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4.3 Geometry & Measurement

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ALEKS Math Placement Study Guide Exam

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1 Chapter 1: ALEKS PPL Overview & Strategy

1.1 What ALEKS PPL Is 1.2 How Placement Works 1.3 The Prep & Learning Module & Retake Strategy

2 Chapter 2: Arithmetic & Number Sense

2.1 Real Numbers & Operations 2.2 Fractions, Decimals & Percents 2.3 Ratios, Proportions & Unit Conversion

3 Chapter 3: Algebra Essentials

3.1 Linear Equations & Inequalities 3.2 Exponents & Polynomials 3.3 Factoring & Rational Expressions 3.4 Quadratic Equations

4 Chapter 4: Functions, Graphs & Geometry

4.1 Lines, Slope & Graphing 4.2 Functions & Function Notation 4.3 Geometry & Measurement 4.4 Coordinate Geometry

5 Chapter 5: Advanced Readiness — Precalculus & Trig

5.1 Exponential & Logarithmic Expressions 5.2 Radicals, Rational Exponents & Complex Numbers 5.3 Trigonometry Basics 5.4 Sequences, Functions Review & Exam Readiness

4.3 Geometry & Measurement

Key Takeaways Explain This Topic AI

Rectangle area = length × width; triangle area = ½ × base × height.

Circle area A = πr² and circumference C = 2πr (or πd), using r = radius.

The Pythagorean theorem a² + b² = c² relates the legs and hypotenuse of a right triangle.

Complementary angles sum to 90°; supplementary angles sum to 180°.

Similar triangles have equal angles and proportional sides; set up a ratio to solve for a missing length.

Last updated: June 2026

Perimeter, Area & Circles

Perimeter is the distance around a flat shape (add the sides); area is the space inside it (square units). Memorize this core table — it appears constantly on placement tests.

  • Shape Perimeter / Circumference Area
  • Rectangle P = 2l + 2w A = l × w
  • Square P = 4s A = s²
  • Triangle P = a + b + c A = ½ × base × height
  • Circle C = 2πr = πd A = πr²
  • Parallelogram P = 2(a + b) A = base × height
  • Trapezoid sum of 4 sides A = ½(b₁ + b₂) × height

For circles, r is the radius (center to edge) and the diameter d = 2r . Use π ≈ 3.14 unless told to leave the answer "in terms of π."

Worked example (rectangle). A garden is 12 m by 5 m. Area = 12 × 5 = 60 m² ; perimeter = 2(12) + 2(5) = 24 + 10 = 34 m .

Worked example (circle). A circle has radius 6 cm. Area = π(6)² = 36π ≈ 113.1 cm² ; circumference = 2π(6) = 12π ≈ 37.7 cm .

Worked example (triangle). A triangle has base 10 in and height 7 in. Area = ½ × 10 × 7 = 35 in² . The height must be perpendicular to the chosen base.

The Pythagorean Theorem & Solids

In a right triangle , the Pythagorean theorem relates the two legs a and b to the hypotenuse c (the side opposite the right angle):

a² + b² = c²

The hypotenuse is always the longest side and always sits alone on the c side of the equation.

Worked example. Legs 6 and 8; find the hypotenuse.

c² = 6² + 8² = 36 + 64 = 100 → c = √100 = 10 .

Worked example (solving for a leg). Hypotenuse 13, one leg 5; find the other leg.

5² + b² = 13² → 25 + b² = 169 → b² = 144 → b = 12 .

For three dimensions, volume measures the space inside a solid (cubic units) and surface area the total outside skin.

  • Solid Volume Surface area
  • Rectangular box V = l × w × h SA = 2(lw + lh + wh)
  • Cube V = s³ SA = 6s²
  • Cylinder V = πr²h SA = 2πr² + 2πrh
  • Sphere V = (4/3)πr³ SA = 4πr²
  • Cone V = (1/3)πr²h SA = πr² + πrℓ

Worked example (cylinder). A can has radius 3 cm and height 10 cm. Volume = π(3)²(10) = 90π ≈ 282.7 cm³ .

Angles & Similar Triangles

Two angle relationships show up repeatedly:

Complementary angles add to 90° . If one is 35°, its complement is 55°.

Supplementary angles add to 180° . If one is 110°, its supplement is 70°.

Also useful: the three interior angles of any triangle sum to 180° , and angles in a straight line sum to 180°.

Worked example. Two complementary angles are x and 2x. Then x + 2x = 90 → 3x = 90 → x = 30°, so the angles are 30° and 60°.

Similar triangles have the same shape but different size : corresponding angles are equal and corresponding sides are proportional . To find a missing side, set up a ratio of corresponding sides and cross-multiply.

Worked example. Triangle ABC is similar to triangle DEF. Side AB = 4 corresponds to DE = 6, and BC = 5 corresponds to EF = x. Set up the proportion:

4/6 = 5/x → 4x = 30 → x = 7.5 .

This is the basis of indirect measurement — e.g., using a shadow and a known height to find a flagpole's height.

Common mistake: Using diameter instead of radius in πr² — if a circle's diameter is 10, the radius is 5, so area = π(5)² = 25π, not π(10)². Another trap: in the Pythagorean theorem, plugging the hypotenuse in for a leg. The longest side is always c.

Unit Conversion & Composite Figures

Placement tests mix in measurement conversions . Keep these handy: 1 ft = 12 in, 1 yd = 3 ft, 1 m = 100 cm, 1 km = 1000 m. When converting areas , the factor is squared, and for volumes it is cubed — a 1 ft² tile is 12 × 12 = 144 in², not 12 in².

Worked example. A floor is 4 ft by 3 ft. Its area is 12 ft². In square inches that is 12 × 144 = 1,728 in² .

Composite figures are built from simpler shapes; find the area by adding or subtracting pieces.

Worked example. A 10 m by 6 m rectangle has a semicircle of diameter 6 m attached to one short end. Rectangle area = 60 m². The semicircle has radius 3, so its area = ½ · π(3)² = 4.5π ≈ 14.1 m². Total ≈ 74.1 m² .

Worked example (shaded region). A circle of radius 4 is inscribed in a square of side 8. The leftover (corner) area is the square minus the circle: 8² − π(4)² = 64 − 16π ≈ 64 − 50.3 = 13.7 square units .

When a problem asks for a shaded portion, identify whether you are adding shapes or subtracting a hole, and keep units consistent throughout. Round only at the final step to avoid accumulating error, and label your answer with the correct square or cubic units.