Grade 8 Finding Distance with the Pythagorean Theorem

Introduction

Finding Distance with the Pythagorean Theorem is an important Grade 8 math skill because students are moving from simple answers toward explaining how the math works.

In this lesson, students use models, real questions, worked examples, practice problems, and two online quizzes to build confidence with finding distance with the pythagorean theorem.

What Is Finding Distance with the Pythagorean Theorem?

Finding Distance with the Pythagorean Theorem means choosing a model, naming what each number means, and explaining the strategy.

The goal is not only to get the answer. Students should be able to show the idea, explain the strategy, and check whether the answer makes sense.

Understanding Finding Distance with the Pythagorean Theorem

Before solving, students should slow down and decide what each number, shape, unit, or label represents.

Read the question carefully and identify what is being asked.
Choose a model, equation, table, or diagram that matches the situation.
Solve one step at a time and keep units or labels attached.
Use the answer explanation to check that the result makes sense.

Visual Models

Visual Model 1

Question: A right triangle is shown on a coordinate grid. The legs of the triangle are horizontal and vertical, with lengths 6 and 8 units. What is the length of the hypotenuse? Which equation correctly uses the Pythagorean Theorem to find the hypotenuse? \Choices{\(c^2 = 6^2 + 8^2 = 100\), so \(c = 10\)}{\(c = 6 + 8 = 14\)}{\(c^2 = 6 + 8 = 14\)}{\(c = \sqrt{6 + 8} = \sqrt{14}\)}

Visual Model 1

Why it works: By the Pythagorean Theorem, \(c^2 = 6^2 + 8^2 = 36 + 64 = 100\). Taking the square root: \(c = \sqrt{100} = 10\).

Answer: \(10\)

Visual Model 2

Question: Two points are plotted on a coordinate grid: \(P = (0,0)\) and \(Q = (5,12)\). What is the distance from \(P\) to \(Q\) rounded to the nearest tenth? \Choices{\(12.9\)}{\(13.0\)}{\(17.0\)}{\(8.5\)}

Visual Model 2

Why it works: Distance \(= \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.0\). (This is a 5-12-13 Pythagorean triple.)

Answer: \(13.0\)

Worked Examples

Example 1

Question: What is the distance between \((-2,3)\) and \((1,7)\)? \Choices{\(5\)}{\(25\)}{\(\sqrt{21}\)}{\(4\)}

Example 1

\(d = \sqrt{(1-(-2))^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\).

Answer: \(5\)

Example 2

Question: On a coordinate grid, point \(A\) is at \((2,5)\) and point \(B\) is at \((8,13)\). Use the distance formula to find the distance from \(A\) to \(B\). \Choices{\(10\)}{\(14\)}{\(\sqrt{90}\)}{\(8\)}

Example 2

\(d = \sqrt{(8-2)^2 + (13-5)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10\).

Answer: \(10\)

Example 3

Question: Which of the following represents the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\)? \Choices{\(d = |x_2 - x_1| + |y_2 - y_1|\)}{\(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)}{\(d = (x_2-x_1)^2+(y_2-y_1)^2\)}{\(d = x_2 - x_1 + y_2 - y_1\)}

Example 3

The distance formula comes from the Pythagorean Theorem applied to the horizontal and vertical legs of a right triangle.

Answer: \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Real-World Word Problems

Problem 1

Question: A student calculates the distance between \((0,0)\) and \((8,6)\) and gets \(\sqrt{100}\). Is this correct? If correct, simplify. If incorrect, what should it be? \Choices{Correct; \(\sqrt{100} = 10\)}{Incorrect; distance is \(14\)}{Incorrect; distance is \(\sqrt{14}\)}{Correct; it stays \(\sqrt{100}\)}

Why it works: Distance \(= \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10\).

Answer: Correct; \(10\)

Problem 2

Question: A delivery truck travels from warehouse \(W = (1,2)\) to store \(S = (9,8)\). What is the distance in units? \Choices{\(10\) units}{\(8\) units}{\(14\) units}{\(\sqrt{90}\) units}

Problem 2

Why it works: \(d = \sqrt{(9-1)^2+(8-2)^2} = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10\) units.

Answer: \(10\) units

Common Mistakes

Rushing before identifying what the numbers represent.
Choosing an operation that does not match the situation.
Dropping labels, units, or context from the answer.
Skipping the estimate or reasonableness check.

Strategy Tips

Underline the question being asked.
Use a model before jumping to computation.
Write an equation that matches the story or picture.
Explain the final answer in a sentence.

Practice Questions

Question 1

What is the distance between the points \((1,2)\) and \((4,6)\)? \Choices{\(5\)}{\(7\)}{\(\sqrt{7}\)}{\(\sqrt{26}\)}

Question 2

A hiker walks from camp at \((0,0)\) to a landmark at \((3,4)\) on a map where each unit represents 1 mile. How many miles did the hiker walk in a straight line? \Choices{\(7\) miles}{\(5\) miles}{\(\sqrt{7}\) miles}{\(12\) miles}

Question 3

A right triangle has legs of 9 units and 12 units. What is the length of the hypotenuse? \Choices{\(15\) units}{\(21\) units}{\(\sqrt{200}\) units}{\(3\) units}

Question 3

Question 4

Two points are plotted on a coordinate grid at \((1,1)\) and \((7,7)\). What is the distance between them, simplified?

Question 4

A. \(6\sqrt{2}\) units
B. \(12\) units
C. \(\sqrt{72}\) units
D. \(36\) units

Question 5

Points \(C\) and \(D\) are on a coordinate grid with \(C = (0,0)\) and \(D = (7,24)\). What is the distance from \(C\) to \(D\)? \Choices{\(25\)}{\(31\)}{\(\sqrt{600}\)}{\(17\)}

Question 6

On a map, a town is located at \((2,2)\) and a city is located at \((5,6)\). If each grid unit represents 2 kilometers, what is the straight-line distance between the town and city? \Choices{\(5\) km}{\(10\) km}{\(20\) km}{\(2.5\) km}

Question 6

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