Chapter 8 — Every Formula You Have to Know

The SAT prints a page of formulas at the start of every maths module. The ACT prints none. Everything you need has to be in your head when you sit down.

This chapter is that list, with the reason each formula is true beside it. Reasons matter: a formula you understand survives a nervous morning, and one you have only memorised often does not. Where a formula can be rebuilt from a simpler one in five seconds, that is said too — it is one fewer thing to carry.

Work through it a section at a time rather than in one sitting, and come back to it between the other chapters. These are the cheapest marks in the book.

The list

Lines and coordinates

formula	why
slope $= \dfrac{y_2 - y_1}{x_2 - x_1}$	rise over run — how far up for each step across
$y = mx + b$	$m$ is the slope, $b$ the height where $x = 0$
distance $= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$	Pythagoras on the horizontal and vertical gaps
midpoint $= \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$	the average of each coordinate
parallel lines	equal slopes
perpendicular lines	slopes multiply to $-1$ (flip it and change the sign)

Do not confuse them: midpoint adds; distance and slope subtract.

Triangles and angles

formula	why
angles in a triangle $= 180°$	tear off the three corners and they form a straight line
angles in an $n$ -sided polygon $= (n-2) \times 180°$	it splits into $n - 2$ triangles
exterior angles of any polygon $= 360°$	one full turn walking round it
exterior angle of a triangle $=$ the two remote interior angles	both equal $180°$ minus the third angle
area of a triangle $= \tfrac{1}{2}bh$	half of the rectangle that encloses it
Pythagoras: $a^2 + b^2 = c^2$	right triangles only, $c$ the hypotenuse

Pythagorean triples worth recognising on sight, because they save all the arithmetic: 3-4-5, 5-12-13, 8-15-17, and any multiple of them (6-8-10, 9-12-15, 10-24-26).

Special right triangles:

$30°\text{–}60°\text{–}90° \; \longrightarrow \; 1 : \sqrt{3} : 2$ $45°\text{–}45°\text{–}90° \; \longrightarrow \; 1 : 1 : \sqrt{2}$

In both, the smallest side faces the smallest angle.

Circles

formula	why
circumference $= 2\pi r$	$\pi$ is defined as circumference over diameter
area $= \pi r^2$
arc length $= \dfrac{\theta}{360} \times 2\pi r$	the same fraction of the way round
sector area $= \dfrac{\theta}{360} \times \pi r^2$	the same fraction of the whole slice
$(x-h)^2 + (y-k)^2 = r^2$	centre $(h,k)\mathclose{,}$ radius $r$ — it is the distance formula

The circle equation is worth seeing that way: it says "every point is a distance $r$ from the centre", which is what a circle is.

Solids

formula	why
box: $V = lwh$
any prism or cylinder: $V = (\text{cross-section}) \times \text{length}$	stacked copies of the end face
cylinder: $V = \pi r^2 h$	the prism rule with a circular end
sphere: $V = \tfrac{4}{3}\pi r^3$
cone: $V = \tfrac{1}{3}\pi r^2 h$	exactly a third of the cylinder around it
box surface area $= 2(lw + lh + wh)$	six faces in three matching pairs

Scaling, which is worth more than any single volume formula: multiply every length by $k$ and areas multiply by $k^2\mathclose{,}$ volumes by $k^3\mathclose{.}$

Algebra

formula	why
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$	solves any quadratic
discriminant $b^2 - 4ac$	positive → 2 roots, zero → 1, negative → none
$(a+b)^2 = a^2 + 2ab + b^2$	there is always a middle term
$(a-b)^2 = a^2 - 2ab + b^2$
$a^2 - b^2 = (a-b)(a+b)$	the middle terms cancel
vertex of $y = ax^2+bx+c$ at $x = -\dfrac{b}{2a}$	halfway between the two roots

Exponents, logs and sequences

formula	why
$x^a x^b = x^{a+b}\mathclose{,}$ $\dfrac{x^a}{x^b} = x^{a-b}\mathclose{,}$ $(x^a)^b = x^{ab}$	count the copies
$x^0 = 1\mathclose{,}$ $x^{-a} = \dfrac{1}{x^a}\mathclose{,}$ $x^{m/n} = \sqrt[n]{x^m}$
$\log_b y = x$ means $b^x = y$	a log is an exponent
$\log(mn) = \log m + \log n$	multiplying inside adds outside
arithmetic: $a + (n-1)d$	$n-1$ steps to reach the $n$ th term
geometric: $a \cdot r^{\,n-1}$	same reason, multiplying instead

Trigonometry

formula	why
SOH-CAH-TOA	sin = opp/hyp, cos = adj/hyp, tan = opp/adj
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$	divide the first two
$\sin^2\theta + \cos^2\theta = 1$	Pythagoras, divided through by $c^2$
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$	law of sines
$c^2 = a^2 + b^2 - 2ab\cos C$	law of cosines — Pythagoras plus a correction
$\pi$ radians $= 180°$
$y = a\sin(bx)\mathclose{:}$ amplitude $\|a\|\mathclose{,}$ period $\dfrac{2\pi}{b}$	outside stretches, inside squeezes

Statistics

formula	why
mean $=\dfrac{\text{total}}{\text{count}}\mathclose{,}$ so total $=$ mean $\times$ count	the rearranged form is the useful one
median	the middle value after sorting
range	largest minus smallest
$P(\text{not }A) = 1 - P(A)$	everything either happens or does not

Section 1 — Using Them From Memory

The questions below give you nothing. Each one needs a formula from the list above, applied once. If you have to look one up, that is the formula to keep practising.

Q1Basic

Topic: A formula you must know — the area of a triangle

A triangular sail has a base measuring 14 units along its foot and a perpendicular height of 9 units from that base to the opposite corner. What is its area?

A) $23$

B) $31.5$

C) $63$

D) $126$

Show the worked solution

Answer: C

Explanation

$\text{area} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 14 \times 9 = 63$

The reason for the half: a triangle is exactly half of the rectangle that encloses it, with the same base and height. Picturing that rectangle is a reliable way to remember which formula carries the $\frac{1}{2}\mathclose{.}$

Why each wrong option is wrong:

A, $23$ — added the two lengths. An area needs two lengths multiplied; adding gives something in the wrong units.
B, $31.5$ — halved twice. The formula carries one half, not two.
D, $126$ — used $bh\mathclose{,}$ the rectangle formula, forgetting the half.

Takeaway: a triangle is half its enclosing rectangle. And the height must be perpendicular to the base — not a slanted side.

Q2Medium

Topic: A formula you must know — the quadratic formula

What are the solutions of $2x^2 + 3x - 2 = 0\mathclose{?}$

F) $- \frac{3}{2}$ and $0$

G) $-1$ and $2$

H) $-2$ and $\frac{1}{2}$

J) $-4$ and $1$

Show the worked solution

Answer: H

Explanation

With $a = 2\mathclose{,}$ $b = 3\mathclose{,}$ $c = -2\mathclose{:}$

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$

$x = \frac{2}{4} = \frac{1}{2} \qquad \text{or} \qquad x = \frac{-8}{4} = -2$

Watch the $-4ac$ term: $c$ is negative, so $-4(2)(-2)$ is $+16\mathclose{,}$ not $-16\mathclose{.}$ That double negative is where the formula usually goes wrong.

Check by factoring, which also works here: $(2x - 1)(x + 2) = 2x^2 + 3x - 2\mathclose{.}$ ✓

Why each wrong option is wrong:

F, $- \frac{3}{2}$ and $0$ — dropped the $-4ac$ term, using $\sqrt{9}$ instead of $\sqrt{25}\mathclose{.}$ Forgetting it entirely, or getting its sign wrong, is the single commonest quadratic-formula error.
G, $-1$ and $2$ — factored as if the leading coefficient were 1, giving $(x+1)(x-2)\mathclose{,}$ which expands to $x^2 - x - 2$ rather than $2x^2 + 3x - 2\mathclose{.}$
J, $-4$ and $1$ — divided by 2 rather than by $2a = 4\mathclose{.}$

Takeaway: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\mathclose{.}$ Substitute $a\mathclose{,}$ $b$ and $c$ with their signs before doing any arithmetic.

Q3Medium

Topic: A formula you must know — the volume of a cone

A conical container has the base radius and the height shown, in units.

What is its volume, in cubic units?

A) $8 \pi$

B) $24 \pi$

C) $72 \pi$

D) $96 \pi$

Show the worked solution

Answer: B

Explanation

$V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times \pi \times 9 \times 8 = 24\pi$

The reason for the third: a cone holds exactly one third of the cylinder with the same base and height. That is worth knowing as a fact in its own right — it means you only really have to remember the cylinder.

Why each wrong option is wrong:

A, $8 \pi$ — forgot to square the radius.
C, $72 \pi$ — gave the cylinder's volume, three times too big.
D, $96 \pi$ — used $\frac{4}{3}\mathclose{,}$ which belongs to the sphere.

Takeaway: cone $= \frac{1}{3} \times$ cylinder. Sphere is $\frac{4}{3}\pi r^3$ and has no height in it — if your sphere formula contains an $h\mathclose{,}$ it is wrong.

Q4Medium

Topic: A formula you must know — perpendicular slopes

A road is to be built perpendicular to an existing one whose equation is $y = \dfrac{2}{5}x + 7\mathclose{.}$ What is the slope of the new road?

F) $- \frac{5}{2}$

G) $- \frac{2}{5}$

H) $\frac{2}{5}$

J) $\frac{5}{2}$

Show the worked solution

Answer: F

Explanation

Perpendicular slopes are negative reciprocals: flip the fraction and change its sign. Both steps, not one.

$\text{flip: } \frac{5}{2} \qquad \text{change sign: } -\frac{5}{2}$

Check with the multiplication test — two perpendicular slopes multiply to $-1\mathclose{:}$

$\frac{2}{5} \times \left(-\frac{5}{2}\right) = -1 \; \checkmark$

The $+7$ is the intercept and plays no part; only the slope matters for perpendicularity.

Why each wrong option is wrong:

H, $\frac{2}{5}$ — that is the parallel slope.
J, $\frac{5}{2}$ and G, $- \frac{2}{5}$ — did one of the two steps.

Takeaway: perpendicular means flip and negate. Multiply your answer by the original slope; if you do not get $-1\mathclose{,}$ you have done only half of it.

Q5Hard

Topic: A formula you must know — the vertex of a parabola

A manufacturer's cost, in thousands of dollars, is modelled by $y = x^2 - 8x + 3\mathclose{,}$ and the production level giving the lowest cost is wanted. At what value of $x$ does the minimum occur?

A) $-8$

B) $-4$

C) $4$

D) $8$

Show the worked solution

Answer: C

Explanation

For $y = ax^2 + bx + c$ the vertex sits at

$x = -\frac{b}{2a}$

Here $a = 1$ and $b = -8\mathclose{:}$

$x = -\frac{-8}{2(1)} = \frac{8}{2} = 4$

The two minus signs cancel, which is exactly where option B goes wrong.

Why the formula is true: a parabola is symmetric, so its vertex lies halfway between its two roots — and the quadratic formula puts those roots at $\frac{-b}{2a}$ plus and minus the same amount. The midpoint is $\frac{-b}{2a}\mathclose{.}$

Check by testing either side: at $x = 3\mathclose{,}$ $y = -12\mathclose{;}$ at $x = 4\mathclose{,}$ $y = -13\mathclose{;}$ at $x = 5\mathclose{,}$ $y = -12\mathclose{.}$ Lowest at 4. ✓

Why each wrong option is wrong:

A, $-8$ — read $b$ off the equation.
B, $-4$ — dropped the minus sign in front of the formula.
D, $8$ — used $b$ without dividing by $2a\mathclose{.}$

Takeaway: $x = -\frac{b}{2a}\mathclose{,}$ and $b$ carries its own sign. Substituting a negative $b$ into a formula that already has a minus gives a positive answer.

Q6Hard

Topic: A formula you must know — the law of cosines

Two sides of a triangular plot measure 5 and 8 metres and the angle between them is $60°\mathclose{.}$ What is the length of the third side?

F) $3$

G) $7$

H) $\sqrt{89}$

J) $13$

Show the worked solution

Answer: G

Explanation

$c^2 = a^2 + b^2 - 2ab\cos C$

With $a = 5\mathclose{,}$ $b = 8\mathclose{,}$ $C = 60°$ and $\cos 60° = \frac{1}{2}\mathclose{:}$

$c^2 = 25 + 64 - 2(5)(8)\left(\tfrac{1}{2}\right) = 89 - 40 = 49$

$c = 7$

Sanity check: the third side must be between $8 - 5 = 3$ and $8 + 5 = 13\mathclose{.}$ It is.

Notice what the formula is doing. Without the last term it would be Pythagoras, giving $\sqrt{89} \approx 9.4\mathclose{.}$ The $-2ab\cos C$ term is the correction for the angle not being $90°$ — and when $C = 90°\mathclose{,}$ $\cos C = 0$ and the correction disappears.

Why each wrong option is wrong:

H, $\sqrt{89}$ — used Pythagoras on a triangle that has no right angle.
J, $13$ and F, $3$ — added or subtracted the sides, which give the absolute limits the third side must lie strictly between.

Takeaway: law of cosines is Pythagoras plus a correction term. Two sides and the angle between them is the signal to use it.

Q7Medium

Topic: A formula you must know — the nth term of a sequence

A savings scheme pays out amounts forming the arithmetic sequence $5, \; 12, \; 19, \; 26, \dots$ What is the 15th term?

A) $75$

B) $98$

C) $103$

D) $110$

Show the worked solution

Answer: C

Explanation

Find the common difference by subtracting:

$12 - 5 = 7 \qquad 19 - 12 = 7 \qquad 26 - 19 = 7$

So $a = 5$ and $d = 7\mathclose{.}$

$a + (n-1)d = 5 + 14 \times 7 = 5 + 98 = 103$

Why $n - 1\mathclose{:}$ you are already standing on the first term, so getting to the 15th takes 14 steps, not 15.

Why each wrong option is wrong:

A, $75$ — multiplied the first term by 15.
B, $98$ — worked out the 14 steps, 98, and forgot to add the term they start from.
D, $110$ — used $a + nd\mathclose{,}$ one difference too many. This is the error the $n-1$ exists to prevent.

Takeaway: $a + (n-1)d$ for arithmetic, $a \cdot r^{\,n-1}$ for geometric. Both have $n - 1\mathclose{,}$ for the same reason: count the gaps, not the terms.

Q8Medium

Topic: A formula you must know — the equation of a circle

A circular region in the $xy$ -plane has its centre at $(-2, 5)$ and a radius of 4 units. Which equation describes this circle?

F) $(x-2)^2 + (y+5)^2 = 16$

G) $(x+2)^2 + (y-5)^2 = 16$

H) $(x+2)^2 + (y-5)^2 = 4$

J) $(x-2)^2 + (y-5)^2 = 16$

Show the worked solution

Answer: G

Explanation

The standard form is

$(x - h)^2 + (y - k)^2 = r^2$

with centre $(h, k)\mathclose{.}$ The formula already contains minus signs, so a negative coordinate produces a plus in the bracket:

$h = -2 \; \Longrightarrow \; (x - (-2))^2 = (x + 2)^2$ $k = 5 \; \Longrightarrow \; (y - 5)^2$

And the right-hand side is $r^2 = 4^2 = 16\mathclose{.}$

$(x+2)^2 + (y-5)^2 = 16$

Check: the point $(2, 5)$ is 4 units right of the centre and should be on the circle. $(2+2)^2 + (5-5)^2 = 16\mathclose{.}$ ✓

Why each wrong option is wrong:

F, $(x-2)^2 + (y+5)^2 = 16$ and J, $(x-2)^2 + (y-5)^2 = 16$ — copied a coordinate's sign straight into the bracket instead of flipping it.
H, $(x+2)^2 + (y-5)^2 = 4$ — put the radius on the right rather than its square.

Takeaway: the signs flip and the right side is $r^2\mathclose{.}$ Write $(x - (-2))$ out in full first if the flip is not automatic yet, then test one point.

Q9Medium

Topic: A formula you must know — the distance formula

What is the distance between the points $(2, -3)$ and $(7, 9)$ in the $xy$ -plane?

A) $\sqrt{17}$

B) $13$

C) $17$

D) $169$

Show the worked solution

Answer: B

Explanation

The gaps are $7 - 2 = 5$ across and $9 - (-3) = 12$ up:

$\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

A 5-12-13 triangle — worth recognising, since it removes the arithmetic.

Why each wrong option is wrong:

A, $\sqrt{17}$ — added before squaring.
C, $17$ — added the gaps, which is the distance along two sides rather than across.
D, $169$ — stopped before the square root.

Takeaway: the distance formula is Pythagoras. Watch the double negative when subtracting a negative coordinate.

Q10Medium

Topic: A formula you must know — the volume of a sphere

A spherical tank has the radius shown, in units.

What is its volume, in cubic units?

F) $9 \pi$

G) $12 \pi$

H) $27 \pi$

J) $36 \pi$

Show the worked solution

Answer: J

Explanation

$V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3} \times \pi \times 27 = 36\pi$

The $\frac{4}{3}$ and the cube are both essential, and there is no height in the formula — a sphere has only one measurement.

Why each wrong option is wrong:

F, $9 \pi$ — used $\pi r^2\mathclose{,}$ the area of a circle.
G, $12 \pi$ — squared the radius instead of cubing it.
H, $27 \pi$ — omitted the $\frac{4}{3}\mathclose{.}$

Takeaway: sphere is $\frac{4}{3}\pi r^3\mathclose{.}$ If your sphere formula has an $h$ in it, it is the wrong formula.

Q11Medium

Topic: A formula you must know — the midpoint

A line segment has endpoints $(-3, 8)$ and $(7, -2)\mathclose{.}$ What is the midpoint?

A) $\left( 2, \ 3\right)$

B) $\left( 5, \ -5\right)$

C) $\left( 4, \ 6\right)$

D) $\left( -5, \ 5\right)$

Show the worked solution

Answer: A

Explanation

Average each coordinate separately:

$x: \frac{-3 + 7}{2} = 2 \qquad y: \frac{8 + (-2)}{2} = 3$

Check that the answer lies between the endpoints in both coordinates: 2 is between $-3$ and 7, and 3 is between $-2$ and 8. ✓

Why each wrong option is wrong:

B, $\left( 5, \ -5\right)$ — subtracted instead of adding, which is the distance formula's first step, not the midpoint's.
C, $\left( 4, \ 6\right)$ — added correctly and forgot to halve.
D, $\left( -5, \ 5\right)$ — sign errors in both coordinates.

Takeaway: midpoint adds and halves; distance and slope subtract. Then check the answer sits between the two endpoints.