Testinar Grade 8 math

Finding Distance with the Pythagorean Theorem

Learn to reason about and solve Grade 8 problems involving Finding Distance with the Pythagorean Theorem.

CH05-07 Geometry and Measurement

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Score: ____ / 8

Directions Solve each problem. Show your reasoning on the page or on separate paper. Check the answer key only after finishing.

Two points are plotted on a coordinate grid at \((1,1)\) and \((7,7)\). What is the distance between them, simplified?
1. \(6\sqrt{2}\) units
2. \(12\) units
3. \(\sqrt{72}\) units
4. \(36\) units
Points \(E = (-1,-2)\) and \(F = (3,1)\) are plotted on a coordinate grid. Calculate the distance between them.
1. \(5\)
2. \(7\)
3. \(\sqrt{26}\)
4. \(3\)
Three points form a right triangle: \((0,0)\), \((5,0)\), and \((5,12)\). What is the length of the hypotenuse (the line segment from \((0,0)\) to \((5,12)\))?
1. \(13\)
2. \(17\)
3. \(\sqrt{170}\)
4. \(12\)
An airplane flies from airport \(A = (3,1)\) to airport \(B = (11,16)\). Both coordinates are in hundreds of kilometers. What is the flight distance in hundreds of kilometers?
1. \(17\) (hundred km)
2. \(24\) (hundred km)
3. \(\sqrt{300}\) (hundred km)
4. \(25\) (hundred km)
A student is finding the distance between points \((3,0)\) and \((0,4)\) and writes: \(d = 3 + 4 = 7\). What is the error?
1. They should square the differences before adding them.
2. The answer is actually correct.
3. They forgot to take the square root.
4. They added instead of subtracting the coordinates.
A right triangle sits on a coordinate plane with vertices at \((0,0)\), \((9,0)\), and \((0,12)\). What is the length of the hypotenuse?
1. \(15\)
2. \(21\)
3. \(\sqrt{200}\)
4. \(3\)
What is the distance between \((1,1)\) and \((13,6)\)?
1. \(7\)
2. \(17\)
3. \(13\)
4. \(\sqrt{119}\)
A hiker's GPS shows the horizontal distance traveled is 24 units and the vertical distance is 7 units. Using the Pythagorean Theorem, what is the total straight-line distance?
1. \(31\) units
2. \(\sqrt{625}\) units
3. \(25\) units
4. \(\sqrt{500}\) units

Teacher and parent reference

Answer key and explanations

A. \(6\sqrt{2}\) units
\(d = \sqrt{(7-1)^2 + (7-1)^2} = \sqrt{6^2+6^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}\) units. (Equivalent: \(\sqrt{72} = 6\sqrt{2}\).)
A. \(5\)
\(d = \sqrt{(3-(-1))^2 + (1-(-2))^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25} = 5\).
A. \(13\)
\(d = \sqrt{(5-0)^2+(12-0)^2} = \sqrt{25+144} = \sqrt{169} = 13\).
A. \(17\) (hundred km)
\(d = \sqrt{(11-3)^2+(16-1)^2} = \sqrt{8^2+15^2} = \sqrt{64+225} = \sqrt{289} = 17\) (hundred km).
A. They should square the differences before adding them.
Correct method: \(d = \sqrt{(3-0)^2+(0-4)^2} = \sqrt{9+16} = \sqrt{25} = 5\) units. The student used addition instead of the Pythagorean Theorem.
A. \(15\)
\(c = \sqrt{9^2+12^2} = \sqrt{81+144} = \sqrt{225} = 15\). (This is a 9-12-15 triple, a multiple of 3-4-5.)
C. \(13\)
\(d = \sqrt{(13-1)^2+(6-1)^2} = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169} = 13\). (This is a 5-12-13 Pythagorean triple.)
C. \(25\) units
\(d = \sqrt{24^2+7^2} = \sqrt{576+49} = \sqrt{625} = 25\) units. (This is a 7-24-25 Pythagorean triple.)