Formulas and Equations from Geometry | Geometry | LevelKite

A formula is an equation that states how quantities are related. For a rectangle with length $\ell$ and width $w$ , the perimeter $P$ is the total distance around it, and $P = 2\ell + 2w$ .

This form makes $P$ easy to find. But suppose you know $P$ and $\ell$ and need $w$ . Then you solve for $w$ : you get $w$ alone on one side, with an expression that does not contain $w$ on the other side. That process is rearranging the formula. The relationship stays the same; only the question it answers directly changes.

You already have the tools for this. Each move you make to both sides uses a property of equality, a rule that lets you do the same operation to both sides and keep the equation true, as long as you never divide by zero. The table solves for $w$ twice: once for a rectangle with perimeter $30$ centimeters and length $9$ centimeters, and once with letters that stand for any rectangle.

The same moves with numbers and with letters

Move	With numbers	With letters	Reason
Start with the formula	$30 = 2(9) + 2w$	$P = 2\ell + 2w$	Perimeter formula for a rectangle
Subtract the length term from both sides	$30 - 2(9) = 2w$	$P - 2\ell = 2w$	Subtraction Property of Equality
Divide both sides by $2$	$\frac{30 - 2(9)}{2} = w$	$\frac{P - 2\ell}{2} = w$	Division Property of Equality, because $2 \neq 0$
Swap the two sides	$w = \frac{30 - 2(9)}{2}$	$w = \frac{P - 2\ell}{2}$	Symmetric Property of Equality: if $a = b$ , then $b = a$

The numbers give $w = \frac{30 - 18}{2} = \frac{12}{2} = 6$ . A width is a length, so it must be positive. Since $6 > 0$ , you keep it: the width is $6$ centimeters.

The letters give you something more. The last row, $w = \frac{P - 2\ell}{2}$ , is a formula for the width of every rectangle. You can reuse it without solving again, as long as you still check that the width it gives is positive.

Worked example

Solve a volume formula for the height

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$ , where $r$ is the radius of its circular base and $h$ is its height. Solve the formula for $h$ . Then find the height of a cone with volume $48\pi$ cubic centimeters and radius $4$ centimeters.

Multiply both sides by 3 to undo the fraction one third. The Multiplication Property of Equality keeps the equation true.
$3V = \pi r^2 h$
Divide both sides by pi times r squared to get h alone. The Division Property of Equality requires this divisor to be nonzero. Since pi is not zero, the condition is that r is not zero, and a radius is a positive length, so it holds. Then swap the two sides by the Symmetric Property of Equality.
$\frac{3V}{\pi r^2} = h, \text{ so } h = \frac{3V}{\pi r^2}, \quad r \neq 0$
Substitute 48 pi for V and 4 for r. The Substitution Property of Equality lets you replace a quantity with an equal one. Then multiply in the numerator and square 4 in the denominator.
$h = \frac{3(48\pi)}{\pi (4)^2} = \frac{144\pi}{16\pi}$
Divide the numerator and the denominator by 16 pi, which is not zero.
$h = \frac{144\pi}{16\pi} = 9$
Check the value. A height is a length, and 9 is positive, so you keep it. Putting 9 back into the original formula gives the stated volume, so the algebra checks too.
$\frac{1}{3}\pi (4)^2 (9) = 48\pi$

Answer

The rearranged formula is $h = \frac{3V}{\pi r^2}$ with $r \neq 0$ , and the height of this cone is $9$ centimeters.

Now you get to build the equation yourself. A geometric description states facts about a figure, and those facts give you the equation. Follow these four steps:

Separate the given, which is what the problem states as true, from what you must find. Use only what is stated. A sketch may suggest that two angles are equal, but that is not a reason.
Name the definition, postulate or theorem that turns the given facts into an equation. A postulate is a statement accepted without proof, and a theorem is a statement that has been proved.
Solve the equation, with a reason for each step and only nonzero divisors.
Substitute to find each length or angle measure, and check it. A length must be positive. An angle measure must be greater than $0^\circ$ and at most $180^\circ$ . Reject any value that breaks these conditions.

Here are the terms you will need:

$AB$ means the length of segment $\overline{AB}$ , and $m\angle A$ means the measure of angle $A$ .
Congruent segments have equal lengths, and congruent angles have equal measures. The symbol is $\cong$ .
Point $B$ is between points $A$ and $C$ when $B$ lies on $\overline{AC}$ and is not an endpoint.
Two intersecting lines form four angles. Two of those angles that share no side are vertical angles.
An isosceles triangle has at least two congruent sides.
In a triangle, the side opposite an angle is the side that does not contain the angle's corner point, its vertex.

Facts that turn a description into an equation

Named fact	What it says	Equation it gives
Vertical Angles Theorem	If two angles are vertical angles, then they are congruent.	$m\angle 1 = m\angle 2$ for vertical angles $\angle 1$ and $\angle 2$ , by the definition of congruent angles
Segment Addition Postulate	If point $B$ is between points $A$ and $C$ , then $AB + BC = AC$ .	$AB + BC = AC$
Isosceles Triangle Theorem	If two sides of a triangle are congruent, then the angles opposite those sides are congruent.	$m\angle B = m\angle C$ when $\overline{AB} \cong \overline{AC}$ , by the definition of congruent angles
Triangle Angle Sum Theorem	The measures of the three angles of a triangle add to $180^\circ$ .	$m\angle A + m\angle B + m\angle C = 180^\circ$

Key concept

Solve with reasons, then test with the geometry

A formula and a geometric description both give you an equation, and you solve both with the same named properties, dividing only by nonzero quantities. A solved value must then pass two tests. It must make the original equation true. It must also give positive lengths and angle measures greater than $0^\circ$ and at most $180^\circ$ . If a value passes the first test but fails the second, you reject it.

Worked example

Vertical angles give equal measures

Two lines intersect, and $\angle 1$ and $\angle 2$ are vertical angles. It is given that $m\angle 1 = (3x + 10)^\circ$ and $m\angle 2 = (5x - 30)^\circ$ . Find the measure of each angle.

Write the equation. The angles are given as vertical, so the Vertical Angles Theorem, used forward, says they are congruent. The definition of congruent angles then says their measures are equal.
$3x + 10 = 5x - 30$
Subtract 3x from both sides, by the Subtraction Property of Equality.
$10 = 2x - 30$
Add 30 to both sides, by the Addition Property of Equality.
$40 = 2x$
Divide both sides by 2, by the Division Property of Equality. The divisor 2 is not zero.
$20 = x$
Substitute 20 for x in each expression, by the Substitution Property of Equality.
$m\angle 1 = 3(20) + 10 = 70^\circ, \quad m\angle 2 = 5(20) - 30 = 70^\circ$
Check the value. The two measures are equal, as the theorem requires. Each one is greater than 0 degrees and at most 180 degrees, so you accept x = 20.
$0^\circ < 70^\circ \leq 180^\circ$

Answer

$\angle 1$ and $\angle 2$ each measure $70^\circ$ .

Worked example

A solution the geometry rejects

Point $B$ is between points $A$ and $C$ . It is given that $AB = 3x - 9$ , $BC = x + 5$ and $AC = 4$ . Find $AB$ and $BC$ , if such lengths exist.

Point B is given as between A and C, so the Segment Addition Postulate, used forward, gives AB + BC = AC. Substitute the given expressions.
$(3x - 9) + (x + 5) = 4$
Combine like terms on the left, by the Commutative, Associative and Distributive Properties.
$4x - 4 = 4$
Add 4 to both sides, by the Addition Property of Equality.
$4x = 8$
Divide both sides by 4, by the Division Property of Equality. The divisor 4 is not zero.
$x = 2$
Substitute 2 for x in each length, by the Substitution Property of Equality. The equation checks, because -3 + 7 = 4, so the algebra is correct.
$AB = 3(2) - 9 = -3, \quad BC = 2 + 5 = 7$
Check the geometry. A segment length must be positive, and AB = -3 is not, so reject x = 2. It is the only solution of the equation, so no value of x fits the description.
$AB = -3 < 0$

Answer

The equation's only solution, $x = 2$ , makes $AB = -3$ , so you reject it, and no such lengths exist.

Worked example

From words to an equation

In $\triangle ABC$ , it is given that $\overline{AB} \cong \overline{AC}$ . It is also given that $m\angle A$ is $30^\circ$ more than $m\angle B$ . Find the measure of each angle.

Let x be the measure of angle B in degrees, so the description gives angle A as x + 30. Segment AB is opposite angle C, and segment AC is opposite angle B. The Isosceles Triangle Theorem, used forward from congruent sides to congruent angles and not as its converse, makes angles B and C congruent. By the definition of congruent angles, their measures are equal.
$m\angle B = x, \quad m\angle C = x, \quad m\angle A = x + 30$
The Triangle Angle Sum Theorem says the three measures add to 180 degrees. Substitute the expressions from the first step.
$x + x + (x + 30) = 180$
Combine like terms, by the Commutative, Associative and Distributive Properties.
$3x + 30 = 180$
Subtract 30 from both sides, by the Subtraction Property of Equality.
$3x = 150$
Divide both sides by 3, which is not zero, by the Division Property of Equality.
$x = 50$
Substitute 50 for x, by the Substitution Property of Equality. Then check: each measure is greater than 0 degrees and at most 180 degrees, and 50 + 50 + 80 = 180, so you accept x = 50.
$m\angle B = 50^\circ, \quad m\angle C = 50^\circ, \quad m\angle A = 50^\circ + 30^\circ = 80^\circ$

Answer

In $\triangle ABC$ , $m\angle A = 80^\circ$ , $m\angle B = 50^\circ$ and $m\angle C = 50^\circ$ .

Takeaway: Get your equation from a formula or a named geometric fact, isolate the variable by dividing only by nonzero quantities, and keep a value only if every length is positive and every angle measure is greater than $0^\circ$ and at most $180^\circ$ .