Plane Geometry

📐 Plane Geometry

Part 1 of 7 — Angles and Lines

Almost every ACT geometry problem, from triangles to circles, eventually comes down to an angle fact or a length fact. This part gives you the angle facts. The Enhanced ACT Math test has 45 questions in 50 minutes, each with 4 answer choices, and a calculator is allowed. You have a little over a minute per question, so these rules need to be automatic.

Read the directions once: ACT figures are not necessarily drawn to scale. Never measure an angle by eye; use the given numbers and the rules below.

Angle Vocabulary

Memory hook: C comes before S in the alphabet, and 90 comes before 180. Complementary = 90°, Supplementary = 180°.

Useful fact: For any acute angle, the supplement is exactly 90° more than the complement, because $(180 - x) - (90 - x) = 90$.

Two Lines That Cross

When two lines intersect, they make four angles: two pairs of vertical angles. If one angle is 70°, its vertical partner is 70°, and each of the other two angles is $180 - 70 = 110°$ because each forms a linear pair with a 70° angle. On the ACT, the answer they want is often the other angle, so reread what the question asks.

Parallel Lines Cut by a Transversal

A transversal is a line that crosses two other lines. When those two lines are parallel, the eight angles formed have only two different measures: the four acute angles are all equal, the four obtuse angles are all equal, and any acute angle plus any obtuse angle equals 180°.

The fast way: decide whether the two angles look both small, both big, or one of each. Both small or both big → set them equal. One small and one big → set their sum = 180. This works only when the lines are parallel; the problem must say so or mark them with arrows.

Parallel line through a vertex: If a line through vertex A of a triangle is drawn parallel to the opposite side BC, alternate interior angles copy angle B and angle C up to line A. The three angles at A (copy of B, angle A, copy of C) form a straight line, which is exactly why every triangle's angles sum to 180°. The ACT sometimes gives you this picture and asks for one of the three angles.

Polygon Angle Sums

From one vertex of an $n$-sided polygon you can draw diagonals that cut it into $n - 2$ triangles, each worth 180°.

$\text{Sum of interior angles} = (n - 2) \times 180°$

Exterior angles: Extend each side of a convex polygon; the exterior angles (one at each vertex) always sum to 360°, no matter how many sides. At each vertex, interior + exterior = 180°.

For a regular polygon (all sides and angles equal): each exterior angle is $\frac{360}{n}$ and each interior angle is $180 - \frac{360}{n}$. To find $n$ from an interior angle, take $180 -$ interior to get the exterior angle, then divide 360 by it.

Worked Examples

Two parallel lines are cut by a transversal. Two same-side interior angles measure $(2x + 10)°$ and $(3x + 20)°$. Find both angles.

1. Same-side interior angles are one small and one big, so they are supplementary.
2. $(2x + 10) + (3x + 20) = 180 \implies 5x + 30 = 180 \implies x = 30$.
3. The angles are $2(30) + 10 = 70°$ and $3(30) + 20 = 110°$. Check: $70 + 110 = 180$.

Trap avoided: setting them equal would give $x = -10$, a negative angle, which tells you the setup was wrong.

The complement of an angle is one-third of its supplement. Find the angle.

1. Translate: complement $= 90 - x$, supplement $= 180 - x$.
2. $90 - x = \frac{1}{3}(180 - x)$. Multiply by 3: $270 - 3x = 180 - x$.
3. $90 = 2x \implies x = 45°$.
4. Check: complement 45°, supplement 135°, and $135 \div 3 = 45$. ✓

Each interior angle of a regular polygon is 150°. How many sides does it have?

1. Exterior angle $= 180 - 150 = 30°$.
2. Exterior angles sum to 360°, so $n = 360 \div 30 = 12$.
3. Check: $(12 - 2) \times 180 = 1{,}800$, and $1{,}800 \div 12 = 150$. ✓

Quick Check: Angle Relationships 🎯

Find the Value 🧮

1. Two lines intersect. A pair of vertical angles measure $(6x - 14)°$ and $(4x + 20)°$. Type the value of $x$.

2. What is the sum of the interior angles of a hexagon, in degrees?

3. What is the supplement of a 47° angle, in degrees?

ACT-Style Practice

Try each in under a minute, then check the answer column.

ACT Tip: After solving for $x$, stop and ask: "Did they want $x$, this angle, or a different angle?" Wrong answer choices are built from exactly those mix-ups.

ACT-Style Questions 📋

Key Takeaways

• Complementary = 90°, supplementary = 180°; a linear pair is supplementary; vertical angles are equal; angles around a point total 360°.
• Translate words to algebra: complement $= 90 - x$, supplement $= 180 - x$.
• Parallel lines + transversal: corresponding, alternate interior, and alternate exterior angles are equal; same-side interior angles add to 180°. Shortcut: small = small, big = big, small + big = 180.
• A line through a vertex parallel to the opposite side copies the two base angles onto a straight line.
• Interior angle sum of an $n$-gon: $(n - 2) \times 180°$. Exterior angles always total 360°; in a regular polygon each is $\frac{360}{n}$.
• Figures are not necessarily drawn to scale, and the answer is often the other angle; check what the question asks.

🔺 Triangle Properties

Part 2 of 7 — Angles, Sides, and Right Triangles

Triangles are the most tested shape in ACT geometry, because nearly every other figure (rectangles, trapezoids, polygons, even circles) gets solved by finding a triangle inside it.

Angle Rules

1. Angle sum. The three interior angles of every triangle add to 180°.

Ratio method: If the angles are in the ratio $2 : 3 : 7$, there are $2 + 3 + 7 = 12$ equal parts, so each part is $180 \div 12 = 15°$ and the angles are 30°, 45°, and 105°.

2. Exterior angle theorem. Extend one side of a triangle. The exterior angle formed equals the sum of the two remote (non-adjacent) interior angles.

$\text{exterior angle} = \text{remote angle 1} + \text{remote angle 2}$

Why: the exterior angle and the adjacent interior angle form a straight line (180°), and so do the adjacent angle plus the other two. In triangle ABC with $\angle A = 50°$ and $\angle B = 60°$, the exterior angle at C is $50 + 60 = 110°$, and the interior angle at C is $180 - 110 = 70°$. The exterior angle is not equal to either remote angle alone.

3. Isosceles triangles. If two sides are equal, the angles opposite those sides (the base angles) are equal, and the reverse is also true. The third angle is the vertex angle.

• Vertex angle 40° → base angles are each $(180 - 40) \div 2 = 70°$.
• Base angle 40° → vertex angle is $180 - 2(40) = 100°$.

Read carefully which angle you are given; both versions appear on the ACT.

4. Equilateral triangles. All three sides equal, so all three angles are 60°.

Side Rules

Triangle inequality. Each side must be shorter than the sum of the other two. For two sides $a$ and $b$, the third side $c$ must satisfy

$|a - b| < c < a + b$

With sides 7 and 12, the third side is between 5 and 19, not including 5 or 19 (those lengths would flatten the triangle into a line segment).

Side-angle order. The longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Right Triangles

Pythagorean theorem (right triangles only): $a^2 + b^2 = c^2$, where $c$ is the hypotenuse, the side opposite the right angle and always the longest side.

• Finding the hypotenuse: add the squares. Legs 6 and 8 → $c = \sqrt{36 + 64} = 10$.
• Finding a leg: subtract the squares. Hypotenuse 17, leg 8 → $\sqrt{289 - 64} = 15$.

Pythagorean triples save time. Know these and their multiples:

Special Right Triangles

• 45-45-90: leg → hypotenuse, multiply by $\sqrt{2}$; hypotenuse → leg, divide by $\sqrt{2}$ (hypotenuse 10 → leg $\frac{10}{\sqrt{2}} = 5\sqrt{2}$).
• 30-60-90: always find the short leg first. Hypotenuse 14 → short leg 7 → long leg $7\sqrt{3}$.

The Equilateral Triangle Shortcut

An altitude of an equilateral triangle with side $s$ bisects the base and the top angle, cutting the triangle into two 30-60-90 triangles with hypotenuse $s$ and short leg $\frac{s}{2}$. So:

$h = \frac{s\sqrt{3}}{2} \qquad \text{Area} = \frac{1}{2} \cdot s \cdot \frac{s\sqrt{3}}{2} = \frac{\sqrt{3}}{4}s^2$

Side 8 → height $4\sqrt{3}$, area $16\sqrt{3}$. If you forget the formula, draw the altitude and rebuild it from the 30-60-90 ratio.

Worked Examples

In isosceles triangle ABC, AB = AC and the vertex angle A measures 40°. Side BC is extended past C to point D. Find angle ACD.

1. Base angles B and C are equal: $(180 - 40) \div 2 = 70°$ each.
2. Angle ACD is the exterior angle at C, so it equals the two remote angles: $40 + 70 = 110°$.
3. Check with the straight line: $180 - 70 = 110°$. ✓

Find the height and area of an equilateral triangle with side 10.

1. The altitude makes a 30-60-90 triangle with hypotenuse 10 and short leg 5.
2. Height = long leg $= 5\sqrt{3}$.
3. Area $= \frac{1}{2}(10)(5\sqrt{3}) = 25\sqrt{3}$. Formula check: $\frac{\sqrt{3}}{4}(100) = 25\sqrt{3}$. ✓

A 13-foot ladder reaches 12 feet up a vertical wall. How far is its foot from the wall?

1. The ladder is the hypotenuse (it is opposite the right angle between wall and ground).
2. Spot the 5-12-13 triple, or compute $\sqrt{169 - 144} = \sqrt{25} = 5$ feet.

Quick Check: Triangle Rules 🎯

Right Triangles and Isosceles Triangles 🧮

1. A right triangle has legs 9 and 12. Type the length of the hypotenuse.

2. In a 30-60-90 triangle, the side opposite the 30° angle is 6. Type the length of the hypotenuse.

3. An isosceles triangle has base angles of 70°. Type the measure of the vertex angle in degrees.

ACT-Style Practice

ACT Tip: When the answer choices contain $\sqrt{2}$ or $\sqrt{3}$, the problem almost certainly hides a 45-45-90 or 30-60-90 triangle. Look for a 45°, 30°, or 60° angle, a square's diagonal, or an equilateral triangle's altitude.

ACT-Style Questions 📋

Key Takeaways

• Triangle angles sum to 180°; for a ratio, divide 180 by the total number of parts.
• Exterior angle = sum of the two remote interior angles (and exterior + adjacent interior = 180°).
• Isosceles: equal sides ↔ equal base angles opposite them. Check whether you were given the vertex angle or a base angle.
• Triangle inequality: $|a - b| < c < a + b$, endpoints excluded. Longest side is opposite the largest angle.
• Pythagorean theorem for right triangles only; add squares for the hypotenuse, subtract for a leg. Know 3-4-5, 5-12-13, 8-15-17, 7-24-25 and their multiples.
• 45-45-90: $x, x, x\sqrt{2}$. 30-60-90: $x, x\sqrt{3}, 2x$; find the short leg first.
• Equilateral triangle: the altitude makes two 30-60-90 triangles; $h = \frac{s\sqrt{3}}{2}$ and Area $= \frac{\sqrt{3}}{4}s^2$.

⬛ Quadrilaterals & Polygons

Part 3 of 7 — Parallelograms, Rectangles, Rhombuses, Squares, Trapezoids

Every quadrilateral's interior angles sum to 360° ($(4 - 2) \times 180$). The special quadrilaterals each add properties on top of that, and ACT questions test whether you know exactly which properties belong to which shape.

The Family Tree

• A parallelogram has both pairs of opposite sides parallel.
• A rectangle is a parallelogram with four right angles.
• A rhombus is a parallelogram with four equal sides.
• A square is both a rectangle and a rhombus, so it has every property below.
• A trapezoid has one pair of parallel sides (the bases); the other two sides (the legs) are not parallel.

Properties Table

Parallelograms

Consecutive angles are supplementary because each pair of neighboring angles is a pair of same-side interior angles between parallel sides. If one angle is 58°, its neighbors are each $180 - 58 = 122°$ and the opposite angle is 58°. Check: $58 + 122 + 58 + 122 = 360$.

Algebra with sides: opposite sides are equal, so set their expressions equal, solve, and then compute what is asked. A perimeter needs all four sides: $P = 2(\text{side}_1 + \text{side}_2)$.

Area $= \text{base} \times \text{height}$, where the height is perpendicular to the base, not the slanted side.

Rectangles and Squares

A diagonal cuts a rectangle into two right triangles, so $d = \sqrt{l^2 + w^2}$. Both diagonals are equal, and they bisect each other, so the four half-diagonals from the center are all equal.

A square with side $s$ has diagonal $d = s\sqrt{2}$ (two 45-45-90 triangles). Going backward, $s = \frac{d}{\sqrt{2}}$ and the area is $s^2 = \frac{d^2}{2}$.

Rhombuses

The diagonals of a rhombus are perpendicular bisectors of each other, so they cut the rhombus into four congruent right triangles whose legs are the half-diagonals and whose hypotenuse is a side.

• Diagonals 16 and 12 → half-diagonals 8 and 6 → side $= \sqrt{64 + 36} = 10$.
• Area $= \frac{d_1 d_2}{2} = \frac{16 \times 12}{2} = 96$.

A rhombus does not need right angles. Four equal sides plus a 70° angle means "rhombus, not a square."

Trapezoids

$\text{Area} = \frac{1}{2}(b_1 + b_2)h$

That is the average of the two bases times the perpendicular height. In an isosceles trapezoid, the legs are equal and the base angles are equal. To find the height, drop perpendiculars from the ends of the shorter base: the two little right triangles each have a horizontal leg (the overhang) of $\frac{b_2 - b_1}{2}$ and a hypotenuse equal to the trapezoid's leg.

Regular Polygons (recap from Part 1)

Each interior angle of a regular $n$-gon is $\frac{(n - 2) \cdot 180}{n}$, each exterior angle is $\frac{360}{n}$, and the two add to 180°. Regular octagon: $\frac{1080}{8} = 135°$ interior, 45° exterior. A regular hexagon splits into six equilateral triangles from its center, which is the fastest way to find its area.

Worked Examples

An isosceles trapezoid has bases 8 and 20 and legs of 10. Find its area.

1. Overhang on each side: $(20 - 8) \div 2 = 6$.
2. Each end is a right triangle with hypotenuse 10 and leg 6, so $h = \sqrt{100 - 36} = 8$.
3. Area $= \frac{1}{2}(8 + 20)(8) = 112$.

Trap avoided: using the slanted leg 10 as the height gives 140.

A rectangle's length is 2 more than its width, and its diagonal is 10. Find the perimeter.

1. The diagonal is a hypotenuse: $w^2 + (w + 2)^2 = 100$.
2. $2w^2 + 4w - 96 = 0 \implies w^2 + 2w - 48 = 0 \implies (w + 8)(w - 6) = 0$, so $w = 6$.
3. Length 8, perimeter $2(6 + 8) = 28$. (A 6-8-10 triangle, the 3-4-5 triple doubled.)

Quick Check: Quadrilateral Properties 🎯

Compute It 🧮

1. One angle of a parallelogram measures 72°. Type the measure of a consecutive (neighboring) angle, in degrees.

2. A rectangle measures 9 by 12. Type the length of its diagonal.

3. A trapezoid has bases 6 and 10 and a height of 7. Type its area.

ACT-Style Practice

ACT Tip: When a question names a specific quadrilateral, list its properties before computing. Many wrong choices come from using a rectangle property (equal diagonals, right angles) on a shape that is only a parallelogram or rhombus.

ACT-Style Questions 📋

Key Takeaways

• Quadrilateral angles sum to 360°.
• Parallelogram: opposite sides and angles equal, consecutive angles supplementary, diagonals bisect each other; area = base × perpendicular height.
• Rectangle: add right angles and equal diagonals; $d = \sqrt{l^2 + w^2}$.
• Rhombus: four equal sides; diagonals are perpendicular bisectors, making four right triangles; area $= \frac{d_1 d_2}{2}$.
• Square: everything above; diagonal $= s\sqrt{2}$, area $= s^2 = \frac{d^2}{2}$.
• Trapezoid: area $= \frac{1}{2}(b_1 + b_2)h$; in an isosceles trapezoid the overhang is $\frac{b_2 - b_1}{2}$, and the leg is the hypotenuse with the height.
• Perimeter means all the sides; a common wrong choice is half the perimeter.

⭕ Circles

Part 4 of 7 — Circumference, Area, Arcs, Sectors, and Angles in Circles

The Core Formulas

Work backward through the radius. Whatever you are given (diameter, circumference, or area), find $r$ first.

• $C = 18\pi \implies 2\pi r = 18\pi \implies r = 9 \implies A = 81\pi$.
• $A = 36\pi \implies r^2 = 36 \implies r = 6 \implies d = 12$.

Answer choices are usually left in terms of $\pi$. If a question says "use 3.14 for $\pi$" or asks for a decimal, multiply at the end.

Arcs and Sectors: Use the Fraction of the Circle

A central angle has its vertex at the center. Its arc is the part of the circle between its sides, and the sector is the pie-slice region. A central angle of $\theta$ degrees takes $\frac{\theta}{360}$ of the whole circle, so:

$\text{Arc length} = \frac{\theta}{360} \cdot 2\pi r \qquad \text{Sector area} = \frac{\theta}{360} \cdot \pi r^2$

• Radius 12, central angle 150°: arc $= \frac{150}{360}(24\pi) = 10\pi$; sector area $= \frac{150}{360}(144\pi) = 60\pi$.
• A pizza with diameter 16 cut into 8 equal slices: $r = 8$, whole area $64\pi$, one slice $8\pi$.
• Sector perimeter = two radii + the arc. Radius 6, angle 60°: arc $= 2\pi$, perimeter $= 12 + 2\pi$.

The same proportion runs backward: an arc of $5\pi$ in a circle with circumference $12\pi$ is $\frac{5}{12}$ of the circle, so its central angle is $\frac{5}{12} \times 360 = 150°$.

Angles in Circles

• Central angle AOB = 96° → an inscribed angle ACB intercepting the same arc is 48°.
• Two inscribed angles that intercept the same arc are equal.
• An angle inscribed in a semicircle is 90°. If a triangle is drawn in a circle with one side a diameter, the angle opposite the diameter is a right angle, so the Pythagorean theorem applies with the diameter as the hypotenuse.

Tangents and Chords

• A tangent line touches the circle at one point and is perpendicular to the radius at that point. A tangent, a radius, and a segment to the center make a right triangle.
• Two tangent segments drawn to a circle from the same outside point are equal in length.
• The perpendicular from the center to a chord bisects the chord. Radius 10 and chord 16 → half-chord 8 → distance from center $= \sqrt{100 - 64} = 6$.

Circles and Squares Together

Similarly, a rectangle inscribed in a circle has its diagonal as a diameter (a 6-by-8 rectangle sits in a circle of diameter 10).

Radian note: The ACT trigonometry questions sometimes measure angles in radians, where arc length is simply $s = r\theta$. In this geometry lesson, angles are in degrees.

Worked Examples

A circle has radius 9. Find the arc length and sector area for a central angle of 80°.

1. Fraction of the circle: $\frac{80}{360} = \frac{2}{9}$.
2. Arc $= \frac{2}{9}(18\pi) = 4\pi$.
3. Sector area $= \frac{2}{9}(81\pi) = 18\pi$.

Trap avoided: arc uses the circumference ($2\pi r$); sector area uses the area ($\pi r^2$).

A square is inscribed in a circle of radius 5. Find the area of the region inside the circle but outside the square.

1. The square's diagonal is a diameter: 10.
2. Side $= \frac{10}{\sqrt{2}} = 5\sqrt{2}$, so square area $= 50$ (or $\frac{d^2}{2} = \frac{100}{2}$).
3. Circle area $= 25\pi$. Shaded region $= 25\pi - 50$.

Triangle ABC is inscribed in a circle, and AB is a diameter of length 20. If AC = 12, find BC.

1. Angle C is inscribed in a semicircle, so it is 90°, and AB is the hypotenuse.
2. $BC = \sqrt{400 - 144} = \sqrt{256} = 16$ (the 3-4-5 triple times 4).

Quick Check: Circle Measurements 🎯

Circle Calculations 🧮

1. A circle has a diameter of 14. Its area is $k\pi$. Type $k$.

2. A sector of a circle with radius 10 has a central angle of 72°. Its area is $k\pi$. Type $k$.

3. A central angle measures 140°. Type the measure, in degrees, of an inscribed angle that intercepts the same arc.

ACT-Style Practice

ACT Tip: Write "$r =$" before anything else. Radius-versus-diameter errors are the single most common way to land on a wrong circle answer, and the wrong choices are built to catch them.

ACT-Style Questions 📋

Key Takeaways

• Find the radius first. $C = 2\pi r$, $A = \pi r^2$.
• Arc length $= \frac{\theta}{360} \cdot 2\pi r$; sector area $= \frac{\theta}{360} \cdot \pi r^2$; sector perimeter = arc + 2 radii.
• Central angle = its arc; inscribed angle = half the central angle on the same arc.
• Angle inscribed in a semicircle = 90°, so a triangle with a diameter side is a right triangle with the diameter as hypotenuse.
• Tangent ⟂ radius at the point of tangency; the perpendicular from the center bisects a chord.
• Circle in a square: diameter = side. Square in a circle: diameter = diagonal ($s\sqrt{2}$).

📏 Area & Perimeter

Part 5 of 7 — Formulas, Composite Figures, Borders, Units, and Scaling

The ACT does not hand you a formula sheet, so the basic area formulas below must be memorized. Perimeter is a length (add the sides); area is the number of unit squares inside (square units).

Formulas to Know

Height means perpendicular height. In a triangle or parallelogram, the height is the perpendicular distance from the base to the opposite vertex or side, never a slanted side. In a right triangle, the two legs are perpendicular, so they serve as base and height: legs 7 and 24 give an area of $\frac{1}{2}(7)(24) = 84$. The hypotenuse is not a height for either leg.

Working backward: If a rectangle has perimeter 46 and length 15, then $46 = 2(15) + 2w$, so $w = 8$ and the area is 120. Solve for the missing dimension before you multiply.

Composite Figures

Break a strange shape into rectangles, triangles, and circle pieces, then add the pieces or subtract a hole.

• Add: A shed wall that is a 30-by-12 rectangle topped by a triangle with base 30 and height 8 has area $360 + 120 = 480$.
• Subtract (shaded region): whole figure minus the unshaded part. A circle of radius 5 inside a 10-by-10 square leaves $100 - 25\pi$.
• Two semicircles = one circle. A track that is a rectangle with a semicircle on each end of diameter 60 has circular area $\pi(30)^2 = 900\pi$, not $1800\pi$.

Composite perimeter: trace only the outside edge. A side that is covered by an attached semicircle is no longer on the boundary, and the semicircle contributes half a circumference, $\pi r$.

Borders, Frames, and Walkways

A border of width $w$ around a rectangle adds $w$ to both ends of each dimension, so each dimension grows by $2w$.

$\text{border area} = (\text{outer rectangle}) - (\text{inner rectangle})$

A 24-by-18 garden with a 3-foot walkway all around: outer rectangle $30 \times 24 = 720$, garden $432$, walkway $288$. Multiplying the garden's perimeter by the width misses the four corner squares.

Units

Convert lengths first, then compute area. Tiles that are 6 inches on a side are 0.5 ft by 0.5 ft, so each covers 0.25 sq ft; a 12-by-15 ft floor (180 sq ft) needs $180 \div 0.25 = 720$ tiles.

Scaling

If every length of a figure is multiplied by $k$:

Doubling the side of a square multiplies its area by 4, not 2. (Part 6 adds volume, which scales by $k^3$.)

Worked Examples

A figure is a 20-by-10 rectangle with a semicircle attached outward along one 10-unit side. Find its area and perimeter.

1. Semicircle radius $= 5$. Area $= 200 + \frac{1}{2}\pi(25) = 200 + 12.5\pi$.
2. Perimeter: the covered 10-unit side is gone. Outside edges: $20 + 10 + 20 = 50$, plus the arc $\frac{1}{2}(2\pi \cdot 5) = 5\pi$.
3. Perimeter $= 50 + 5\pi$.

An 8-by-10 inch photo has a 2-inch frame on all sides. Find the frame's area.

1. Outer dimensions: $8 + 4 = 12$ by $10 + 4 = 14$, area 168.
2. Frame $= 168 - 80 = 88$ square inches.

Trap avoided: adding only 2 to each dimension (10 by 12) gives $120 - 80 = 40$.

Quick Check: Area Formulas 🎯

Composite Areas 🧮

1. A trapezoid has bases 5 and 11 and a height of 4. Type its area.

2. A 4-by-2 rectangle is cut out of a 10-by-6 rectangle. Type the area that remains.

3. A figure is a square with side 6 topped by a triangle whose base is the square's top side and whose height is 4. Type the figure's total area.

ACT-Style Practice

ACT Tip: For shaded-region questions, write "shaded = whole − unshaded" before touching numbers. The answer choices usually include the unshaded area by itself as a trap.

ACT-Style Questions 📋

Key Takeaways

• Memorize the area formulas; the ACT gives no formula sheet. Height is always perpendicular; in a right triangle, the legs are the base and height.
• Solve for a missing dimension first (from a perimeter or a diagonal), then compute the area.
• Composite figures: add pieces or subtract holes; two semicircles make one circle.
• Composite perimeter: outside edges only; a semicircle adds $\pi r$.
• Borders: each dimension grows by twice the border width; border = outer − inner.
• Units: convert lengths before multiplying; 1 sq ft = 144 sq in.
• Scaling by $k$: lengths × $k$, areas × $k^2$.

🔍 Similar Triangles & 3-D Solids

Part 6 of 7 — Proportional Figures, Volume, and Surface Area

Similar Triangles

Two triangles are similar when they have the same shape: all corresponding angles are equal and all corresponding sides are in the same ratio $k$ (the scale factor).

AA similarity: If two angles of one triangle equal two angles of another, the triangles are similar. (The third angles must then match, since each triangle totals 180°.) You never need to check the sides to prove similarity on the ACT; you need two matching angles.

Where similar triangles hide:

Setting up the proportion. Match sides by the angles they are opposite, then write

$\frac{\text{small side}}{\text{matching big side}} = \frac{\text{other small side}}{\text{its matching big side}}$

Nested-triangle trap: In triangle ABC with DE parallel to BC (D on AB, E on AC), the small triangle is ADE and the big one is ABC. Compare AD to the whole side AB, not to the leftover piece DB. With AD = 4, DB = 6, and DE = 5: AB = 10, so $\frac{4}{10} = \frac{5}{BC}$ and BC = 12.5.

Congruent triangles are similar with $k = 1$ (same shape and same size). The ACT sometimes uses the congruence shortcuts SSS, SAS, ASA, and AAS to justify that two lengths or angles are equal.

Scale Factor Rules

If two similar figures have length ratio $k$:

Similar triangles with sides in ratio $2 : 5$ have areas in ratio $4 : 25$. Similar solids with lengths in ratio $2 : 3$ have volumes in ratio $8 : 27$.

Volume and Surface Area

Volume is the space inside (cubic units). Surface area is the total area of all the outside faces (square units).

What to memorize: box, cube, prism, and cylinder formulas. The ACT has no formula sheet, but when a question needs the cone, pyramid, or sphere formula, it typically states the formula in the question. Your job is to plug in correctly: use the radius (not the diameter), square or cube it before multiplying, and keep the $\frac{1}{3}$ or $\frac{4}{3}$.

Prism idea: Every prism and cylinder is "base area × height." A cylinder is a prism with a circular base, so $V = (\pi r^2) h$.

Displacement: When an object is fully submerged in a tank, the volume of the object equals the volume of the water rise (base of the tank × rise in height).

Cubic units: 1 cu ft = $12^3$ = 1,728 cu in.

Worked Examples

A 6-foot person casts a 4-foot shadow at the same time a tree casts a 30-foot shadow. How tall is the tree?

1. Person and tree each make a right angle with the ground, and the sun's angle is the same, so the triangles are similar (AA).
2. $\frac{\text{height}}{\text{shadow}}$: $\frac{6}{4} = \frac{h}{30}$.
3. $h = 45$ feet.

A cylinder has radius 3 and height 10. Find its volume and total surface area.

1. Volume $= \pi(3^2)(10) = 90\pi$.
2. Two circular ends: $2\pi(9) = 18\pi$. Curved side (unrolls to a rectangle with width $2\pi r$ and height $h$): $2\pi(3)(10) = 60\pi$.
3. Surface area $= 78\pi$.

Quick Check: Similar Triangles 🎯

Volume and Surface Area 🧮

1. A box measures 4 by 5 by 6. Type its volume.

2. A cube has edges of length 3. Type its total surface area.

3. A cylinder has radius 2 and height 5. Its volume is $k\pi$. Type $k$.

ACT-Style Practice

ACT Tip: Before computing a volume, check the units in the question and the answer choices. If dimensions are in feet and the answer is in cubic inches, convert each length to inches first.

ACT-Style Questions 📋

Key Takeaways

• Similar triangles: equal angles, proportional sides. Two equal angles (AA) are enough.
• Look for similarity in parallel segments inside a triangle, shadows, hourglass figures, and altitudes to a hypotenuse.
• In nested triangles, compare the small side to the whole big side, not the leftover piece.
• Scale factor $k$: lengths × $k$, areas × $k^2$, volumes × $k^3$.
• Memorize box $lwh$, cube $s^3$, prism $Bh$, and cylinder $\pi r^2 h$; cone, pyramid, and sphere formulas are usually given, so plug in carefully.
• Surface area = sum of all face areas; a cylinder's side unrolls into a $2\pi r$-by-$h$ rectangle.
• A submerged object's volume = tank base area × water rise.

🧩 Review & Applications

Part 7 of 7 — Multi-Step ACT Geometry Problems

Harder ACT geometry questions rarely test one rule. They chain two or three rules from Parts 1–6: a diagonal becomes a diameter, a height comes from a right triangle, a ratio comes from similar triangles. With 45 questions in 50 minutes, you need a routine that gets you from the figure to the answer without false starts.

A Five-Step Routine

1. Draw or redraw. If the problem describes a figure in words, sketch it. If a figure is given, copy the numbers onto it. Remember the figure is not necessarily to scale.
2. Label everything you can. Fill in every angle and length that follows directly from a rule (vertical angles, base angles, half-diagonals, radii).
3. Find the link. Ask, "What shape connects what I know to what I want?" Usually it is a right triangle, a pair of similar triangles, or a fraction of a circle.
4. Write one equation in one unknown and solve it.
5. Answer the question asked. Radius or diameter? Larger or smaller angle? $x$ or the angle? Area or perimeter? Then check that the size is sensible.

Trigger → Tool

Hidden Right Triangles

Many "impossible" problems become routine once you draw one segment:

• Isosceles triangle: the altitude from the vertex angle bisects the base. Legs 13, base 10 → half-base 5 → height 12 → area $\frac{1}{2}(10)(12) = 60$.
• Chord: draw the radius to the chord's endpoint and the perpendicular from the center. Radius 13, chord 24 → half-chord 12 → distance 5.
• Regular hexagon: six equilateral triangles from the center. Side 4 → area $6 \cdot \frac{\sqrt{3}}{4}(16) = 24\sqrt{3}$.
• Segment of a circle (between a chord and its arc): sector area − triangle area.

The Trap List

Pacing: If a geometry problem has no clear first step after about 20 seconds, mark it and move on. Come back after the questions you can solve quickly; every question is worth the same.

Worked Examples

A 6-by-8 rectangle is inscribed in a circle. Find the area of the region inside the circle but outside the rectangle.

1. Link: the rectangle's diagonal is a diameter. Diagonal $= \sqrt{36 + 64} = 10$, so $r = 5$.
2. Circle area $= 25\pi$; rectangle area $= 48$.
3. Shaded region $= 25\pi - 48$.

In triangle ABC, angle A $= (x + 15)°$, angle B $= (2x)°$, and the exterior angle at C is $(4x - 25)°$. Find angle C.

1. Exterior angle = sum of remote angles: $4x - 25 = (x + 15) + 2x \implies x = 40$.
2. Angle A $= 55°$, angle B $= 80°$, exterior angle at C $= 135°$.
3. Angle C $= 180 - 135 = 45°$. Check: $55 + 80 + 45 = 180$. ✓

Mixed Practice: Two-Step Problems 🎯

Pick the First Tool 🔍

ACT-Style Practice

ACT Tip: On a multi-step problem, the wrong answers are the numbers you pass through on the way (the radius, the half-chord, $x$). If your answer matches an intermediate value, reread the question before you bubble it.

ACT-Style Questions 📋

Key Takeaways

• Routine: draw, label, find the link, one equation, answer what was asked.
• The most common links: a right triangle (often hidden), similar triangles, a diagonal that is a diameter, a fraction of a circle.
• Drop a perpendicular whenever a height is missing in an isosceles triangle or trapezoid; draw radii to chords and tangent points.
• Shaded regions are whole − unshaded; a circle segment is sector − triangle.
• Wrong answers are usually intermediate values or one classic slip ($r$ vs $d$, the missing $\frac{1}{2}$, $k$ vs $k^2$).
• Keep moving: skip a geometry problem with no clear first step and return to it.