Geometry on the ISEE is mostly one big game: find the missing angle or the missing side. The good news is there are only a handful of rules, and almost every problem uses two or three of them in a row. Learn the rules below, learn to spot which ones a picture is begging you to use, and you'll chew through these.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch on a question to type your answer instead: a number of degrees like 47, a decimal like 10.5, a root like 7√3, a side like BC, or a word like obtuse. When the answer has to be one of the choices (a set of sides, a range, a triangle from a list), that version stays multiple choice, just like the real ISEE.
First, angles come in three sizes, and the whole section leans on telling them apart.
An angle less than $90^\circ$ is an acute angle, an angle of exactly $90^\circ$ is a right angle, and an angle between $90^\circ$ and $180^\circ$ is an obtuse angle.
Two special totals show up over and over: $90^\circ$ (a corner) and $180^\circ$ (a straight line). Each one gets a name for the two angles that make it.
Two angles are complementary if they sum to $90^\circ$.
Complementary comes before supplementary, and $90$ comes before $180$. That's honestly the only reason I remember which is which.
And the straight-line version:
Two angles are supplementary if they sum to $180^\circ$.
Watch what happens when one ray swings around inside a corner, or along a straight line. One piece grows and the other shrinks by exactly the same amount, so the total never moves.
So if you know one piece, the other is just the total minus it. Try one.
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When two lines cross, they make four angles, and those come in pairs too.
When two lines intersect, two opposite (non-adjacent) angles form a vertical pair.
The angles right next to each other along one of the lines are a different story.
Two adjacent angles that together form a straight line make a linear pair.
Here's what each kind of pair does.
Turn one of the lines and keep an eye on the purple pair. They change together, and they're always equal.
That's a vertical pair: across from each other, equal. Its orange neighbor fills up the rest of the straight line. Your turn.
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Sometimes the ISEE doesn't give you either angle. It just tells you how the two relate. Call the smaller one $x$ and write the other one in terms of it.
Angles $A$ and $B$ are supplementary, and one is three times the other. Find both. Then: $\alpha$ and $\beta$ are complementary, and one is $18^\circ$ less than the other. Find both.
Call the smaller one $x$: $x + 3x = 180$, so $x = 45$. The angles are $45^\circ$ and $135^\circ$.
Call the bigger one $\alpha$: $\alpha + (\alpha - 18) = 90$, so $2\alpha = 108$ and $\alpha = 54$. The angles are $54^\circ$ and $36^\circ$.
One letter, one equation, then answer what they asked for (the smaller one, the bigger one, or both).
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Now add a second line, parallel to the first, and cut through both.
A transversal is a line that intersects other lines.
A transversal makes two crossings, one on each line, and each crossing has four angles. The angles sitting in the same spot at the two crossings have a name.
When a transversal cuts two lines, corresponding angles are in the same relative position at each intersection.
When the two lines are parallel, the transversal crosses both at exactly the same tilt, so the two crossings are copies of each other. Slide one onto the other and see.
A transversal makes eight angles and there are only two different sizes. Eight angles. Two numbers. Lazy.
Every angle lands on its twin, so corresponding angles are equal. Two more pairs get their own names, because the ISEE uses them. The space between the two parallel lines is the inside, or interior, and everything above and below them is the outside, or exterior.
When a transversal cuts two lines, two angles on opposite sides of the transversal are alternate interior angles if both are between the lines, and alternate exterior angles if both are outside the lines.
And one pair on the same side:
When a transversal cuts two lines, two angles between the lines on the same side of the transversal are same-side interior angles.
Put all of those together with vertical pairs and linear pairs, and you get the whole trick:
When a transversal cuts two parallel lines:
That last line is the one you'll actually use. Here it is on a real picture.
Two horizontal lines are parallel and cut by a slanted line, as shown. Find $a$, $b$, and $c$.
$a$ is in the same spot at the lower crossing as the $52^\circ$ is at the upper one, so they're corresponding: $a = 52^\circ$.
$b$ forms a linear pair with the $52^\circ$: $b = 180 - 52 = 128^\circ$.
$c$ is vertical to $a$: $c = 52^\circ$. (Or just notice it's acute, so it's $52^\circ$ like every other acute angle.)
Notice we never needed the names to get the numbers. Acute or obtuse was enough.
Before you calculate anything, look: is the angle you want acute or obtuse? If it matches the given angle, it's equal. If one is acute and the other obtuse, they add to $180^\circ$. Picking the wrong one of those two is the whole trap.
Tilt the transversal and watch all eight angles at once: four green, four orange, never a third color.
Now use it: one angle given, one asked for, and only two sizes to choose from.
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Some problems call the pairs by name instead. Click an angle to see what each of its partners is called.
Same spot at each crossing: corresponding. Between the lines on opposite sides: alternate interior. Put a name on a pair.
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The ISEE loves to dress all of these pairs up with algebra. Every one of them comes down to one question: are the two angles equal, or do they add up to something?
Once you write the equation, it's just a linear equation like the ones you solved in Section 3.
First the equal kind.
$m\angle 1 = (2x + 10)^\circ$ and $m\angle 2 = (5x - 20)^\circ$ are vertical angles. Find $m\angle 1 + m\angle 2$.
Vertical angles are equal: $2x + 10 = 5x - 20$, so $30 = 3x$ and $x = 10$. Each angle is $2(10) + 10 = 30^\circ$, and together they make $60^\circ$.
Read that question one more time before you celebrate.
Don't stop at $x$! The question above asks for the sum of the angles, and $x = 10$ is sitting right there as a wrong answer. Always reread what they actually asked for.
Vertical pairs, corresponding angles, alternate angles: if the two are equal, set the expressions equal.
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Now the adding kind. A linear pair, a same-side interior pair, or two angles filling a corner: add the expressions and set them equal to the total.
Two angles form a linear pair. They measure $(3x + 10)^\circ$ and $(x + 30)^\circ$. What is the larger angle?
A linear pair adds to $180^\circ$: $(3x + 10) + (x + 30) = 180$, so $4x + 40 = 180$, $4x = 140$, and $x = 35$.
The angles are $3(35) + 10 = 115^\circ$ and $35 + 30 = 65^\circ$. Check: $115 + 65 = 180$. The larger one is $115^\circ$.
Same setup every time: decide the total, add, solve, then plug back in if they want an angle.
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Three sides, three angles, and one rule that never breaks.
A triangle is a polygon formed by three points that aren't all on one line, joined by three line segments.
Some triangles have special names, and each name is a clue about equal sides or angles.
The tick marks in the picture mean "these sides are the same length." First the one with two ticks.
An isosceles triangle has two equal sides, called its legs. The two angles across from the legs are its base angles, and the angle where the legs meet is its vertex angle.
Next, the one with the little box in its corner.
A right triangle has one $90^\circ$ angle.
Then the most symmetric triangle there is.
An equilateral triangle has all three sides equal.
And the triangle with nothing special about it at all.
A scalene triangle has no equal sides.
One more piece before the facts: what happens just outside a corner.
Stretch one side of a triangle past a corner, and the angle between that extension and the other side at the corner is an exterior angle of the triangle. The two angles inside the triangle that don't touch it are its remote interior angles.
Here are the facts the ISEE actually tests about triangles.
Start with the angle sum. Pull the corners around. However weird you make it, the angles still add to exactly $180^\circ$.
Here's why it works: tear off the three corners and line them up, and they make a straight line.
Every polygon can be chopped into triangles, and that's exactly how you'll find the angle sum of a pentagon or hexagon in Section 7.
The ISEE's favorite version hides two triangles in one picture: a bowtie, where two triangles meet at a point.
Segments $\overline{AD}$ and $\overline{BC}$ cross at $O$, as shown. In $\triangle ABO$, $m\angle A = 48^\circ$ and $m\angle B = 67^\circ$. In $\triangle CDO$, $m\angle C = 72^\circ$. Find $x$, the angle at $D$.
The left triangle's angle at $O$ is what's left of $180^\circ$: $180 - 48 - 67 = 65^\circ$.
The two angles at $O$ are a vertical pair, so the right triangle has a $65^\circ$ angle too. Then $x = 180 - 65 - 72 = 43^\circ$.
Two angles known means the third is free. Find a missing angle.
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Now the isosceles fact. Why do equal legs force equal base angles? Fold one in half and watch.
The two base angles land right on top of each other, so they share whatever's left after the vertex angle. Next, the exterior angle fact: slide the two far angles over to the corner outside and see what they fill.
Both facts in one problem:
In isosceles $\triangle ABC$ where $AB = AC$, the vertex angle is $34^\circ$, and side $CB$ is extended past $B$, as shown. Find the two base angles and the exterior angle at vertex $B$.
The base angles are equal and share what's left of $180^\circ$: $\frac{180 - 34}{2} = 73^\circ$ each.
The exterior angle at $B$ equals the two remote angles, $A$ and $C$: $34 + 73 = 107^\circ$. (Check: it also forms a linear pair with $\angle B$, and $180 - 73 = 107$.)
Halve what's left after the vertex angle, and you have each base angle. Try it.
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The exterior angle has one trap worth knowing before you try it.
An exterior angle equals the two angles far away from it, not the one right next to it. The angle next door is its linear-pair partner, so those two add to $180^\circ$.
Far angles added, or the neighbor subtracted from $180^\circ$. Either way works.
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The ISEE likes to put algebra on these too. Same facts as before, then solve for $x$.
In isosceles $\triangle ABC$, $AB = AC$. The base angles measure $(3x - 5)^\circ$ and $(2x + 15)^\circ$. What is the vertex angle?
Base angles are equal: $3x - 5 = 2x + 15$, so $x = 20$, and each base angle is $3(20) - 5 = 55^\circ$. The vertex angle gets what's left: $180 - 55 - 55 = 70^\circ$.
Equal base angles means equal expressions. Set them equal and go.
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An exterior angle with $x$ in it works the same way: the far angles add up to it.
Side $BC$ of $\triangle ABC$ is extended past $C$, making an exterior angle of $125^\circ$. The two remote interior angles measure $x^\circ$ and $(2x - 10)^\circ$. What is $x$?
The exterior angle equals the two remote angles added: $x + (2x - 10) = 125$, so $3x = 135$ and $x = 45$.
Add the two far angles, set them equal to the outside one, solve.
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Sometimes the angles are only described by how they relate to each other. Put everything in terms of one angle.
In $\triangle XYZ$, $\angle X$ is half the measure of $\angle Y$ and $30^\circ$ more than $\angle Z$. What is the largest angle?
Write everything with $X$: $Y = 2X$ and $Z = X - 30$. Then $X + 2X + (X - 30) = 180$, so $4X = 210$ and $X = 52.5$. The angles are $52.5^\circ$, $105^\circ$, and $22.5^\circ$, so the largest is $105^\circ$.
Pick the angle everything else is described by, call it one letter, and the rest falls out.
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A ratio is the tidiest version of this. "In the ratio $2:3:7$" means the angles are $2$ shares, $3$ shares, and $7$ shares of the same size, and all the shares together fill $180^\circ$.
The angles of a triangle are in the ratio $2:3:7$. Find all three angles.
There are $2 + 3 + 7 = 12$ shares, so each share is $\frac{180}{12} = 15^\circ$.
The angles are $2 \cdot 15 = 30^\circ$, $3 \cdot 15 = 45^\circ$, and $7 \cdot 15 = 105^\circ$. Check: $30 + 45 + 105 = 180$.
That's the same ratio sharing you did in Section 1. The only new part is that the whole is always $180^\circ$.
Add the parts, split $180^\circ$, multiply back.
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The last fact on the list turns angles into a ranking of the sides, without measuring anything.
In $\triangle JKL$, $m\angle J = 47^\circ$ and $m\angle K = 68^\circ$. Which side is the longest?
First the missing angle: $m\angle L = 180 - 47 - 68 = 65^\circ$. The largest angle is $\angle K$, so the longest side is the one across from it. That's the side that doesn't touch $K$ at all: $JL$.
That third angle came out close, which is exactly why you can't skip it.
Don't skip the third angle. Here it lost by just $3^\circ$, but if it had come out bigger than $68^\circ$, the answer would have been a different side.
Drag a corner and watch the two rankings stay lined up: each side wears the color of the angle across from it.
Biggest angle, longest side. Smallest angle, shortest side. Your turn.
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You can't make a triangle out of just any three lengths. The two shorter sides have to be able to reach.
In any triangle, each side is shorter than the other two sides added together: $a + b > c$, $b + c > a$, and $a + c > b$.
If the two short sides can't reach, it's not a triangle. It's just three sad sticks lying on the floor.
In practice you only need one check: the two shorter sides must add up to more than the longest one. When you know two sides, the third has to land between their difference and their sum.
A triangle has sides of length $5$ and $9$. What lengths are possible for the third side? And if the third side has to be a whole number, how many lengths could it be?
More than $9 - 5 = 4$ and less than $9 + 5 = 14$. So the third side is strictly between $4$ and $14$.
The whole numbers in there are $5, 6, 7, \ldots, 13$. Count them: $13 - 5 + 1 = 9$ lengths. ($4$ and $14$ don't count, because they'd lie flat.)
Those two edge numbers are where people slip.
"Equal" is not good enough. Sides $5$, $9$, and $14$ don't make a skinny triangle. They make a flat line. The ISEE puts exactly that number in the choices.
Find the smallest and largest third stick for yourself.
More than the difference, less than the sum. Use it.
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Now test your eye: guess first, then watch the sticks swing.
The isosceles version of this question hides a second step.
Watch for isosceles triangles with two given sides, like $4$ and $9$. The third side has to repeat one of them, but $4, 4, 9$ can't close ($4 + 4 = 8$, which is less than $9$). So it has to be $4, 9, 9$.
Short sides added, compared with the long one. Which sets close?
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Two triangles can be the same shape without being the same size, like a photo and its enlargement.
Two triangles are similar if they have the same shape: matching angles are equal and matching sides are in proportion. Each one is a scaled copy of the other, and the number every length gets multiplied by is the scale factor.
Watch a copy grow and shrink. The sides change, the angles never do.
You don't have to check every angle and every side to know two triangles are similar. Two angles are enough, because the third angle is whatever's left of $180^\circ$.
If two angles of one triangle match two angles of another, the triangles are similar.
Sometimes the matching angle is the one they didn't tell you.
A triangle has angles of $40^\circ$ and $75^\circ$. Is a triangle with angles of $65^\circ$ and $75^\circ$ similar to it?
The first triangle's third angle is $180 - 40 - 75 = 65^\circ$. So both triangles have a $65^\circ$ angle and a $75^\circ$ angle: two matches, so yes, they're similar.
Find all three angles first, then look for two matches.
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Now the sides. "In proportion" means every side of one triangle is the same multiple of its partner.
"In proportion" is the same ratio idea from Section 1: set two fractions equal and cross-multiply.
Written out with letters, it looks like this:
If $\triangle ABC \sim \triangle DEF$, then $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$.
The letters do the matching for you, even when the picture is turned around.
The order of the letters is the matching. In $\triangle ABC \sim \triangle DEF$, $A$ goes with $D$, $B$ with $E$, $C$ with $F$, so side $BC$ goes with side $EF$. Don't match sides by how they look in the picture.
Try it on a copy that's bigger, then turned, then flipped. The angle marks tell you which corner is which.
The ISEE's favorite similar triangles are made by the sun. A post and a tree cast shadows at the same time, so the sun's rays hit them at the same angle.
A $5$-foot post casts a $2$-foot shadow. At the same time, a tree casts a $14$-foot shadow. How tall is the tree?
Each object and its shadow make a right triangle, and the two triangles are similar. Matching sides: $\dfrac{\text{tree}}{5} = \dfrac{14}{2}$.
The tree's shadow is $7$ times the post's, so the tree is $7$ times as tall: $5 \cdot 7 = 35$ feet.
Drag the sun and watch that "$7$ times" hold no matter where it is.
If a side is labeled $x$ and its match is labeled $\frac{3}{2}x$, don't panic about the $x$. The scale factor is $\frac{\frac{3}{2}x}{x} = \frac{3}{2}$: the $x$'s cancel, and you never needed its value. A side of $4$ becomes $6$, not "$6x$".
Match the letters, set up the ratio, solve.
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The letters match angles too, and matching angles are equal.
$\triangle ABC \sim \triangle DEF$. If $m\angle A = 50^\circ$ and $m\angle E = 70^\circ$, what is $m\angle F$?
$D$ matches $A$, so $m\angle D = 50^\circ$. Then the angles of $\triangle DEF$ add to $180^\circ$: $m\angle F = 180 - 50 - 70 = 60^\circ$.
Copy the matching angles over, then use the angle sum.
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The sneakiest version: a segment drawn parallel to one side cuts off a smaller triangle, and that smaller triangle is automatically similar to the big one. (The parallel segment makes corresponding angles, so two angles match.)
Measure everything from the shared corner, whole sides, not the leftover pieces.
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Double every side of a triangle and its area doesn't double. It grows four times as big. Section 7 shows why, and what happens to volume.
Right triangles get their own superpower. First, the sides need names.
In a right triangle, the hypotenuse is the side opposite the right angle. The other two sides, the ones that make the right angle, are the legs.
The hypotenuse sits across from the biggest angle, the $90^\circ$ one, so it's always the longest side. Here's how all three connect.
If $a$ and $b$ are the legs and $c$ is the hypotenuse, then $a^2 + b^2 = c^2$.
Common triples: $(3, 4, 5)$, $(5, 12, 13)$, $(8, 15, 17)$, and their multiples, like $(6, 8, 10)$.
Remember the distance formula? It's this theorem: the horizontal and vertical gaps are the legs, and the distance is the hypotenuse.
The first step is always the same: find the hypotenuse. It's the side across from the right angle, wherever the letters put it.
In $\triangle ABC$, $\angle A$ is a right angle, $AB = 7$, and $AC = 11$. What is $BC$?
$BC$ is across from the right angle, so it's the hypotenuse: $BC^2 = 7^2 + 11^2 = 49 + 121 = 170$, so $BC = \sqrt{170}$. Not every answer is a nice triple.
Both legs known: add their squares.
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Going the other way, from the hypotenuse back to a leg, flips one sign.
Finding the hypotenuse? Add the squares. Finding a leg? Subtract: $b^2 = c^2 - a^2$. If your "leg" comes out longer than the hypotenuse, you added when you should have subtracted.
The ISEE loves to hide that leg inside an isosceles triangle: the height splits it into two right triangles.
An isosceles triangle has two sides of length $13$ and a base of $10$. What is its height?
The height drops to the middle of the base, so each half is a right triangle with hypotenuse $13$ and one leg $10 \div 2 = 5$.
The height is the other leg: $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.
Hypotenuse squared, minus the leg you know.
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Often the hypotenuse is just a stepping stone, like when they ask for the distance all the way around.
A right triangle has legs of length $9$ and $12$. What is its perimeter?
First the hypotenuse: $\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$. Then add all three sides: $9 + 12 + 15 = 36$.
Missing side first, then add up all three (or all four, for a rectangle).
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The theorem teams up with similar triangles, too. Sometimes the side you need to scale isn't given yet, and Pythagoras finds it first.
For a right triangle $\triangle ABC$, $AB = 8$ and $AC = 6$, where $BC$ is the hypotenuse. $\triangle DEF \sim \triangle ABC$ with scale factor $\frac{1}{2}$. What is the length of $EF$?
$EF$ matches $BC$, so find $BC$ first: it's a $6$-$8$-$10$ triangle, so $BC = 10$. Then $EF = \frac{1}{2} \cdot 10 = 5$.
Missing side of the original first, then scale it.
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Where does the theorem come from? Build a square on each side and look at the areas.
That shows it works for the triangles you build. This shows why it works for every right triangle. Four copies of one right triangle fit inside a big square, and the space they leave is a square of side $c$.
Now slide the same four triangles around. Nothing gets added or taken away, so the empty space can't change, only its shape.
The theorem also works backwards: it can tell you what kind of triangle you have from its sides alone. Hold two sides at $6$ and $8$ and open the angle between them.
So $c^2$ compared with $a^2 + b^2$ tells you the angle across from $c$.
Compare the two smaller squared sides with the biggest one: $a^2 + b^2 = c^2$ means right, $a^2 + b^2 > c^2$ means acute, and $a^2 + b^2 < c^2$ means obtuse. Skip the square roots: squared lengths are easier and exact.
Sort a few before you try one.
Square, add the two small ones, compare.
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Put that together with the distance formula and you can name a triangle from nothing but its corners on a grid.
Triangle $ABC$ has vertices $A(2, 1)$, $B(5, 3)$, and $C(7, 0)$. Which type of triangle is it: acute, isosceles, obtuse, or scalene?
Use the distance formula, but keep the sides squared: $AB^2 = 3^2 + 2^2 = 13$, $BC^2 = 2^2 + 3^2 = 13$, $AC^2 = 5^2 + 1^2 = 26$.
$AB = BC$, so it's isosceles. And $13 + 13 = 26$, so it's also a right triangle, which rules out acute and obtuse.
Drag the corners yourself and let the squared sides name what you built.
Three squared sides, two checks. That's every rule in this section.
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Every question so far came right after its idea, which is a hint the real test never gives. Down here the hint is gone. First, the quantitative comparison: two columns, four fixed choices, and any idea from this section fair game. Watch out for "cannot be determined": a picture that isn't labeled, or two sides that don't match, often isn't enough to settle a comparison.
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Next, the mixed set. Every question type on this page except the quantitative comparison goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the ones that need their choices stay multiple choice.
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Alright, if you've made it this far, it's time for something with a little heat. This one chains parallel lines, an angle bisector, and a triangle together. Mark every angle you find on the picture as you go.
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