sum of roots = -b / a
- •Relates roots to coefficients
Why learn this
It relates the roots to the coefficients without solving the equation.
💡 Memory trick
Product of roots = c / a; sum = -b / a.
MathematicsIntroduction to TrigonometryeasyWhat is the complement of a 30 degree angle? Slide the angle to explore.
Reveal answer ↓
What it is
Two angles are complementary when they add up to 90 degrees.
Answer
Complement = 90 - angle = 90 - 30 = 60 degrees. Complementary angles add to a right angle, which is why the sine of an angle equals the cosine of its complement: sin(30) = cos(60).
complement = 90 - angle
- •Complement = 90 - angle
- •The two add to 90 degrees
- •sin(theta) = cos(90 - theta)
Why learn this
Complementary angles link sine and cosine: sin(theta) = cos(90 - theta).
💡 Memory trick
Complement = 90 - angle.
MathematicsAreas Related to CirclesmediumFind the perimeter of a semicircle of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The perimeter of a semicircle is the curved half of the circumference plus the straight diameter.
Answer
Perimeter = pi x r + 2r = 3.14159 x 7 + 2 x 7 = 21.99 + 14 = 35.99 units. The curved part is half the circumference (pi x r), and we must add the straight diameter (2r) that closes the shape.
perimeter = pi x r + 2r
- •Perimeter = pi x r + 2r
- •Curved half + straight diameter
- •Do not forget the diameter
Why learn this
A common trap is to forget the straight edge and halve the whole circumference.
💡 Memory trick
Perimeter = pi x r + 2r (curved half + diameter).
MathematicsAreas Related to CircleseasyFind the area of a semicircle of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The area of a semicircle is half the area of the full circle.
Answer
Area = 1/2 x pi x r^2 = 1/2 x 3.14159 x 7^2 = 1/2 x 3.14159 x 49 = 76.97 square units. A semicircle is exactly half a circle, so we take half of pi x r^2.
area = (1/2) x pi x r^2
- •Area = 1/2 x pi x r^2
- •Half of a full circle
- •Unit: square units
Why learn this
It sizes half-round windows, arches and protractors.
💡 Memory trick
Area = 1/2 x pi x r^2.
MathematicsIntroduction to TrigonometrymediumIn a right triangle the hypotenuse is 5 and the opposite side is 3. Find cosec of the angle. Slide to explore.
Reveal answer ↓
What it is
The cosecant of an angle is the hypotenuse divided by the opposite side, the reciprocal of sine.
Answer
cosec(theta) = hypotenuse / opposite = 5 / 3 = 1.667. It is the reciprocal of the sine ratio, so cosec(theta) = 1 / sin(theta).
cosec(theta) = hypotenuse / opposite
- •cosec = hypotenuse / opposite
- •cosec = 1 / sin
- •A reciprocal ratio
Why learn this
The reciprocal ratios complete the trigonometric toolkit.
💡 Memory trick
cosec = hypotenuse / opposite = 1 / sin.
MathematicsIntroduction to TrigonometrymediumIn a right triangle the hypotenuse is 5 and the adjacent side is 4. Find sec of the angle. Slide to explore.
Reveal answer ↓
What it is
The secant of an angle is the hypotenuse divided by the adjacent side, the reciprocal of cosine.
Answer
sec(theta) = hypotenuse / adjacent = 5 / 4 = 1.25. It is the reciprocal of the cosine ratio, so sec(theta) = 1 / cos(theta).
sec(theta) = hypotenuse / adjacent
- •sec = hypotenuse / adjacent
- •sec = 1 / cos
- •A reciprocal ratio
Why learn this
It appears whenever the cosine sits in a denominator.
💡 Memory trick
sec = hypotenuse / adjacent = 1 / cos.
MathematicsIntroduction to TrigonometrymediumIn a right triangle the adjacent side is 4 and the opposite side is 3. Find cot of the angle. Slide to explore.
Reveal answer ↓
What it is
The cotangent of an angle is the adjacent side divided by the opposite side, the reciprocal of tangent.
Answer
cot(theta) = adjacent / opposite = 4 / 3 = 1.333. It is the reciprocal of the tangent ratio, so cot(theta) = 1 / tan(theta).
cot(theta) = adjacent / opposite
- •cot = adjacent / opposite
- •cot = 1 / tan
- •A reciprocal ratio
Why learn this
It completes the six trigonometric ratios.
💡 Memory trick
cot = adjacent / opposite = 1 / tan.
MathematicsSome Applications of TrigonometrymediumFrom 50 m away, the top of a tower has an angle of elevation of 30 degrees. Find its height. Slide to explore.
Reveal answer ↓
What it is
The height of an object is the horizontal distance to it times the tangent of the angle of elevation.
Answer
Height = distance x tan(theta) = 50 x tan(30) = 50 x 0.577 = 28.87 m. The horizontal distance is the adjacent side and the height is the opposite side, so tan links them.
height = distance x tan(angle of elevation)
- •Height = distance x tan(angle)
- •Uses tan = opposite / adjacent
- •Measures unreachable heights
Why learn this
It's how we measure the height of towers, trees and hills from the ground.
💡 Memory trick
Height = distance x tan(angle of elevation).
MathematicsAreas Related to CircleseasyFind the area of a quadrant of radius 7 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
A quadrant is a quarter of a circle, so its area is a quarter of the circle's area.
Answer
Area = 1/4 x pi x r^2 = 1/4 x 3.14159 x 7^2 = 1/4 x 3.14159 x 49 = 38.48 square units. A quadrant is a 90 degree sector, which is one quarter of the whole circle.
area = (1/4) x pi x r^2
- •Quadrant area = 1/4 x pi x r^2
- •A 90 degree sector
- •One quarter of the circle
Why learn this
It sizes quarter-round corners, fans and garden beds.
💡 Memory trick
Quadrant area = 1/4 x pi x r^2.
MathematicsArithmetic ProgressionseasyFind 1 + 3 + 5 + 7 + 9. Slide how many odd numbers to explore.
Reveal answer ↓
What it is
The sum of the first n odd numbers is always n squared.
Answer
The sum of the first n odd numbers = n^2. For n = 5 that is 1 + 3 + 5 + 7 + 9 = 25 = 5^2. It is an arithmetic progression with first term 1 and common difference 2.
sum = n^2
- •Sum of first n odd numbers = n^2
- •1 + 3 + 5 + 7 + 9 = 25
- •An AP with a = 1, d = 2
Why learn this
It's a neat pattern that connects odd numbers to perfect squares.
💡 Memory trick
1 + 3 + 5 + ... (n terms) = n^2.
MathematicsArithmetic ProgressionseasyFind 2 + 4 + 6 + 8 + 10. Slide how many even numbers to explore.
Reveal answer ↓
What it is
The sum of the first n even numbers is n times n plus one.
Answer
The sum of the first n even numbers = n(n + 1). For n = 5 that is 2 + 4 + 6 + 8 + 10 = 30 = 5 x 6. It is an arithmetic progression with first term 2 and common difference 2.
sum = n(n + 1)
- •Sum of first n even numbers = n(n + 1)
- •2 + 4 + 6 + 8 + 10 = 30
- •An AP with a = 2, d = 2
Why learn this
It's a quick pattern that mirrors the odd-number sum.
💡 Memory trick
2 + 4 + 6 + ... (n terms) = n(n + 1).
MathematicsQuadratic EquationshardFind a root of x^2 - 5x + 6 = 0 using the quadratic formula. Slide a, b, c to explore.
Reveal answer ↓
What it is
The quadratic formula gives the roots of any quadratic equation from its coefficients.
Answer
x = (-b + sqrt(b^2 - 4ac)) / 2a = (5 + sqrt(25 - 24)) / 2 = (5 + 1) / 2 = 3. Using the minus sign gives the other root, 2. If b^2 - 4ac is negative there are no real roots.
x = (-b + sqrt(b^2 - 4ac)) / 2a
- •x = (-b +/- sqrt(b^2 - 4ac)) / 2a
- •The + sign gives the larger root
- •No real root when b^2 - 4ac < 0
Why learn this
It solves every quadratic, even those that do not factorise neatly.
💡 Memory trick
x = (-b +/- sqrt(b^2 - 4ac)) / 2a.
MathematicsReal NumbersmediumThe HCF of 12 and 18 is 6. Find their LCM. Slide the values to explore.
Reveal answer ↓
What it is
For two numbers, the product of the HCF and LCM equals the product of the numbers.
Answer
Since HCF x LCM = a x b, LCM = (a x b) / HCF = (12 x 18) / 6 = 216 / 6 = 36. This identity holds for any pair of positive integers.
LCM = (a x b) / HCF
- •HCF x LCM = a x b
- •LCM = (a x b) / HCF
- •For 12 and 18, LCM = 36
Why learn this
It's a fast way to find the LCM once you know the HCF.
💡 Memory trick
HCF x LCM = a x b, so LCM = (a x b) / HCF.
MathematicsAreas Related to CirclesmediumFind the area of a ring with outer radius 10 and inner radius 6 units (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The area of a ring is the area of the outer circle minus the area of the inner circle.
Answer
Ring area = pi x (R^2 - r^2) = 3.14159 x (10^2 - 6^2) = 3.14159 x (100 - 36) = 3.14159 x 64 = 201.06 square units. We take the big circle's area and subtract the hole in the middle.
area = pi x (R^2 - r^2)
- •Ring area = pi x (R^2 - r^2)
- •Outer circle minus inner circle
- •Unit: square units
Why learn this
It measures washers, pipe cross-sections and circular tracks.
💡 Memory trick
Ring area = pi x (R^2 - r^2).
MathematicsAreas Related to CirclesmediumA circle has an area of 154 square units. Find its radius (pi = 3.14159). Slide to explore.
Reveal answer ↓
What it is
The radius of a circle is the square root of its area divided by pi.
Answer
Radius = sqrt(area / pi) = sqrt(154 / 3.14159) = sqrt(49.02) = 7.0 units. This reverses the area formula area = pi x r^2 to make the radius the subject.
r = sqrt(area / pi)
- •r = sqrt(area / pi)
- •Reverses area = pi x r^2
- •Area 154 -> radius 7
Why learn this
It reverses the area formula to find the radius.
💡 Memory trick
Since area = pi x r^2, r = sqrt(area / pi).
MathematicsTrianglesmediumFind the mean proportional between 4 and 9. Slide the values to explore.
Reveal answer ↓
What it is
The mean proportional between two numbers is the square root of their product.
Answer
Mean proportional = sqrt(a x b) = sqrt(4 x 9) = sqrt(36) = 6. It is the value x for which a : x = x : b, so x^2 = a x b.
mean proportional = sqrt(a x b)
- •Mean proportional = sqrt(a x b)
- •The geometric mean
- •Between 4 and 9 it is 6
Why learn this
It appears in similar triangles and the geometric mean.
💡 Memory trick
Mean proportional between a and b = sqrt(a x b).
MathematicsArithmetic ProgressionseasyFind the average term of the AP from 2 to 14. Slide the first and last terms to explore.
Reveal answer ↓
What it is
The average of all terms of an arithmetic progression equals the average of its first and last term.
Answer
Average term = (first + last) / 2 = (2 + 14) / 2 = 16 / 2 = 8. Because the terms are evenly spaced, the middle value is exactly the average of the two ends.
average term = (first + last) / 2
- •Average term = (first + last) / 2
- •Terms are evenly spaced
- •Sum = number of terms x average
Why learn this
It's why the AP sum equals the number of terms times this average.
💡 Memory trick
Average term = (first + last) / 2.
MathematicsArithmetic ProgressionsmediumHow many terms are in the AP 2, 5, 8, 11, 14? Slide the values to explore.
Reveal answer ↓
What it is
The number of terms in an arithmetic progression is found from the first term, last term and common difference.
Answer
Number of terms n = (last - first) / d + 1 = (14 - 2) / 3 + 1 = 12 / 3 + 1 = 4 + 1 = 5. We count the steps of size d from the first to the last term, then add one for the starting term.
n = (last - first) / d + 1
- •n = (last - first) / d + 1
- •Count the steps, then add 1
- •2, 5, 8, 11, 14 has 5 terms
Why learn this
It tells how long a sequence of evenly spaced values is.
💡 Memory trick
n = (last - first) / d + 1.
MathematicsCoordinate GeometryeasyFind the x-coordinate of the midpoint of x = 2 and x = 8. Slide the values to explore.
Reveal answer ↓
What it is
The midpoint of two points is found by averaging their x-coordinates and their y-coordinates.
Answer
Midpoint x = (x1 + x2) / 2 = (2 + 8) / 2 = 10 / 2 = 5. The full midpoint formula averages both coordinates: ((x1 + x2)/2, (y1 + y2)/2).
midpoint x = (x1 + x2) / 2
- •Midpoint x = (x1 + x2) / 2
- •Average both coordinates
- •Midpoint of 2 and 8 is 5
Why learn this
It locates the exact centre of a line segment.
💡 Memory trick
Midpoint x = (x1 + x2) / 2 (and the same for y).
MathematicsStatisticsmediumA data set has a median of 40 and a mean of 38. Estimate the mode. Slide the values to explore.
Reveal answer ↓
What it is
An empirical relation estimates the mode from the mean and the median.
Answer
Mode = 3 x Median - 2 x Mean = 3 x 40 - 2 x 38 = 120 - 76 = 44. This empirical formula connects the mean, median and mode for moderately skewed data.
mode = 3 x median - 2 x mean
- •Mode = 3 x Median - 2 x Mean
- •Links the three averages
- •Median 40, mean 38 -> mode 44
Why learn this
It links the three measures of central tendency.
💡 Memory trick
Mode = 3 x Median - 2 x Mean.
MathematicsSome Applications of TrigonometrymediumA 50 m tower casts a shadow when the sun is at 45 degrees. Find the shadow length. Slide to explore.
Reveal answer ↓
What it is
The length of a shadow is the object's height divided by the tangent of the sun's angle of elevation.
Answer
Shadow = height / tan(angle) = 50 / tan(45) = 50 / 1 = 50 m. As the sun gets lower, the angle shrinks, tan gets smaller, and the shadow grows much longer.
shadow = height / tan(angle of elevation)
- •Shadow = height / tan(angle)
- •Lower sun -> longer shadow
- •At 45 degrees, shadow equals the height
Why learn this
It's why shadows are long at sunrise and short at noon.
💡 Memory trick
Shadow = height / tan(sun's angle).
MathematicsReal NumbersmediumThe HCF of two numbers is 6 and their product is 216. Find their LCM.
Reveal answer ↓
What it is
For two numbers, the product of their HCF and LCM equals the product of the numbers.
Answer
For any two numbers, HCF x LCM = product of the two numbers. So LCM = product / HCF = 216 / 6 = 36. Therefore the LCM of the two numbers is 36.
HCF x LCM = product of the two numbers
- •HCF x LCM = product of the numbers
- •LCM = product / HCF
- •= 216 / 6
- •LCM = 36
Why learn this
It lets you find the LCM quickly once you know the HCF, and vice versa.
💡 Memory trick
HCF x LCM = product of the two numbers.
MathematicsReal NumbersmediumState the fundamental theorem of arithmetic and use it to find the HCF and LCM of 12 and 18.
Reveal answer ↓
What it is
Every composite number can be expressed as a unique product of prime numbers.
Answer
The fundamental theorem of arithmetic states that every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique except for the order of the factors. Prime factorising: 12 = 2^2 x 3 and 18 = 2 x 3^2. The HCF is the product of the smallest powers of common primes = 2^1 x 3^1 = 6. The LCM is the product of the greatest powers of all primes = 2^2 x 3^2 = 36.
HCF = product of least powers ; LCM = product of greatest powers
- •Every composite number = unique product of primes
- •12 = 2^2 x 3, 18 = 2 x 3^2
- •HCF = smallest powers of common primes = 6
- •LCM = greatest powers of all primes = 36
Why learn this
It is used to find the HCF and LCM by prime factorisation.
💡 Memory trick
Break a number into primes - the factorisation is unique (order aside).
MathematicsReal NumbershardHow is sqrt(2) proved to be irrational (outline)?
Reveal answer ↓
What it is
A number that cannot be written as p/q (with q not zero) is irrational, and can be proved so by contradiction.
Answer
We prove sqrt(2) is irrational by contradiction. Assume that sqrt(2) is rational, so sqrt(2) = p/q where p and q are co-prime integers (no common factor) and q is not zero. Squaring gives 2 = p^2/q^2, so p^2 = 2q^2, which means p^2 is even, and hence p is even. Let p = 2m; then 2q^2 = 4m^2, so q^2 = 2m^2, meaning q^2 is even and q is even too. But then p and q have a common factor 2, contradicting our assumption that they are co-prime. Hence our assumption is wrong, and sqrt(2) is irrational.
Irrational: cannot be written as p/q
- •Proof by contradiction
- •Assume sqrt(2) = p/q in lowest terms
- •Show both p and q are even -> contradiction
- •So sqrt(2) is irrational
Why learn this
It shows that numbers like sqrt(2) are not fractions, deepening the number system.
💡 Memory trick
To prove sqrt(2) irrational: assume it is p/q in lowest terms, then reach a contradiction.
MathematicsPolynomialsmediumFind the sum and product of the zeroes of the polynomial x^2 - 5x + 6.
Reveal answer ↓
What it is
For a quadratic ax^2 + bx + c, the sum of zeroes is -b/a and the product is c/a.
Answer
For a quadratic polynomial ax^2 + bx + c, the sum of the zeroes is -b/a and the product of the zeroes is c/a. Here a = 1, b = -5, c = 6. Sum of zeroes = -b/a = -(-5)/1 = 5. Product of zeroes = c/a = 6/1 = 6. (Indeed the zeroes are 2 and 3, whose sum is 5 and product is 6.)
Sum = -b/a ; Product = c/a
- •Sum of zeroes = -b/a = 5
- •Product of zeroes = c/a = 6
- •a = 1, b = -5, c = 6
- •Zeroes are 2 and 3
Why learn this
It lets us check zeroes or form a quadratic from its zeroes without solving.
💡 Memory trick
Sum of zeroes = -b/a ; product of zeroes = c/a.
MathematicsPair of Linear Equations in Two VariablesmediumSolve by substitution: x + y = 10 and x - y = 4.
Reveal answer ↓
What it is
A pair of linear equations can be solved by expressing one variable in terms of the other and substituting.
Answer
From the first equation, x = 10 - y. Substitute this into the second equation: (10 - y) - y = 4, so 10 - 2y = 4, which gives 2y = 6 and y = 3. Then x = 10 - y = 10 - 3 = 7. So the solution is x = 7 and y = 3. Check: 7 + 3 = 10 and 7 - 3 = 4, both correct.
Substitute one variable to solve the pair
- •Express x = 10 - y
- •Substitute: 10 - 2y = 4
- •y = 3, then x = 7
- •Solution: x = 7, y = 3
Why learn this
It is a reliable algebraic method to find a unique solution.
💡 Memory trick
Make one variable the subject, substitute into the other equation, then solve.
MathematicsPair of Linear Equations in Two VariableshardState the conditions for a pair of linear equations to have a unique solution, no solution and infinitely many solutions.
Reveal answer ↓
What it is
The ratios of coefficients decide whether a pair of linear equations has one solution, no solution or infinitely many.
Answer
For a pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: if a1/a2 is not equal to b1/b2, the lines intersect at one point and there is a unique solution (consistent). If a1/a2 = b1/b2 = c1/c2, the lines are coincident and there are infinitely many solutions (consistent and dependent). If a1/a2 = b1/b2 but not equal to c1/c2, the lines are parallel and there is no solution (inconsistent).
Compare a1/a2, b1/b2, c1/c2
- •a1/a2 not equal b1/b2: unique solution (intersecting lines)
- •a1/a2 = b1/b2 = c1/c2: infinitely many (coincident)
- •a1/a2 = b1/b2 not equal c1/c2: no solution (parallel)
- •Ratios of coefficients decide the type
Why learn this
It tells us in advance the kind of solution and how the lines are related.
💡 Memory trick
a1/a2 not equal b1/b2 -> one solution; all three equal -> infinite; first two equal but not third -> none.
MathematicsQuadratic EquationsmediumSolve the quadratic equation x^2 - 5x + 6 = 0 by factorisation.
Reveal answer ↓
What it is
A quadratic equation can be solved by splitting the middle term and factorising into two brackets.
Answer
To factorise x^2 - 5x + 6, split the middle term -5x into two terms whose product is 1 x 6 = 6 and whose sum is -5; these are -2 and -3. So x^2 - 2x - 3x + 6 = 0, giving x(x - 2) - 3(x - 2) = 0, that is (x - 2)(x - 3) = 0. Setting each factor to zero, x = 2 or x = 3. So the roots are 2 and 3.
Split middle term to factorise
- •Split -5x into -2x and -3x (product 6, sum -5)
- •x(x-2) - 3(x-2) = 0
- •(x - 2)(x - 3) = 0
- •Roots: x = 2 and x = 3
Why learn this
It is the quickest method when the quadratic factorises neatly.
💡 Memory trick
Split the middle term into two numbers that add to b and multiply to a x c.
MathematicsQuadratic EquationsmediumSolve x^2 - 4x + 1 = 0 using the quadratic formula.
Reveal answer ↓
What it is
Any quadratic ax^2 + bx + c = 0 can be solved using the quadratic formula.
Answer
Here a = 1, b = -4, c = 1. The quadratic formula is x = (-b +/- sqrt(b^2 - 4ac)) / (2a). The discriminant b^2 - 4ac = (-4)^2 - 4(1)(1) = 16 - 4 = 12. So x = (4 +/- sqrt(12)) / 2 = (4 +/- 2sqrt(3)) / 2 = 2 +/- sqrt(3). So the roots are 2 + sqrt(3) and 2 - sqrt(3).
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
- •x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
- •a=1, b=-4, c=1; discriminant = 12
- •x = (4 +/- 2sqrt(3))/2
- •Roots: 2 + sqrt(3) and 2 - sqrt(3)
Why learn this
It works for every quadratic, even those that do not factorise easily.
💡 Memory trick
x = (-b +/- sqrt(b^2 - 4ac)) / (2a).
MathematicsQuadratic EquationsmediumWhat is the discriminant? Find the nature of the roots of x^2 + 4x + 4 = 0.
Reveal answer ↓
What it is
The discriminant D = b^2 - 4ac tells the nature of the roots of a quadratic equation.
Answer
The discriminant of a quadratic equation ax^2 + bx + c = 0 is D = b^2 - 4ac. If D > 0 the roots are real and distinct; if D = 0 the roots are real and equal; and if D < 0 the equation has no real roots. For x^2 + 4x + 4 = 0, a = 1, b = 4, c = 4, so D = 4^2 - 4(1)(4) = 16 - 16 = 0. Since D = 0, the roots are real and equal (both equal to -2).
D = b^2 - 4ac
- •Discriminant D = b^2 - 4ac
- •D > 0: real and distinct roots
- •D = 0: real and equal roots
- •For x^2+4x+4: D = 0, roots equal (-2)
Why learn this
It reveals whether the roots are real and distinct, equal or not real without solving.
💡 Memory trick
D > 0 real and distinct; D = 0 real and equal; D < 0 no real roots.
MathematicsArithmetic ProgressionseasyFind the 15th term of the AP: 3, 7, 11, 15, ...
Reveal answer ↓
What it is
The nth term of an arithmetic progression is a + (n - 1)d, where a is the first term and d the common difference.
Answer
In this AP, the first term a = 3 and the common difference d = 7 - 3 = 4. The nth term is given by an = a + (n - 1)d. For the 15th term, a15 = 3 + (15 - 1) x 4 = 3 + 14 x 4 = 3 + 56 = 59. So the 15th term is 59.
an = a + (n - 1)d
- •an = a + (n - 1)d
- •a = 3, d = 4
- •a15 = 3 + 14 x 4
- •15th term = 59
Why learn this
It lets us find any term of a sequence without listing all the terms.
💡 Memory trick
an = a + (n - 1)d. Find d by subtracting any term from the next.
MathematicsArithmetic ProgressionsmediumFind the sum of the first 20 terms of the AP: 2, 5, 8, 11, ...
Reveal answer ↓
What it is
The sum of the first n terms of an AP is (n/2)[2a + (n - 1)d].
Answer
Here a = 2 and d = 5 - 2 = 3, and n = 20. The sum is Sn = (n/2)[2a + (n - 1)d] = (20/2)[2 x 2 + (20 - 1) x 3] = 10[4 + 57] = 10 x 61 = 610. So the sum of the first 20 terms is 610.
Sn = (n/2)[2a + (n - 1)d]
- •Sn = (n/2)[2a + (n - 1)d]
- •a = 2, d = 3, n = 20
- •= 10[4 + 57] = 10 x 61
- •Sum = 610
Why learn this
It quickly adds a long list of evenly spaced numbers.
💡 Memory trick
Sn = (n/2)[2a + (n-1)d], or (n/2)(first term + last term).
MathematicsTrianglesmediumState the criteria for the similarity of two triangles.
Reveal answer ↓
What it is
Two triangles are similar if their corresponding angles are equal and corresponding sides are proportional.
Answer
Two triangles are similar if they have the same shape, that is, their corresponding angles are equal and their corresponding sides are in the same ratio. The criteria for similarity are: AA (or AAA), when two angles of one triangle equal two angles of the other; SSS, when the three pairs of corresponding sides are in the same ratio; and SAS, when one angle equals the corresponding angle and the two sides including these angles are in the same ratio. Similar triangles are written with the symbol for similarity.
AA, SSS, SAS similarity criteria
- •Corresponding angles equal
- •Corresponding sides proportional
- •Criteria: AA, SSS, SAS
- •Similar = same shape, may differ in size
Why learn this
Similarity is used to find unknown lengths and to prove geometric results.
💡 Memory trick
Similarity rules: AA, SSS (proportional), SAS. Similar triangles have the same shape, not size.
MathematicsTrianglesmediumState the basic proportionality theorem (Thales theorem).
Reveal answer ↓
What it is
A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.
Answer
The basic proportionality theorem, also called Thales theorem, states that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides those two sides in the same ratio. For a triangle ABC with a line DE parallel to BC cutting AB at D and AC at E, this means AD/DB = AE/EC. The converse is also true: if a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
AD/DB = AE/EC (DE parallel to BC)
- •A line parallel to one side divides the other two proportionally
- •AD/DB = AE/EC
- •Converse is also true
- •Used to prove similarity
Why learn this
It is used to prove similarity and to find lengths of divided sides.
💡 Memory trick
Parallel line inside a triangle splits the two sides proportionally: AD/DB = AE/EC.
MathematicsTrianglesmediumThe two shorter sides of a right triangle are 6 cm and 8 cm. Find the hypotenuse.
Reveal answer ↓
What it is
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
Answer
By the Pythagoras theorem, in a right-angled triangle, hypotenuse^2 = base^2 + height^2. Here hypotenuse^2 = 6^2 + 8^2 = 36 + 64 = 100, so the hypotenuse = sqrt(100) = 10 cm. Therefore the hypotenuse is 10 centimetres.
hypotenuse^2 = base^2 + height^2
- •hypotenuse^2 = base^2 + height^2
- •= 6^2 + 8^2 = 36 + 64 = 100
- •hypotenuse = sqrt(100)
- •Hypotenuse = 10 cm
Why learn this
It is used to find distances and lengths in countless real problems.
💡 Memory trick
hypotenuse^2 = base^2 + height^2 (only for a right angle).
MathematicsCoordinate GeometrymediumFind the distance between the points (1, 2) and (4, 6).
Reveal answer ↓
What it is
The distance between two points is found from the differences of their coordinates using the Pythagoras theorem.
Answer
The distance between two points (x1, y1) and (x2, y2) is given by sqrt((x2 - x1)^2 + (y2 - y1)^2). Here = sqrt((4 - 1)^2 + (6 - 2)^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. So the distance between the two points is 5 units.
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
- •Distance = sqrt((x2-x1)^2 + (y2-y1)^2)
- •= sqrt(3^2 + 4^2)
- •= sqrt(25)
- •Distance = 5 units
Why learn this
It measures the straight-line distance between any two points on a graph.
💡 Memory trick
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2).
MathematicsCoordinate GeometryhardFind the coordinates of the point dividing the line joining (2, 3) and (6, 7) in the ratio 1:1 (the midpoint).
Reveal answer ↓
What it is
The section formula gives the coordinates of a point dividing a line segment in a given ratio.
Answer
The section formula for a point dividing the join of (x1, y1) and (x2, y2) in the ratio m:n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For the ratio 1:1 this becomes the midpoint formula ((x1 + x2)/2, (y1 + y2)/2). Here midpoint = ((2 + 6)/2, (3 + 7)/2) = (8/2, 10/2) = (4, 5). So the midpoint is (4, 5).
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
- •Section formula for ratio m:n
- •Midpoint = ((x1+x2)/2, (y1+y2)/2)
- •= ((2+6)/2, (3+7)/2)
- •Midpoint = (4, 5)
Why learn this
It is used to find dividing points and midpoints of segments.
💡 Memory trick
Point = ((m x2 + n x1)/(m+n), (m y2 + n y1)/(m+n)) for ratio m:n.
MathematicsIntroduction to TrigonometrymediumDefine sin, cos and tan of an acute angle in a right-angled triangle.
Reveal answer ↓
What it is
In a right triangle, the trigonometric ratios sine, cosine and tangent relate an angle to pairs of its sides.
Answer
In a right-angled triangle, for an acute angle A: the sine of A is the ratio of the side opposite to A to the hypotenuse (sin A = opposite/hypotenuse); the cosine of A is the ratio of the side adjacent to A to the hypotenuse (cos A = adjacent/hypotenuse); and the tangent of A is the ratio of the side opposite to A to the side adjacent to A (tan A = opposite/adjacent). Also, tan A = sin A / cos A. The reciprocals are cosec A, sec A and cot A.
sin = opp/hyp ; cos = adj/hyp ; tan = opp/adj
- •sin A = opposite / hypotenuse
- •cos A = adjacent / hypotenuse
- •tan A = opposite / adjacent
- •tan A = sin A / cos A
Why learn this
They connect angles to side lengths and are the basis of all trigonometry.
💡 Memory trick
SOH-CAH-TOA: sin = Opp/Hyp, cos = Adj/Hyp, tan = Opp/Adj.
MathematicsIntroduction to TrigonometrymediumEvaluate sin 30 + cos 60 and tan 45.
Reveal answer ↓
What it is
The trig ratios of 0, 30, 45, 60 and 90 degrees have fixed standard values.
Answer
Using the standard values: sin 30 = 1/2 and cos 60 = 1/2, so sin 30 + cos 60 = 1/2 + 1/2 = 1. Also tan 45 = 1. So sin 30 + cos 60 = 1 and tan 45 = 1.
sin 30 = 1/2 ; cos 60 = 1/2 ; tan 45 = 1
- •sin 30 = 1/2, cos 60 = 1/2
- •sin 30 + cos 60 = 1
- •tan 45 = 1
- •Learn the standard-angle table
Why learn this
They are used constantly in solving triangles and applications.
💡 Memory trick
sin: 0, 1/2, 1/sqrt2, sqrt3/2, 1 for 0, 30, 45, 60, 90. cos is the reverse.
MathematicsIntroduction to TrigonometrymediumState the three trigonometric identities. If sin A = 3/5, find cos A.
Reveal answer ↓
What it is
The fundamental trigonometric identity is sin^2 A + cos^2 A = 1.
Answer
The three fundamental trigonometric identities are: sin^2 A + cos^2 A = 1; 1 + tan^2 A = sec^2 A; and 1 + cot^2 A = cosec^2 A. Given sin A = 3/5, use sin^2 A + cos^2 A = 1: cos^2 A = 1 - sin^2 A = 1 - (3/5)^2 = 1 - 9/25 = 16/25. So cos A = sqrt(16/25) = 4/5 (taking the positive value for an acute angle).
sin^2 A + cos^2 A = 1
- •sin^2 A + cos^2 A = 1
- •1 + tan^2 A = sec^2 A
- •1 + cot^2 A = cosec^2 A
- •cos A = 4/5 when sin A = 3/5
Why learn this
It lets us find one ratio from another and simplify expressions.
💡 Memory trick
sin^2 + cos^2 = 1 always. The other two: 1 + tan^2 = sec^2 ; 1 + cot^2 = cosec^2.
MathematicsSome Applications of TrigonometrymediumThe angle of elevation of the top of a tower from a point 30 m away is 45 degrees. Find the height of the tower.
Reveal answer ↓
What it is
Trigonometry finds unknown heights and distances using an angle of elevation or depression.
Answer
Let the height of the tower be h. The point is 30 m from the base, and the angle of elevation of the top is 45 degrees. Using tan(angle) = opposite/adjacent = height/distance: tan 45 = h/30. Since tan 45 = 1, we get 1 = h/30, so h = 30 m. Therefore the height of the tower is 30 metres.
tan(angle of elevation) = height / distance
- •tan(angle) = height / distance
- •tan 45 = h/30
- •tan 45 = 1
- •Height h = 30 m
Why learn this
It measures tall or far objects without physically reaching them.
💡 Memory trick
Angle of elevation looks up; use tan(angle) = height / distance.
MathematicsCirclesmediumWhat is a tangent to a circle? State the relationship between a tangent and the radius at the point of contact.
Reveal answer ↓
What it is
A tangent to a circle touches it at exactly one point and is perpendicular to the radius at that point.
Answer
A tangent to a circle is a straight line that touches the circle at exactly one point, called the point of contact. A key theorem states that the tangent at any point of a circle is perpendicular to the radius drawn to the point of contact; that is, the radius and the tangent make a right angle (90 degrees) at the point of contact. A line that cuts the circle at two points is called a secant, not a tangent.
Radius is perpendicular to the tangent at the point of contact
- •Tangent touches the circle at one point
- •Point of contact
- •Tangent is perpendicular to the radius there
- •A secant cuts the circle at two points
Why learn this
It is the key property used in all tangent-related problems.
💡 Memory trick
Radius meets tangent at 90 degrees at the point of contact.
MathematicsCirclesmediumHow many tangents can be drawn from an external point to a circle, and what is special about them?
Reveal answer ↓
What it is
The lengths of the two tangents drawn from an external point to a circle are equal.
Answer
From a point lying outside a circle, exactly two tangents can be drawn to the circle. A key theorem states that the lengths of these two tangents drawn from an external point to a circle are equal. From a point on the circle, only one tangent can be drawn, and from a point inside the circle, no tangent can be drawn. These equal-tangent lengths are used in many geometry proofs and constructions.
Tangents from an external point are equal in length
- •Two tangents from an external point
- •The two tangent lengths are equal
- •One tangent from a point on the circle
- •No tangent from a point inside
Why learn this
It is used to find tangent lengths and to prove results about circles.
💡 Memory trick
From one outside point, the two tangents to a circle are always equal in length.
MathematicsAreas Related to CirclesmediumFind the area of a sector of a circle of radius 7 cm with a central angle of 90 degrees. (Take pi = 22/7.)
Reveal answer ↓
What it is
The area of a sector of a circle is a fraction of the whole circle's area, proportional to its central angle.
Answer
The area of a sector = (central angle / 360) x pi r^2 = (90/360) x (22/7) x 7^2 = (1/4) x (22/7) x 49 = (1/4) x 22 x 7 = (1/4) x 154 = 38.5 cm^2. So the area of the sector is 38.5 square centimetres.
Area of sector = (angle/360) x pi r^2
- •Area of sector = (angle/360) x pi r^2
- •= (90/360) x (22/7) x 49
- •= (1/4) x 154
- •Area = 38.5 cm^2
Why learn this
It is used to find areas of pie slices, fan shapes and portions of circular fields.
💡 Memory trick
Area of sector = (angle/360) x pi r^2.
MathematicsStatisticsmediumThe class marks and frequencies of a data set are: (10, 2), (20, 3), (30, 5). Find the mean.
Reveal answer ↓
What it is
The mean of grouped data is found using the class marks and their frequencies.
Answer
For grouped data, the mean = sum of (frequency x class mark) / sum of frequencies. Compute the products: 2 x 10 = 20, 3 x 20 = 60, 5 x 30 = 150. Sum of products = 20 + 60 + 150 = 230. Sum of frequencies = 2 + 3 + 5 = 10. Mean = 230 / 10 = 23. So the mean of the data is 23.
Mean = sum(fi xi) / sum(fi)
- •Mean = sum(f x x) / sum(f)
- •Products: 20, 60, 150; sum = 230
- •Total frequency = 10
- •Mean = 230/10 = 23
Why learn this
It gives a single average value that represents grouped data.
💡 Memory trick
Mean = sum of (frequency x class mark) / sum of frequencies.
MathematicsProbabilityeasyA die is rolled once. Find the probability of getting an even number.
Reveal answer ↓
What it is
Theoretical probability of an event is the number of favourable outcomes divided by the total number of equally likely outcomes.
Answer
When a die is rolled, the total number of equally likely outcomes is 6 (the numbers 1 to 6). The even numbers are 2, 4 and 6, so the number of favourable outcomes is 3. Therefore the probability of getting an even number = favourable outcomes / total outcomes = 3/6 = 1/2. So the probability is 1/2.
P(E) = number of favourable outcomes / total number of outcomes
- •P(E) = favourable / total outcomes
- •Total outcomes = 6
- •Even numbers: 2, 4, 6 (3 outcomes)
- •P(even) = 3/6 = 1/2
Why learn this
It predicts the chance of an event in games and experiments.
💡 Memory trick
P(E) = favourable outcomes / total outcomes, always between 0 and 1.
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