PSAT 10 Math — Geometry & Right-Triangle Trigonometry (Learn-from-Scratch Notes)

Area and volume

Geometry often asks you to measure how much space something takes up. That can mean space on a flat surface (area) or space inside a 3D object (volume). On the PSAT 10, these questions usually test whether you can (1) recognize the shape (or break a complicated shape into simpler ones), (2) choose the right formula, and (3) keep units and scaling straight.

A big idea that connects this entire topic is that length, area, and volume scale differently. If you double a length, you do not double the area or volume—you square or cube the scale factor. This shows up in problems about similar figures, resizing, maps, and models.

Area: what it measures and how to think about it

Area is the amount of 2D space inside a boundary, measured in square units like $\text{cm}^2$ or $\text{ft}^2$. It matters because many real situations—painting a wall, tiling a floor, fencing a garden bed—are about covering surfaces.

A reliable way to avoid mistakes is to connect each formula to a mental picture:

• Rectangles cover space by rows and columns.
• Triangles are “half of a rectangle/parallelogram.”
• Circles are measured relative to their radius.

Core area formulas (and what the variables mean)

• Rectangle: if length is $l$ and width is $w$, then
  $A = l w$
• Parallelogram: base $b$ times _perpendicular_ height $h$ (not the slanted side)
  $A = b h$
• Triangle: base $b$ and perpendicular height $h$
  $A = \frac{1}{2} b h$
• Trapezoid: parallel bases $b_1$ and $b_2$ with height $h$
  $A = \frac{1}{2} (b_1 + b_2) h$
• Circle: radius $r$
  $A = \pi r^2$

A common “what goes wrong” moment: students use the wrong height. In a triangle or parallelogram, the height must be perpendicular to the base you chose.

Composite area: breaking shapes apart (or subtracting)

PSAT problems often use figures that aren’t one “named” shape. The strategy is to decompose:

• Split the figure into rectangles/triangles you can handle.
• Or compute a larger “easy” area and subtract a cut-out.

This matters because it tests geometry reasoning more than memorization: can you see structure?

Worked example: composite rectangle minus triangle

A rectangle is $12\,\text{m}$ by $7\,\text{m}$. A right triangle with legs $3\,\text{m}$ and $4\,\text{m}$ is cut out of one corner. Find the remaining area.

1) Rectangle area:
$A_{\text{rect}} = 12 \times 7 = 84$
So $A_{\text{rect}} = 84\,\text{m}^2$.

2) Triangle area:
$A_{\text{tri}} = \frac{1}{2} \times 3 \times 4 = 6$
So $A_{\text{tri}} = 6\,\text{m}^2$.

3) Subtract:
$A_{\text{remain}} = 84 - 6 = 78$
So $A_{\text{remain}} = 78\,\text{m}^2$.

Circumference and arc ideas (when they appear)

Sometimes a problem involves distance around a circle, not the area inside. Circumference is the circle’s perimeter:
$C = 2\pi r$
If you see a “fraction of a circle” (like a sector), you typically take the same fraction of the full circumference or full area. For example, a $90^{\circ}$ sector is $\frac{1}{4}$ of the circle.

Volume: what it measures and why formulas look the way they do

Volume is the amount of 3D space an object occupies, measured in cubic units like $\text{cm}^3$. A helpful way to understand most volume formulas is:

Volume is “area of the base” times “how many layers (height)” for prisms and cylinders.

That’s why so many formulas have the pattern $V = B h$, where $B$ is the area of the base.

Core volume formulas

• Rectangular prism: length $l$, width $w$, height $h$
  $V = l w h$
• Cylinder: radius $r$, height $h$
  $V = \pi r^2 h$
• Cone: radius $r$, height $h$
  $V = \frac{1}{3} \pi r^2 h$
• Sphere: radius $r$
  $V = \frac{4}{3} \pi r^3$

A frequent error is mixing up height and slant height in cones. Volume uses the perpendicular height.

Surface area: area of the outside “skin”

Surface area is the total area of all outer faces of a 3D object. It matters in contexts like painting, wrapping, or materials cost.

• Rectangular prism (faces come in pairs):
  $SA = 2(lw + lh + wh)$
• Cylinder (two circles plus a rectangle wrapped around):
  $SA = 2\pi r^2 + 2\pi r h$

The term $2\pi r h$ comes from the “unrolled” side: it becomes a rectangle with one side $h$ and the other side equal to the circumference $2\pi r$.

Units and scaling (a high-value PSAT idea)

Units are not decoration—they’re a built-in error checker.

• Area units are squared.
• Volume units are cubed.

If a figure is scaled by a factor $k$ (every length multiplied by $k$):

• Perimeter scales by $k$.
• Area scales by $k^2$.
• Volume scales by $k^3$.

Worked example: scale factor and area

A rectangular poster is enlarged so that every length is multiplied by $1.5$. By what factor does area change?

Area scale factor:
$k^2 = (1.5)^2 = 2.25$
So the area becomes $2.25$ times the original.

Exam Focus

• Typical question patterns:
  • Find area/volume of a figure after decomposing it into simpler shapes (add/subtract parts).
  • Use a real-world context (paint, fencing, capacity) to decide whether you need perimeter, area, surface area, or volume.
  • Apply a scale factor to determine how area or volume changes.
• Common mistakes:
  • Using slant height instead of vertical height in volume problems (especially cones).
  • Forgetting to square or cube units (writing $\text{cm}$ instead of $\text{cm}^2$ or $\text{cm}^3$).
  • Using a non-perpendicular “height” in triangle/parallelogram area.

Lines, angles, and triangles

This topic is about structure: how straight paths relate, how turns are measured, and how triangles behave. On the PSAT 10, you’re often asked to use a few core facts—angle relationships, triangle sum, and similarity/congruence reasoning—to find unknown measures.

Lines and angle relationships

An angle measures a turn, typically in degrees on the PSAT. The most important relationships come from how lines intersect.

Key definitions (and why you care)

• Intersecting lines form vertical angles (the opposite angles). Vertical angles are always equal. This matters because it gives you an equation immediately.
• Linear pair: two adjacent angles that form a straight line. They are supplementary, meaning they add to $180^{\circ}$.
• Parallel lines cut by a transversal create angle pairs that repeat. If lines are parallel:
  • Corresponding angles are equal.
  • Alternate interior angles are equal.
  • Same-side interior angles are supplementary (sum to $180^{\circ}$).

These facts matter because they let you turn a picture into algebra—unknown angles become variables, and the relationships become equations.

Worked example: parallel lines with a transversal

Two parallel lines are cut by a transversal. One angle is $68^{\circ}$. Find the measure of the alternate interior angle.

Alternate interior angles are equal when lines are parallel, so the angle is also
$68^{\circ}$.

A common mistake here is choosing the wrong “partner angle” because the diagram is rotated. The relationship depends on position, not on whether angles “look” similar.

Coordinate geometry of lines (when diagrams are on a grid)

Some PSAT geometry questions place figures on the coordinate plane. Then line relationships become about slope.

The slope of a line measures its steepness. For two points $(x_1, y_1)$ and $(x_2, y_2)$:
$m = \frac{y_2 - y_1}{x_2 - x_1}$

• Lines are parallel if they have the same slope.
• Lines are perpendicular if their slopes multiply to $-1$:
  $m_1 m_2 = -1$

Two more tools that show up:

• Distance formula (from the Pythagorean theorem):
  $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
• Midpoint of a segment:
  $\left(\frac{x_1 + x_2}{2},\,\frac{y_1 + y_2}{2}\right)$

Worked example: distance on the coordinate plane

Find the distance between $A(-1, 2)$ and $B(5, -6)$.

1) Compute differences:
$x_2 - x_1 = 5 - (-1) = 6$
$y_2 - y_1 = -6 - 2 = -8$

2) Apply distance formula:
$d = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100}$
So
$d = 10$

A frequent error is forgetting to square the negative change in $y$, or mixing up subtraction order inconsistently.

Triangles: the most testable polygon

A triangle is the simplest rigid polygon—once you know enough information, it’s determined. Triangles are everywhere in geometry because complicated shapes can often be broken into triangles.

Triangle angle sum

The interior angles of any triangle add to $180^{\circ}$:
$A + B + C = 180^{\circ}$

This matters constantly: if you know two angles, you know the third.

Types of triangles (useful for interpreting problems)

• Acute: all angles less than $90^{\circ}$.
• Right: one angle equals $90^{\circ}$.
• Obtuse: one angle greater than $90^{\circ}$.
• Isosceles: at least two equal sides; the angles opposite those sides are equal.
• Equilateral: all sides equal; all angles equal to $60^{\circ}$.

A key connection: in an isosceles triangle, the “equal sides” and “equal angles” are linked by the idea that the larger side is opposite the larger angle.

Worked example: isosceles triangle angles

An isosceles triangle has two equal angles. The third angle is $40^{\circ}$. Find the other two angles.

Let the equal angles be $x$ and $x$. Then
$x + x + 40^{\circ} = 180^{\circ}$
$2x = 140^{\circ}$
$x = 70^{\circ}$
So the equal angles are
$70^{\circ}$

Common mistake: assuming the $40^{\circ}$ must be one of the equal angles without reading carefully.

Triangle congruence vs. similarity

These ideas are about comparing triangles.

• Congruent triangles have the same size and shape; corresponding sides and angles are equal.
• Similar triangles have the same shape but may be different sizes; corresponding angles are equal and corresponding sides are proportional.

Similarity is especially important on the PSAT because it supports ratio reasoning, scale drawings, and indirect measurement.

Similarity and proportional sides

If triangles are similar with a scale factor $k$ from triangle 1 to triangle 2, then each corresponding side in triangle 2 is $k$ times the matching side in triangle 1.

Example proportional relationship:
$\frac{a_2}{a_1} = \frac{b_2}{b_1} = \frac{c_2}{c_1} = k$

A classic source of errors is mismatching corresponding sides. You must match sides opposite equal angles.

Worked example: using similar triangles to find a missing side

Two similar triangles have corresponding sides $6$ and $9$ (small to large). A different side is $10$ on the small triangle. Find the corresponding side on the large triangle.

1) Scale factor from small to large:
$k = \frac{9}{6} = \frac{3}{2}$

2) Apply scale factor:
$10 \times \frac{3}{2} = 15$
So the missing side is
$15$

The Pythagorean theorem (bridge to the next section)

In a right triangle (one angle is $90^{\circ}$), the side opposite the right angle is the **hypotenuse**. If the legs are $a$ and $b$ and the hypotenuse is $c$:
$a^2 + b^2 = c^2$

This theorem matters because it powers distance on the coordinate plane, checks whether a triangle is right, and sets up trigonometry.

Worked example: checking for a right triangle

A triangle has side lengths $7$, $24$, and $25$. Is it a right triangle?

Check whether
$7^2 + 24^2 = 25^2$
Compute:
$49 + 576 = 625$
And
$25^2 = 625$
So the equality holds, and the triangle is a right triangle.

A common mistake is forgetting the hypotenuse must be the longest side.

Exam Focus

• Typical question patterns:
  • Use parallel line angle relationships to find missing angles (often with variables and equations).
  • Use triangle angle sum or isosceles properties to solve for angles.
  • Use slope/distance in coordinate geometry to prove lines are parallel/perpendicular or to find lengths.
• Common mistakes:
  • Mixing up corresponding vs. alternate interior angles because the diagram is rotated.
  • Treating any side as the hypotenuse in the Pythagorean theorem (it must be opposite $90^{\circ}$).
  • In similarity problems, pairing the wrong sides in a proportion.

Right triangles and right triangle trigonometry

Right triangles are a “sweet spot” in geometry: they are simple enough to compute with, but powerful enough to model real situations like ramps, ladders, shadows, roof pitches, and navigation. On the PSAT 10, right-triangle problems often boil down to two tools:

1) The Pythagorean theorem (relates side lengths).
2) Trigonometric ratios (relate angles to side ratios).

Anatomy of a right triangle: legs, hypotenuse, and reference angles

In a right triangle:

• The hypotenuse is the longest side, opposite the $90^{\circ}$ angle.
• The other two sides are the legs.

When you use trigonometry, you choose one of the acute angles as the reference angle (call it $\theta$). Then the legs are labeled relative to $\theta$:

• Opposite: the leg across from $\theta$.
• Adjacent: the leg next to $\theta$ (but not the hypotenuse).

A common confusion is that “opposite” and “adjacent” are not fixed labels—they depend on which angle you chose.

Special right triangles (fast exact values)

Some right triangles appear so often that it’s worth knowing their side ratios. These let you avoid approximation.

$45^{\circ}$-$45^{\circ}$-$90^{\circ}$ triangle

This comes from cutting a square along a diagonal.

• Legs are equal.
• Hypotenuse is $\sqrt{2}$ times a leg.

If each leg is $x$, then hypotenuse is
$x\sqrt{2}$

$30^{\circ}$-$60^{\circ}$-$90^{\circ}$ triangle

This comes from splitting an equilateral triangle in half.

• Short leg (opposite $30^{\circ}$): $x$
• Long leg (opposite $60^{\circ}$): $x\sqrt{3}$
• Hypotenuse: $2x$

A memory aid many students use is “$1$, $\sqrt{3}$, $2$” (short, long, hypotenuse), scaled by $x$.

Worked example: special triangle scaling

A $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ triangle has hypotenuse $14$. Find the short and long legs.

In this triangle, hypotenuse $= 2x$. So:
$2x = 14$
$x = 7$
Short leg is
$7$
Long leg is
$7\sqrt{3}$

Common mistake: swapping which leg gets $\sqrt{3}$. The long leg is opposite $60^{\circ}$.

Trigonometric ratios: connecting angles to side ratios

Right-triangle trigonometry uses ratios of side lengths. The three main ratios are:

Why these matter: they let you solve triangles when you know an angle and a side—very common in “angle of elevation/depression” word problems.

A widely used mnemonic is SOH-CAH-TOA:

• Sine: Opposite over Hypotenuse
• Cosine: Adjacent over Hypotenuse
• Tangent: Opposite over Adjacent

Solving for a side using trig (given an angle and a side)

To solve for a missing side:
1) Choose the reference angle $\theta$.
2) Label opposite/adjacent/hypotenuse relative to $\theta$.
3) Pick the trig ratio that uses the side you know and the side you want.
4) Write an equation and solve.

Worked example: finding a height (angle of elevation)

A person stands $20\,\text{m}$ from the base of a tree. The angle of elevation to the top is $35^{\circ}$. Approximate the tree’s height (ignore the person’s height).

1) The height is opposite $35^{\circ}$, and $20\,\text{m}$ is adjacent.

2) Use tangent:
$\tan(35^{\circ}) = \frac{h}{20}$

3) Solve:
$h = 20\tan(35^{\circ})$

Using a calculator,
$h \approx 20 \times 0.700 = 14.0$
So the height is about
$14\,\text{m}$

Common mistakes:

• Using $\sin$ or $\cos$ instead of $\tan$ when neither side is the hypotenuse.
• Swapping opposite and adjacent because the triangle is drawn in an unusual orientation.

Solving for an angle (using inverse trig)

Sometimes you know side lengths and need the angle. Then you use an inverse trig function.

For example, if you know opposite and hypotenuse:
$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$
Then
$\theta = \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)$

Worked example: finding an angle from sides

In a right triangle, the opposite side to angle $\theta$ is $9$ and the adjacent side is $12$. Find $\theta$.

1) Use tangent:
$\tan(\theta) = \frac{9}{12} = \frac{3}{4}$

2) Inverse tangent:
$\theta = \tan^{-1}\left(\frac{3}{4}\right)$

Using a calculator,
$\theta \approx 36.9^{\circ}$

A common error is forgetting your calculator mode (degrees vs. radians). PSAT trig work is in degrees.

Choosing between Pythagorean theorem and trig

A good instinct is:

• If you have two sides (or can get two sides), use the Pythagorean theorem to find the third side.
• If you have an angle (not $90^{\circ}$) and a side, use trig.

Many multi-step problems use both: trig to find a leg, then Pythagorean theorem to find another length, or vice versa.

Worked example: combine trig and Pythagorean theorem

A right triangle has hypotenuse $13$ and one acute angle $\theta = 40^{\circ}$. Find the length of the adjacent leg and the opposite leg (approximate).

1) Adjacent leg uses cosine:
$\cos(40^{\circ}) = \frac{a}{13}$
$a = 13\cos(40^{\circ})$
Approximate:
$a \approx 13 \times 0.766 = 9.96$
So
$a \approx 10.0$

2) Opposite leg uses sine:
$\sin(40^{\circ}) = \frac{o}{13}$
$o = 13\sin(40^{\circ})$
Approximate:
$o \approx 13 \times 0.643 = 8.36$
So
$o \approx 8.4$

Common mistake: mixing the ratios (using $\sin$ for adjacent or $\cos$ for opposite). Always label sides relative to $\theta$ first.

Real-world modeling: elevation, depression, and “line of sight”

Right-triangle trig is a modeling tool. The most common contexts:

• Angle of elevation: you look up from horizontal.
• Angle of depression: you look down from horizontal.

These angles are measured from a horizontal line, which helps you identify the reference angle in the triangle.

One subtle trap: diagrams sometimes include extra lines (like the height of a building and the line of sight). The right angle is usually between a vertical height and a horizontal ground.

Exam Focus

• Typical question patterns:
  • Use special right triangles to find exact lengths involving $\sqrt{2}$ or $\sqrt{3}$.
  • Use $\sin$, $\cos$, or $\tan$ in an angle-of-elevation word problem to find a missing side.
  • Use inverse trig to find an angle from given side lengths.
• Common mistakes:
  • Labeling “opposite” and “adjacent” without first choosing the reference angle $\theta$.
  • Using radians on the calculator instead of degrees.
  • Confusing hypotenuse with a leg when selecting a trig ratio (only sine/cosine involve the hypotenuse).