Plane Geometry (Angles, Triangles, Circles)

Master angle relationships, triangle properties, circle theorems, and area formulas for ACT Math with examples and practice.

Plane geometry is the single largest content area on the ACT Math section, accounting for roughly 10–14 questions out of 60 — that is nearly 20–25% of the entire test. Unlike some other standardized exams, the ACT tests geometry heavily and directly, covering angle relationships, triangle properties, circle theorems, polygon angle sums, and area and perimeter formulas. These questions span the full difficulty range, from straightforward angle calculations in the first quarter of the test to complex multi-step problems combining multiple geometric concepts in the final quarter.

The ACT gives you 60 questions in 60 minutes, a calculator is allowed on every question, and there is no penalty for wrong answers — so always answer every question, even if you must guess. Geometry questions appear throughout the entire test, with simpler angle and area problems early on and more sophisticated circle, similarity, and combined-figure problems later.

Here is the critical fact: the ACT does not provide a formula sheet or reference page. This is different from the SAT, which gives you a formula reference. On the ACT, you must have all geometry formulas memorized before test day. This includes area formulas for every common shape, the Pythagorean theorem, circle formulas, and special right triangle ratios. There is no way around this requirement — it is simply part of being prepared for the ACT.

This comprehensive guide covers every essential plane geometry concept tested on the ACT: angle relationships and parallel lines, triangle properties (including area, similarity, congruence, and special triangles), circle theorems (arcs, sectors, inscribed angles, tangent lines), polygon angle formulas, and area and perimeter formulas for all common shapes. Master this content, and you will be well-prepared for the largest single topic area on the test.

Core Concepts

Angle Fundamentals

Types of angles:

  • Acute: less than 90°90°
  • Right: exactly 90°90° (marked with a small square)
  • Obtuse: between 90°90° and 180°180°
  • Straight: exactly 180°180° (a straight line)
  • Reflex: between 180°180° and 360°360°

Angle pair relationships:

  • Complementary angles sum to 90°90°. If one angle is 35°35°, its complement is 55°55°.
  • Supplementary angles sum to 180°180°. If one angle is 110°110°, its supplement is 70°70°.
  • Vertical angles are the non-adjacent angles formed when two lines intersect. Vertical angles are always equal. If two lines cross and one angle is 65°65°, the angle directly across is also 65°65°, and the other two angles are each 115°115°.

Angles with Parallel Lines and Transversals

When a transversal (a line that crosses two parallel lines), it creates eight angles with specific, predictable relationships. This is one of the most frequently tested concepts on the ACT.

  • Corresponding angles are equal (same position at each intersection — both upper-left, both lower-right, etc.)
  • Alternate interior angles are equal (opposite sides of the transversal, between the parallel lines — these form a "Z" shape)
  • Alternate exterior angles are equal (opposite sides, outside the parallel lines)
  • Co-interior angles (same-side interior) are supplementary (sum to 180°180° — same side of transversal, between the lines)

Key insight: If you know any ONE of the eight angles, you can determine ALL eight. Each angle is either equal to the known angle or supplementary to it.

Triangle Properties

Angle sum theorem: The three interior angles of any triangle always sum to exactly 180°180°. This is perhaps the single most-used geometry fact on the ACT.

Exterior angle theorem: An exterior angle of a triangle (formed by extending one side) equals the sum of the two non-adjacent interior angles. If the interior angles are 40°40°, 60°60°, and 80°80°, the exterior angle adjacent to the 80°80° angle is 40°+60°=100°40° + 60° = 100°.

Types of triangles by sides:

  • Equilateral: All three sides equal → all three angles are 60°60°
  • Isosceles: Two sides equal → the base angles (opposite the equal sides) are equal
  • Scalene: All sides different → all angles different

Types of triangles by angles:

  • Acute: All angles less than 90°90°
  • Right: One angle equals 90°90°
  • Obtuse: One angle greater than 90°90°

Triangle inequality theorem: The sum of any two sides must be greater than the third side. For sides aa, bb, cc: a+b>ca + b > c, a+c>ba + c > b, and b+c>ab + c > a. The ACT uses this to ask which set of side lengths can form a triangle.

The Pythagorean Theorem

For a right triangle with legs aa and bb and hypotenuse cc:

a2+b2=c2a^2 + b^2 = c^2

Common Pythagorean triples (memorize these — they save enormous time):

  • 3,4,53, 4, 5 (and multiples: 6,8,106, 8, 10; 9,12,159, 12, 15; 12,16,2012, 16, 20; 15,20,2515, 20, 25)
  • 5,12,135, 12, 13 (and multiples: 10,24,2610, 24, 26)
  • 8,15,178, 15, 17
  • 7,24,257, 24, 25

Recognizing a Pythagorean triple instantly eliminates the need for square root calculations. When you see a right triangle with legs 9 and 12, you should immediately know the hypotenuse is 15 (the 33-44-55 triple scaled by 3).

Converse: If a2+b2=c2a^2 + b^2 = c^2 for three sides, the triangle is a right triangle. If a2+b2>c2a^2 + b^2 > c^2, it is acute. If a2+b2<c2a^2 + b^2 < c^2, it is obtuse.

Special Right Triangles

These appear on virtually every ACT and must be memorized:

45-45-90 triangle: Sides in ratio 1:1:21 : 1 : \sqrt{2}.

Both legs are equal, and the hypotenuse is 2\sqrt{2} times a leg. This triangle results from cutting a square along its diagonal. If each leg is ss, the hypotenuse is s2s\sqrt{2}. If the hypotenuse is hh, each leg is h2=h22\frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}.

30-60-90 triangle: Sides in ratio 1:3:21 : \sqrt{3} : 2.

The side opposite 30°30° is the shortest. The side opposite 60°60° is 3\sqrt{3} times the shortest side. The hypotenuse (opposite 90°90°) is exactly twice the shortest side. This triangle results from cutting an equilateral triangle in half.

Similar Triangles

Two triangles are similar if their corresponding angles are equal. This can be established by:

  • AA (Angle-Angle): If two angles of one triangle equal two angles of another, the triangles are similar. (The third angles must also be equal, since angles sum to 180°180°.)

When triangles are similar:

  • Corresponding sides are proportional: a1a2=b1b2=c1c2=k\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k (the scale factor)
  • The ratio of their perimeters equals the scale factor kk
  • The ratio of their areas equals the square of the scale factor: A1A2=k2\frac{A_1}{A_2} = k^2

This last point is crucial: if the sides are in ratio 2:32:3, the areas are in ratio 4:94:9.

Congruence

Two triangles are congruent if they have the same shape AND size. Methods to prove congruence: SSS, SAS, ASA, AAS, HL (for right triangles). The ACT tests congruence less frequently than similarity.

Area Formulas

Shape Area Formula Notes
Triangle A=12bhA = \frac{1}{2}bh hh is the perpendicular height
Rectangle A=lwA = lw
Square A=s2A = s^2 Also A=d22A = \frac{d^2}{2} where dd is diagonal
Parallelogram A=bhA = bh bb is base, hh is height
Trapezoid A = rac{1}{2}(b_1 + b_2)h Average of parallel sides imes imes height
Circle A=pir2A = pi r^2 Radius rr

Perimeter and Circumference

  • Perimeter of a polygon: Sum of all side lengths.
  • Circumference of a circle: C=2pirC = 2pi r or C=pidC = pi d.

Note: The ACT often asks for the perimeter of complex shapes composed of rectangles and semicircles. Break them down part by part.

Circle Theorems

Central angle: An angle formed by two radii with the vertex at the center. The measure of the central angle equals the measure of its intercepted arc.

Inscribed angle: An angle formed by two chords with the vertex on the circle. The measure of the inscribed angle is half the measure of its intercepted arc. ext{Inscribed Angle} = rac{1}{2} imes ext{Intercepted Arc}

Thales's Theorem: Any triangle inscribed in a circle where one side is a diameter is a right triangle. The angle opposite the diameter is 90°90°.

Tangent line: A line tangent to a circle is perpendicular to the radius drawn to the point of tangency. This creates right triangles, which allows you to use the Pythagorean theorem.

Arc length: A portion of the circumference. ext{Arc Length} = rac{ heta}{360°} imes 2pi r where heta heta is the central angle in degrees.

Sector area: A "slice of pie" portion of the circle's area. ext{Sector Area} = rac{ heta}{360°} imes pi r^2

Polygons

Sum of interior angles: For a polygon with nn sides, the sum of the interior angles is: extSum=(n2)imes180° ext{Sum} = (n-2) imes 180°

Regular polygon: A polygon where all sides and all angles are equal. Each interior angle measures: ext{Angle} = rac{(n-2) imes 180°}{n}

Exterior angles: The sum of the exterior angles of any convex polygon is always 360°360°.

Strategy Tips

Tip 1: Draw and Label Everything

Geometry is visual. If a problem describes a shape without showing it, draw it immediately. If a figure is provided, label every known length and angle directly on the drawing. Mark right angles with squares and equal sides with tick marks. This reveals relationships that are impossible to see in your head.

Tip 2: Look for Hidden Triangles

Complex shapes are often just triangles in disguise. A trapezoid can be split into a rectangle and two right triangles. A regular hexagon is composed of six equilateral triangles. When in doubt, draw lines to create triangles — especially right triangles, because then you can use the Pythagorean theorem.

Tip 3: Trust the Figure (Mostly)

On the ACT (unlike the SAT), figures are drawn to scale unless explicitly stated "Not drawn to scale." You can use this to your advantage. If an angle looks obtuse, it probably is. If one line looks twice as long as another, it probably is. Use this to estimate and eliminate obviously wrong answer choices.

Tip 4: Memorize the Formulas

Since there is no formula sheet, you must memorize the area formulas (especially trapezoid and circle) and the Pythagorean triples. Writing them down at the start of the section can help relieve anxiety.

Tip 5: Work Backwards from the Answer

In geometry problems asking for a specific length or angle, you can often plug in the answer choices. If the question asks for the width of a rectangle, try the values in the choices to see which one produces the correct area or perimeter given in the problem.

Worked Example: Example 1

Problem

Two angles of a triangle measure 47°47° and 68°68°. What is the measure of the third angle?

Solution

The sum of angles is 180°180°. 180°47°68°=65°180° - 47° - 68° = 65° The third angle is 65°65°.

Worked Example: Example 2

Problem

A circle has a radius of 6. What is the area of a sector defined by a 120°120° central angle?

Solution

Use the sector area formula: ext{Area} = rac{120}{360} imes pi (6)^2 ext{Area} = rac{1}{3} imes 36pi extArea=12pi ext{Area} = 12pi

Worked Example: Example 3

Problem

In the figure below, lines mm and nn are parallel. Transversal tt intersects both lines. If one of the acute angles is 55°55°, what is the measure of the obtuse alternate exterior angle?

Solution

All acute angles in a parallel line system are equal (55°55°) and all obtuse angles are equal. Acute and obtuse angles are supplementary. extObtuseangle=180°55°=125° ext{Obtuse angle} = 180° - 55° = 125°

Worked Example: Example 4

Problem

A right triangle has legs of length 9 and 12. What is the length of the hypotenuse?

Solution

Recognize the 3453-4-5 triple scaled by 3: 3imes3=93 imes 3 = 9 4imes3=124 imes 3 = 12 So the hypotenuse is 5imes3=155 imes 3 = 15. Alternatively: c=sqrt92+122=sqrt81+144=sqrt225=15c = sqrt{9^2 + 12^2} = sqrt{81 + 144} = sqrt{225} = 15.

Worked Example: Example 5

Problem

Two similar triangles have corresponding sides in the ratio 2:52:5. If the area of the smaller triangle is 20, what is the area of the larger triangle?

Solution

The ratio of lengths is k = rac{5}{2} = 2.5. The ratio of areas is k2=(2.5)2=6.25k^2 = (2.5)^2 = 6.25 (or rac{25}{4}). Area_{large} = Area_{small} imes rac{25}{4} = 20 imes rac{25}{4} = 5 imes 25 = 125

Worked Example: Example 6

Problem

Point AA is on a circle with center OO. Point BB is outside the circle such that ABAB is tangent to the circle at AA. If OA=5OA = 5 and OB=13OB = 13, what is the length of tangent segment ABAB?

Solution

A tangent is perpendicular to the radius, so riangleOAB riangle OAB is a right triangle with hypotenuse OBOB. OA2+AB2=OB2OA^2 + AB^2 = OB^2 52+AB2=1325^2 + AB^2 = 13^2 This is a 512135-12-13 triangle, so AB=12AB = 12.

Practice Problems

  1. Problem 1

    What is the sum of the interior angles of a regular octagon (8-sided polygon)?

    A) 720°720° quad B) 900°900° quad C) 1080°1080° quad D) 1260°1260° quad E) 1440°1440°

    Answer: C) 1080°1080°. Sum =(82)imes180°=6imes180°=1080°= (8-2) imes 180° = 6 imes 180° = 1080°.

    Problem 2

    The area of a trapezoid is 60 square units. If the height is 6 and one base is 8, what is the length of the other base?

    A) 8 quad B) 10 quad C) 12 quad D) 14 quad E) 20

    Answer: C) 12. A = rac{1}{2}(b_1 + b_2)h Rightarrow 60 = rac{1}{2}(8 + x)(6) Rightarrow 60 = 3(8+x) Rightarrow 20 = 8+x Rightarrow x = 12.

    Problem 3

    In circle OO, a central angle of 60°60° intercepts arc ABAB. If the radius is 12, what is the length of arc ABAB?

    A) 2pi2pi quad B) 4pi4pi quad C) 6pi6pi quad D) 12pi12pi quad E) 24pi24pi

    Answer: B) 4pi4pi. Length = rac{60}{360} imes 2pi(12) = rac{1}{6} imes 24pi = 4pi.

    Problem 4

    A square has a diagonal of length 8sqrt28sqrt{2}. What is the area of the square?

    A) 32 quad B) 64 quad C) 128 quad D) 64sqrt264sqrt{2} quad E) 16

    Answer: B) 64. Diagonal d=ssqrt2=8sqrt2d = ssqrt{2} = 8sqrt{2}, so side s=8s = 8. Area =s2=64= s^2 = 64.

    Problem 5

    Triangle ABCABC is isosceles with AB=ACAB = AC. If angleA=40°angle A = 40°, what is the measure of angleBangle B?

    A) 40°40° quad B) 50°50° quad C) 70°70° quad D) 80°80° quad E) 100°100°

    Answer: C) 70°70°. The vertex angle is 40°40°. The base angles share the remaining 140°140° equally. 140°div2=70°140° div 2 = 70°.

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Common Mistakes

  • Confusing area and perimeter. Read carefully — finding the area when the question asks for perimeter is a frustrating way to lose points.
  • Forgetting (n2)(n-2) in the polygon formula. Students often just multiply nimes180°n imes 180°, which is incorrect. A triangle (n=3n=3) has 180°180°, not 540°540°.
  • Misidentifying the height. In a parallelogram or triangle, the height must be perpendicular to the base, not the slanted side length.
  • Assuming lines are parallel. Do not assume lines are parallel just because they look like it. Look for arrows on the lines or a statement in the text.
  • Mixing up arc measure and arc length. Arc measure is in degrees (equal to the central angle); arc length is a distance (in cm, inches, etc.).
  • Squaring the wrong ratio. For similar triangles, the ratio of areas is the square of the side ratio. If sides are 3:13:1, areas are 9:19:1, not 3:13:1.

Frequently Asked Questions

How many geometry questions are on the ACT?

Plane geometry accounts for roughly 10–14 questions (20–23%), and coordinate geometry another 9–10 questions (15%). Together, geometry makes up over a third of the math section. It is the single most important subject to review.

Do I really need to memorize the formulas?

Yes. Unlike the SAT, the ACT gives you zero formulas. If you do not know the area of a trapezoid or the circumference of a circle, you will likely miss easy questions.

Are proofs tested on the ACT?

No. You will never be asked to write a proof. You only need to apply the theorems to solve for missing values.

What is the hardest geometry topic on the ACT?

Multi-step problems involving circles and shaded regions often challenge students. For example, finding the area of a square minus the area of an inscribed circle. These require combining multiple formulas.

Can I use my calculator to solve geometry problems?

Yes, but geometry is mostly about setting up the right equation. The calculation part is usually simple arithmetic. The calculator helps, but it cannot tell you which formula to use.

Key Takeaways

  • Triangles are key. The sum of angles is always 180°180°. The area is rac{1}{2}bh. These two facts solve nearly half of all geometry problems.

  • Memorize Pythagorean triples. 3453-4-5, 512135-12-13, 815178-15-17. These appear constantly and save calculation time.

  • Know your circles. A=pir2A = pi r^2 and C=2pirC = 2pi r. Remember that an inscribed angle is half the central angle.

  • Parallel lines create equal angles. Look for the "Z" shape for alternate interior angles.

  • Draw it out. Never try to solve a geometry problem entirely in your head. A sketched diagram prevents errors.

  • Similarity involves squared ratios. If a shape is doubled in size (sides imes2 imes 2), its area quadruples (area imes4 imes 4).

  • Geometry is a high-reward topic. Because there are so many geometry questions, a few hours of studying formulas can translate to a significant score increase.

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