Overview

This chapter studies right triangles and quadrilaterals. You will learn how to use the Pythagorean theorem to find distances, how to work with common right triangles (special ratios and triples), the main types of quadrilaterals and their properties, and how to compute areas. You will also practice turning geometric descriptions into equations (modeling).

Pythagorean theorem (PT)

Statement: In a right triangle with legs a and b and hypotenuse c (opposite the right angle),

c² = a² + b²

Use this to find the third side, to test whether a triangle is right (the converse), and to derive the distance formula in the coordinate plane.

Worked example — find the hypotenuse

1. Given legs 6 and 8, compute c² = 6² + 8² = 36 + 64 = 100.
2. So c = √100 = 10. The hypotenuse is 10.

Converse (test if triangle is right)

1. Sides 7, 24, 25 — check largest side 25: is 25² = 7² + 24²? 625 = 49 + 576 = 625.
2. Yes — the triangle is right (with hypotenuse 25).

Distance formula (derivation from PT)

For two points (x1,y1) and (x2,y2), form a right triangle with horizontal leg |x2-x1| and vertical leg |y2-y1|. By PT, the distance d is

d = √((x2 - x1)² + (y2 - y1)²)

Distance example

Distance between (1,2) and (5,5): dx = 4, dy = 3, so distance = √(4² + 3²) = √(16 + 9) = √25 = 5.

Pythagorean triples

A Pythagorean triple is three positive integers (a,b,c) with a² + b² = c². Common triples you should know: (3,4,5), (5,12,13), (7,24,25), (8,15,17). If you multiply a triple by any positive integer k, you get another triple (e.g., 6,8,10 from 3,4,5).

Brief note: primitive triples (not multiples of smaller triples) can be generated by formulas, but for most problems you just memorize the common ones and know how to scale them.

Special right triangles

Two standard types have fixed side ratios, which lets you find missing sides quickly without PT calculations.

45°‑45°‑90° triangle

This is an isosceles right triangle. If each leg = s, then hypotenuse = s√2.

Example: legs 7 and 7 → hypotenuse = 7√2.

30°‑60°‑90° triangle

Ratios (short leg : long leg : hypotenuse) = 1 : √3 : 2. The short leg is opposite 30°, the long leg opposite 60°, and the hypotenuse opposite 90°.

Example: If hypotenuse = 10, short leg = 10/2 = 5, long leg = 5√3.

Types of quadrilaterals and key properties

A quadrilateral is a four-sided polygon. Main types and what to remember:

• Parallelogram: opposite sides parallel and equal; opposite angles equal; diagonals bisect each other. Area = base × height.
• Rectangle: parallelogram with right angles; diagonals equal. Area = length × width.
• Square: rectangle and rhombus — all sides equal and all angles 90°; area = side².
• Rhombus: all sides equal; opposite angles equal; diagonals are perpendicular bisectors of each other. Area = (d1 × d2)/2 where d1,d2 are diagonals.
• Trapezoid (US) / trapezium (some regions): one pair of parallel sides (bases). Area = (b1 + b2)/2 × height.
• Kite: two pairs of consecutive equal sides; diagonals are perpendicular; area = (d1 × d2)/2.

Quadrilateral area methods

Common formulas:

• Rectangle: A = l × w
• Parallelogram: A = base × height
• Trapezoid: A = (b1 + b2)/2 × h
• Rhombus / Kite: A = (d1 × d2)/2
• General quadrilateral: split into two triangles along a diagonal, find the two triangle areas, then add. In coordinate geometry you can also use the shoelace formula.

Worked example — trapezoid

1. Bases b1 = 8, b2 = 12, height = 5.
2. Area = (8 + 12)/2 × 5 = (20/2) × 5 = 10 × 5 = 50.

Worked example — rhombus from diagonals

1. Diagonals are 10 and 12, area = (10 × 12)/2 = 60.

Using Pythagorean theorem to find missing heights or diagonals

Often you need PT inside a quadrilateral problem. Example: For a rectangle 6 by 8, diagonal = √(6² + 8²) = 10. For an isosceles trapezoid you can drop perpendiculars to form right triangles and use PT to find the height.

Translating geometry into equations (A‑CED.1) — modeling examples

Geometry problems often become equations you can solve. Practice writing the equation before solving.

Ladder problem (write an equation then solve)

A ladder leans against a wall. Its foot is 5 ft from the wall and its top reaches 12 ft high. How long is the ladder?

1. Set up with PT: length² = 5² + 12² → L² = 25 + 144 = 169.
2. Solve: L = √169 = 13 ft.

Area leads to an equation

A trapezoid's area is 60, bases are 8 and x, height is 4. Find x.

1. Area formula: (8 + x)/2 × 4 = 60.
2. Solve: (8 + x) × 2 = 60 → 8 + x = 30 → x = 22.

How this maps to the standards

• 8.G.5 — Use the Pythagorean theorem to find distances between points in the plane: this is exactly the PT and distance formula material and the coordinate examples.
• 8.G.7 — Apply the Pythagorean theorem to solve problems involving right triangles and use reasoning about right triangles in proofs: examples include using PT to find heights/diagonals and using the converse to determine right triangles.
• A‑CED.1 — Translate situations into equations: ladder, area, and length problems in this chapter are modeled by equations (see the ladder and trapezoid examples).

Study tips and common pitfalls

• Always identify which side is the hypotenuse before applying PT (it is the longest side).
• For special triangles, remember which side corresponds to 30°, 45°, or 60° (shortest opposite 30°, equals opposite 45°s).
• When a quadrilateral is not one of the standard types, try splitting it into triangles — area often becomes easier that way.
• Practice converting word problems into an equation first; label lengths on a diagram to avoid mistakes.

If you want, I can give you practice problems with step-by-step solutions or a short quiz that focuses on the parts you find hardest. Which would you prefer?