The whole General Aptitude formula sheet, to print

A past action with a present result, or a period that runs up to now. Never with a finished past time (yesterday, ago, last year, in 2019).

An action that began in the past and is still going on: 'She has been teaching here since June' / 'for three months'. It is also used for an activity that has only just stopped, leaving a visible result: 'The roads are wet; it has been raining.' Use since or for here, not from.

Marks an action completed before another past action or a stated past time: 'The train had left when we reached the platform.' Not for a lone past event: 'I went there yesterday', not 'I had gone there yesterday'.

An action that will be complete before a stated future point: 'By March, the team will have tested every unit.'

These clauses (also before, after, once) take the present simple, or the present perfect for an action that will be complete, even for future time; 'will' goes only in the main clause: 'I will call you when I reach the station', not 'when I will reach the station'; 'You may leave once you have signed the register.' The rule covers clauses of time and condition only. A when- or if-clause that is the object of a verb such as know, ask, tell or wonder is a noun clause and takes 'will' as usual: 'Nobody knows when the results will be declared.' 'If you will' is also correct where 'will' means willingness, as in a polite request: 'If you will wait here, I will fetch the form.'

A general truth or a rule that always holds: 'If you mix blue and yellow paint, you get green.'

A real, possible future: 'If the supply fails, the backup will start.'

An unreal present or an unlikely future. Formal English uses 'were' for every person: 'If I were the organiser, I would start earlier.'

An unreal past. 'Would have' never goes in the if-clause. Inverted form without 'if': 'Had I known, I would have helped.' A mixed conditional joins an unreal past condition to a present result: 'If I had taken that job, I would be living in Pune now.'

'She said that she was tired.' In reported speech after a past reporting verb, each tense moves one step back: is → was, is doing → was doing, has done → had done, did → had done (often left as did when the order of events is clear), will → would, can → could, may → might. A universal truth keeps the present: 'The teacher said that light travels faster than sound.' After a present or future reporting verb nothing shifts: 'She says that she is tired.'

Only transitive verbs have a passive. Happen, occur, arrive, die, seem and other intransitive verbs cannot be made passive: 'The accident happened', never 'was happened'. Not every transitive verb has one either: have (in the sense of own), resemble, lack and suit are not normally passive ('Ravi resembles his father', never 'His father is resembled by Ravi').

Also for plain 'X or Y'. The verb agrees with the nearer subject: 'Neither the lecturer nor the students were informed.' 'Either the students or the lecturer is responsible.'

Also 'in addition to', 'accompanied by', 'like', 'including'. 'The captain, along with his players, was felicitated.' Of the joining words, only 'and' makes a plural subject, and even 'and' does not when the two nouns name one person or one idea (see the list below).

'One of the valves is faulty.' But in 'one of the + plural noun + who/that + verb', the relative clause refers to the plural noun, so its verb is plural: 'She is one of the engineers who have designed the dam.' With 'the only one of ... who', it is singular again.

'The number of applicants has doubled.' 'A number of applicants have complained.'

'Each of / Either of / Neither of + plural noun' is also singular in formal English: 'Neither of the answers is correct.' Exception: 'each' AFTER a plural subject keeps the verb plural: 'The boys each have a locker.'

Articles, possessives and numbers come first. 'a beautiful old Kashmiri woollen shawl' (opinion, age, origin, material); 'a large round wooden table' (size, shape, material).

These take 'to', never 'than', and never 'more' in front ('more superior' and 'more preferable' are wrong): 'She is senior to me by two years.' Likewise the verb 'prefer X to Y': 'He prefers trains to buses.' With infinitives, use 'rather than': 'He prefers to walk rather than wait.'

Each half stands directly before matching words: noun with noun, verb with verb. 'She is not only a singer but also a dancer.' 'He not only designs circuits but also tests them.' Also whether ... or, neither ... nor.

Inverted order after the negative opener. 'No sooner had the lecture begun than the lights went out.' 'Hardly had we sat down when the bell rang.' 'Before' may replace 'when' after hardly and scarcely, and the past simple form 'No sooner did + subject + V1 ... than' is also correct: 'No sooner did the lecture begin than the lights went out.' Never 'No sooner ... when' or 'Hardly ... than'.

Here $\leftrightarrow$ is read 'means exactly the same as'. For 'can be inferred', 'is implied' and 'must be true', an option passes only if it holds in every situation in which all the passage's sentences hold. 'Probably true' and 'true in real life' do not pass. Ordinary word meanings may be used in applying the test; facts about the topic that the passage does not give may not.

Of the three standard rearrangements of an if-then statement (converse, inverse, contrapositive), only the contrapositive keeps the meaning. The converse (if B then A) and the inverse (if not A then not B) do NOT follow; they are equivalent to each other, not to the original.

The part that follows 'if' is sufficient: it guarantees the other part. 'Only if', below, points the other way. 'A if and only if B' asserts both directions: if B then A, and if A then B.

B is necessary for A. Knowing that B holds tells you nothing about A; knowing that B fails tells you that A fails.

The group named after 'only' becomes the one that every member of the other group belongs to. It does not say that every X is a Y.

Replace 'unless' with 'if not'. Its contrapositive, 'if not A then B', is the same statement. On its own, the sentence says nothing about the case where B occurs.

'Some' means at least one and possibly all. It does not by itself mean 'not all'. 'At least' and 'at most' are read inclusively.

The opposite of 'all' is 'not all', not 'none'. You need this when you apply the negation test to an option that contains 'all' or 'every'.

The opposite of 'at least one' is 'none'. Read the other way, the opposite of 'no X is a Y' is 'at least one X is a Y'.

Denying 'both' says only that at least one of the two fails. It does not say that both fail.

Denying 'either' (inclusive or) says that both fail. This is what 'neither A nor B' means.

Here $\to$ is read 'leads to'. State the mildest opposite of the option (the opposite of 'all' is 'not all'). If the conclusion then no longer follows from the evidence, the option is an assumption the argument needs. If the argument survives, it is not an assumption.

For parts of one whole (a pie, a stacked bar, a column of a table with its total). The shares of all the parts add to 100%, apart from rounding in the display.

The base is the old (earlier) value unless the question names another. A positive result is an increase and a negative one a decrease. For a quantity that cannot go below zero, a decrease can never exceed 100%; an increase can (a rise of 100% is a doubling).

For a change in a quantity that is itself a percentage (a share, a pass rate, an interest rate). From 20% to 25% is a rise of 5 percentage points, but a 25% increase in the rate, because $\frac{25-20}{20}=0.25$.

For positive $A$ and $B$ with $A>B$. The base is the quantity after 'than'. The two percentages differ: with $A=125$ and $B=100$, $A$ is 25% more than $B$ but $B$ is 20% less than $A$.

The first holds for any $r\ge 0$; the second for $0\le r<100$. A 25% rise is undone by a 20% fall, and a 20% fall needs a 25% rise to recover.

Each change acts on the value left by the previous one. Take a decrease as negative. The order of the changes does not affect the final value.

For changes of $a\%$ then $b\%$, decreases negative. A rise of 20% followed by a fall of 20% gives $20-20-\frac{400}{100}=-4$, a net fall of 4%. The same result gives the change in a product: price up 10% and quantity sold down 10% changes revenue by $10-10-1=-1$, a 1% fall.

When the value after a change of $r\%$ is known ($r$ negative for a decrease, with $r>-100$). A value of 132 after a 10% rise was $\frac{132}{1.1}=120$, not $132-13.2=118.8$.

The base is the expenditure (the cost) unless the question names another; a negative result is a loss. Rearranged, Income $=$ Expenditure $\times \left(1+\frac{p}{100}\right)$ for a profit of $p\%$. An income of Rs. 69 lakh on an expenditure of Rs. 60 lakh is a profit of $\frac{9}{60}\times 100=15\%$.

For nested data: if 40% of a firm's staff are engineers and 25% of the engineers are women, women engineers are $\frac{40\times 25}{100}=10\%$ of the staff.

When $A$ and $B$ are $p_A\%$ and $p_B\%$ of the SAME total, their ratio needs no total. It fails in general across two different totals (two pies, two years): convert to amounts first.

Each part is its term of the ratio over the sum of the terms. Rs. 9 lakh divided in the ratio $2:3:4$ gives Rs. 2 lakh, Rs. 3 lakh and Rs. 4 lakh.

$p$ is the part's percentage share. 1% corresponds to $3.6^{\circ }$, 10% to $36^{\circ }$, 25% to $90^{\circ }$ and 50% to $180^{\circ }$.

Needs the total of THAT pie. The angles of all the sectors add to $360^{\circ }$ and their percentages to 100.

Only within one pie. Across two pies, convert each sector to an amount using its own pie's total first. The difference between two sectors is $\frac{{\theta }_1-{\theta }_2}{360^{\circ }}\times \text{Total}$.

The bottom segment's lower edge is zero. The upper edge of any higher segment is a running total, not that segment's value. The top of the whole stack is the total of all segments.

The bar totals must come from elsewhere (a table or a label). Without them, segments of different bars can be compared only as shares.

In units per period (for example tonnes per year). On a line graph it is the slope of the segment joining the two points, computed from the axis values. How steep a segment looks depends on the scales chosen for the axes.

The percentage change over one period, on the earlier value as base. A falling growth rate does not mean a falling value: a line that keeps rising but flattens has positive, shrinking growth rates.

Growth of $r\%$ in each of $n$ periods, each period's growth applied to the value at the start of that period. For a steady decline, $r$ is negative.

For positive $V_0$ and $V_n$. $n$ is the number of periods BETWEEN the two values: from 2020 to 2025, $n=5$, not 6. CAGR is the constant rate that would carry $V_0$ to $V_n$; it ignores the path in between.

The CAGR comes from the geometric mean of the growth factors. Provided every yearly rate is above $-100\%$ (no value falls to zero), it never exceeds the arithmetic mean of the yearly rates $g_1,\dots ,g_n$, and the two are equal only when every yearly rate is the same.

The base period has index 100. The difference $I_2-I_1$ is a percentage change only when the earlier period is the base ($I_1=100$). Indices of 125 and 150 differ by 25 points but mean a rise of $\frac{25}{125}\times 100=20\%$.

When a graph or table shows running totals $C_t$ ('total sales up to month $t$'), the value for period $t$ is the difference of consecutive readings. A running total of non-negative amounts never falls, so a line that falls anywhere is not a cumulative line of such amounts.

Each item counts equally. Rearranged, sum of values $=n\bar{x}$, which recovers a missing value when the mean of all the items is given.

The weights $w_i$ are the group sizes (students, units sold, workers). With positive weights, the weighted mean always lies between the smallest and largest $x_i$, pulled towards the values with larger weights. For a frequency table the weights are the frequencies; for grouped classes, taking each class mid-point as $x_i$ gives an estimate of the mean, not its exact value.

$p_i$ is group $i$'s percentage and $n_i$ its size. The plain mean of the $p_i$ equals $p_{\text{overall}}$ whenever all the $n_i$ are equal; otherwise it generally differs.

The combined mean divides the gap between the two group values in the inverse ratio of the weights. The second form finds the ratio of group sizes from a known combined mean and needs $x_1<{\bar{x}}_w<x_2$. Groups with means 60 and 80 in sizes $3:1$ have combined mean $60+\frac{1}{4}(80-60)=65$.

An overall rate (revenue per unit sold, output per worker, marks per student) is total divided by total. In general the mean of the individual ratios is a different number; the two agree when all the $b_i$ are equal.

A bar of height 6 per unit over the class 20 to 30 stands for $6\times 10=60$ items. Comparing the heights of bars of different widths compares densities, not frequencies.

Holds only when the plotted origin really is $(0,0)$. If either axis is offset (starts above zero), compute $\frac{y}{x}$ from the values read off instead.

Both in the same unit. The horizontal distance is the ground distance, so convert the map distance with the scale first. Often written as '1 in $N$', meaning a rise of 1 unit for every $N$ units horizontally.

Convert units last: 1 km $=1,000$ m $=1,00,000$ cm.

Areas scale with the SQUARE of the linear scale. On a $1:50,000$ map, 1 cm represents 0.5 km, so 1 cm² represents $(0.5{)}^2=0.25$ km².

Scale both ratios so that the common term $b$ has the same value ($qr$) in each.

The shares add back to $N$. If only the difference between two shares is given, divide it by the difference of their ratio terms to get the value of one ratio unit.

Needs $b,d\ne 0$, and $a\ne b$ for the second form. The mean proportional of two positive numbers $a$ and $b$ is $\sqrt{ab}$; for $a\ne 0$, the third proportional to $a$ and $b$ is $\frac{b^2}{a}$ and the fourth proportional to $a$, $b$ and $c$ is $\frac{bc}{a}$ (so that $a:b=c:\frac{bc}{a}$).

Positive for an increase, negative for a decrease. 'A is $x\%$ less than B' uses B as the base; the matching statement 'B is $y\%$ more than A' uses A, so $y=\frac{100x}{100-x}$ for $0<x<100$, which is larger than $x$.

Each change is taken on the value left by the previous step; write an increase as positive and a decrease as negative. For three or more changes multiply the factors: $\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)\left(1+\frac{c}{100}\right)$.

Expenditure = price × consumption. If instead the price falls by $r\%$ ($r<100$), consumption can rise by $\frac{r}{100-r}\times 100\%$ for the same expenditure. The same algebra covers any product held fixed (speed × time for a fixed distance).

$p$ is the profit percentage on CP (negative for a loss); $d$ is the discount percentage on MP. Successive discounts combine like successive changes, so 20% and 10% make a single discount of $28\%$, not $30\%$.

$P$ principal, $R$ rate in per cent per annum, $T$ time in years. Interest is earned on the principal only, so the amount grows by the same sum every year (an AP).

Interest is added to the principal at the end of each period, so the amount grows by a constant factor (a GP). $CI=A-P$. For different rates in successive years multiply the factors $\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\cdots$. The same formula gives population growth at $R\%$ a year, and a value that depreciates by $R\%$ a year becomes $P{\left(1-\frac{R}{100}\right)}^T$.

Same $P$ and $R$, compounded annually.

Adding $k$ to every value adds $k$ to the average. Replacing one value changes the sum by (new − old), so the average changes by $\frac{\text{new}-\text{old}}{n}$. Work with sums, not averages, when items join or leave a group.

Quantity $q_1$ at price (or concentration) $c_1$ mixed with $q_2$ at $c_2$ gives the mean $m$. It is the weighted-average formula rearranged, so it works for prices, percentages of a component and marks. It also works for average speeds, provided the quantities are the TIMES spent at each speed, not the distances. Water counts as price 0.

A container is full with $V$ litres of pure liquid A. $x$ litres are drawn off and replaced by liquid B, and this is done $n$ times, with complete mixing each time. Each operation multiplies the amount of A by $\left(1-\frac{x}{V}\right)$, so if the container instead starts with $Q_0$ litres of A in a mixture of total volume $V$, the amount left is $Q_0{\left(1-\frac{x}{V}\right)}^n$.

Put every quantity in one unit system before calculating. For a fixed distance, time is inversely proportional to speed.

Over equal distances at speeds $u$ and $v$ the average speed is their harmonic mean, which is less than the arithmetic mean unless $u=v$.

Time to close a gap or to pass = (length to be covered) ÷ (relative speed). A train of length $L$ passes a pole or a standing person in $\frac{L}{v}$ and a platform of length $P$ in $\frac{L+P}{v}$; two trains pass each other in $\frac{L_1+L_2}{v_{\text{rel}}}$.

$b$ is the speed in still water, $s$ the speed of the stream, $d$ the downstream and $u$ the upstream speed ($b>s$ for the boat to make headway upstream).

Rates add when workers act together; an outlet pipe, or anyone undoing the job, has a negative rate. Taking the total work as the LCM of the given times keeps every rate a whole number of units per day (or hour).

Valid for all real exponents when $a,b>0$; for integer exponents any non-zero $a,b$ will do. A tower is read from the top: $a^{m^n}=a^{(m^n)}$, which is not ${\left(a^m\right)}^n$.

For $a>0$, $\sqrt[q]{a}$ is the positive $q$-th root.

For $a,b\ge 0$, with $a\ne b$ in the second. Multiply by the conjugate to clear a root from the denominator. $\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$.

The logarithm exists only for a positive argument and a positive base other than 1. ${\log }_{a}1=0$ and ${\log }_{a}a=1$ for every valid base. For $a>1$, ${\log }_{a}x$ increases as $x$ increases; for $0<a<1$ it decreases, so taking logarithms to such a base reverses an inequality.

For $x,y>0$. There is no law for ${\log }_a(x+y)$.

Any valid base $b$ may be used; choose 10, $e$, or whichever base makes the numbers simplest. The last form needs $k\ne 0$.

For $x,y>0$. To verify the second, take ${\log }_a$ of both sides: each becomes ${\log }_{a}x\cdot {\log }_{a}y$.

$\lfloor \cdot \rfloor$ is the greatest integer not exceeding the value. For a non-integer $N\ge 1$ the same expression counts the digits before the decimal point. For $0<N<1$, the number of zeros between the decimal point and the first non-zero digit is $\lceil -{\log }_{10}N\rceil -1$.

For an integer $a$ and a positive integer $b$, the quotient $q$ and remainder $r$ are unique. A reported remainder is never negative: a 'remainder of $-1$' on division by 7 is only a working shortcut and means 6.

Replace each number by its remainder before adding or multiplying; remainders of powers follow by repeated multiplication. $a\equiv b(\mathrm{mod}m)$ means $a$ and $b$ leave the same remainder on division by $m$.

For a prime $p$ that does not divide $a$. For the remainder of a large power on division by $p$, reduce the exponent modulo $p-1$.

True for exactly two positive integers, not for three or more. The HCF takes the lowest power of each common prime; the LCM takes the highest power of every prime present.

$p_1,\dots ,p_k$ are distinct primes. $\tau (N)$ counts all positive factors including 1 and $N$; $\sigma (N)$ is their sum. $N$ is a perfect square exactly when $\tau (N)$ is odd. $N$ can be written as a product of two positive factors (order ignored) in $\frac{\tau (N)}{2}$ ways, or $\frac{\tau (N)+1}{2}$ ways when $N$ is a perfect square. Example: $720=2^4\cdot 3^2\cdot 5$ has $5\times 3\times 2=30$ factors, so it is a product of two factors in 15 ways.

Stop when $p^k>n$. The number of trailing zeros of $n!$ is this sum with $p=5$, since there are always at least as many 2s as 5s. Example: $100!$ ends in $20+4=24$ zeros.

$a$ first term, $d$ common difference, $l$ last term. The number of terms from $a$ to $l$ is $\frac{l-a}{d}+1$. Take three terms in AP as $a-d,a,a+d$ so that their sum is simply $3a$. For any sequence, not only an AP, $t_1=S_1$ and $t_n=S_n-S_{n-1}$ for $n\ge 2$.

If $r=1$, $S_n=na$. An infinite GP has a finite sum only when $|r|<1$. Take three terms in GP as $\frac{a}{r},a,ar$ so that their product is $a^3$.

Equality holds only when $a=b$. A harmonic progression is a sequence whose reciprocals form an AP. AM ≥ GM also holds for $n$ positive numbers: with the sum fixed, the product is greatest when the numbers are equal, and with the product fixed, the sum is least when they are equal.

Also: the sum of the first $n$ odd numbers is $n^2$ and the sum of the first $n$ even numbers is $n(n+1)$.

Split each term into a difference so that neighbouring pieces cancel; only the first and last pieces survive.

$0\le r\le n$, no repetition. ${}^n{P}_n=n!$, ${}^n{C}_0={}^n{C}_n=1$ and $0!=1$.

If ${}^n{C}_x={}^n{C}_y$ then $x=y$ or $x+y=n$. The number of ways to select at least one item from $n$ distinct items is $2^n-1$.

For a positive integer $n$. There are $n+1$ terms and $T_{r+1}$ is the general term. Putting $x=y=1$ gives ${\sum }_{r=0}^n{}^n{C}_r=2^n$.

Anything counted twice is subtracted once, and so on. Among $1,2,\dots ,N$ exactly $\lfloor \frac{N}{k}\rfloor$ numbers are multiples of $k$. Example: from 1 to 100, the numbers divisible by 2 or 3 number $50+33-16=67$ (16 being the multiples of 6).

$r$ positions each filled from $n$ distinct kinds in unlimited supply give $n^r$ ordered arrangements. When only how many of each kind are chosen matters, the number of selections of $r$ items from $n$ kinds is ${}^{n+r-1}{C}_r$.

$n$ objects of which $p_1$ are alike of one kind, $p_2$ alike of another, and so on ($p_1+p_2+\cdots +p_k=n$; an object with no duplicate counts as a kind with $p=1$). Selecting at least one item from $p$ identical items of one kind, $q$ of a second kind and $t$ of a third can be done in $(p+1)(q+1)(t+1)-1$ ways.

For $n$ distinct objects only the relative order counts. The halved count ($n\ge 3$) applies to necklaces, garlands and rings of beads that can be turned over, never to people seated round a table. If the seats are numbered, or one position is otherwise marked, the arrangement is linear: $n!$.

Equivalently, $x_1+x_2+\cdots +x_r=n$ has ${}^{n+r-1}{C}_{r-1}$ solutions in non-negative integers and ${}^{n-1}{C}_{r-1}$ in positive integers (needs $n\ge r$). The count $r^n$ for distinct objects allows empty boxes; with no box empty it is $r^n-{}^r{C}_1(r-1{)}^n+{}^r{C}_2(r-2{)}^n-\cdots$, by inclusion and exclusion. Example: 4 distinct objects into 3 boxes with none empty: $81-3\times 16+3\times 1=36$.

Lines and triangles are for $n$ points with no three collinear. If $m$ of them ($m\ge 3$) lie on one line and no other three are collinear, subtract ${}^m{C}_2-1$ lines and ${}^m{C}_3$ triangles. Grid paths counts the shortest routes made of $m$ unit steps right and $n$ unit steps up in any order.

Good when $|nx|$ is a few per cent; the neglected term is about $\frac{n(n-1)}{2}{x}^2$. Example: $1.02^5\approx 1.10$ (exact value $1.1041$).

Example: $\sqrt{1604}\approx 40+\frac{4}{80}=40.05$; indeed $40.05^2=1604.0025$.

For small relative changes. A 2% rise in length and a 3% rise in width raise the area by about 5% (exactly $5.06\%$).

Any simple polygon with $n$ sides (sides do not cross), convex or not: $180^{\circ }$ for a triangle, $360^{\circ }$ for a quadrilateral, $540^{\circ }$ for a pentagon.

One exterior angle taken at each vertex of a convex polygon; the sum is $360^{\circ }$ whatever the value of $n$. At every vertex, interior angle $+$ exterior angle $=180^{\circ }$.

Regular polygon only (all sides equal and all angles equal).

A diagonal joins two vertices that are not neighbours. Each vertex joins $n-3$ others by a diagonal (not itself and not its two neighbours); dividing by 2 stops each diagonal being counted twice.

$c$ is the side opposite angle $C$. Apply the test to the longest side: the triangle is acute, right or obtuse according to the angle opposite that side.

$h$ is the perpendicular height onto the base $b$ (in an obtuse triangle it can fall outside the triangle, on the extended base); $C$ is the angle between sides $a$ and $b$.

Any triangle whose three sides are known. If any of $s-a$, $s-b$, $s-c$ is zero or negative, the three lengths do not form a triangle.

All four centres coincide at the centroid, which divides each height in the ratio $2:1$ from the vertex. So $R=\frac{2}{3}h$ and $r=\frac{1}{3}h$, giving $R=2r$ and $h=3r=\frac{3}{2}R$.

Any triangle: $r$ is the radius of the incircle (touches all three sides), $R$ the radius of the circumcircle (passes through all three vertices) and $s$ the semi-perimeter.

$h_c$ is the altitude from the right angle to the hypotenuse. If it cuts the hypotenuse into parts $p$ (next to leg $a$) and $q$ (next to leg $b$), then $h_c^2=pq$, $a^2=pc$ and $b^2=qc$.

$AD$ is the internal bisector of angle $A$, meeting $BC$ at $D$: it divides the opposite side in the ratio of the two sides that form the angle, and the second relation gives its length. When $AB\ne AC$, the external bisector of angle $A$ meets $BC$ produced at a point $E$ with $BE/EC=AB/AC$.

Parallelogram: $\theta$ is the angle between sides $a$ and $b$, and $h$ the height onto base $b$. Trapezium: $a$ and $b$ are the parallel sides and $h$ the distance between them. Any quadrilateral: $\theta$ is the angle between the diagonals ($90^{\circ }$ for a rhombus, a square or a kite).

A path of width $w$ running all round the outside or the inside of a rectangular plot $l\times b$. The crossing roads are two roads of width $w$ through the plot, one parallel to each pair of sides; the square where they cross is counted once, which is why $w^2$ is subtracted. A circular path of width $w$ round the outside of a circle of radius $r$ has area $\pi \left[(r+w{)}^2-r^2\right]=\pi w(2r+w)$.

Radius $r$, diameter $d$. A wheel of radius $r$ rolling without slipping moves forward $2\pi r$ in one complete revolution.

$\theta$ is the angle at the centre, in degrees. With $\theta$ in radians the same results read $l=r\theta$ and $A=\frac{1}{2}{r}^2\theta$.

Minor segment cut off by a chord that subtends $\theta$ (degrees) at the centre: sector minus triangle. $d$ is the perpendicular distance of the chord from the centre. Major segment $=\pi r^2-$ minor segment.

$PT$: tangent from a point at distance $d$ from the centre of a circle of radius $r$. For two circles of radii $r_1,r_2$ with centres $d$ apart: a direct common tangent has both circles on the same side of it and exists only if $d\ge |r_1-r_2|$; a transverse common tangent passes between them and exists only if $d\ge r_1+r_2$. Each length is measured between the two points of contact.

Between $A(x_1,y_1)$ and $B(x_2,y_2)$. The distance of $(x,y)$ from the origin is $\sqrt{x^2+y^2}$.

$P$ divides $AB$ in the ratio $m:n$, that is $AP:PB=m:n$, from $A(x_1,y_1)$ towards $B(x_2,y_2)$. External division needs $m\ne n$. The ratio $1:1$ gives the mid-point.

Triangle $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ with $a=BC$, $b=CA$, $c=AB$: in the incentre, each vertex is weighted by the length of the side opposite it.

Take the absolute value: the expression inside is negative when the vertices are listed clockwise. $\Delta =0$ means the three points are collinear. For a polygon, list the vertices in order round the boundary and use the shoelace sum $\frac{1}{2}\left|\sum (x_i{y}_{i+1}-x_{i+1}{y}_i)\right|$, returning to the first vertex at the end.