Geometry: Explaining and Applying the Pythagorean Theorem (CCSS.8.G.6)
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Common Core 8th Grade Math › Geometry: Explaining and Applying the Pythagorean Theorem (CCSS.8.G.6)
Two points on a coordinate plane are (0, 0) and (3, 4). What is the distance between the two points?
7
5
$\sqrt{7}$
25
Explanation
The horizontal leg is 3 and the vertical leg is 4, so distance $= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. Adding gives 7 (wrong), skipping the square makes 25 (wrong), and $\sqrt{7}$ comes from not squaring first.
Points $(-2, 5)$ and $(4, 1)$ are plotted on a coordinate plane. What is the distance between them?
10
$\sqrt{52}$
$2\sqrt{13}$
52
Explanation
Horizontal distance is $|4 - (-2)| = 6$ and vertical distance is $|1 - 5| = 4$. Distance $= \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$. Choices 10 (added legs), $\sqrt{52}$ (not simplified), and 52 (skipped the square root) are common errors.
Two points on a coordinate plane are (0, 0) and (3, 4). What is the distance between the two points?
7
$\sqrt{7}$
5
25
Explanation
The horizontal and vertical legs are $\Delta x=|3-0|=3$ and $\Delta y=|4-0|=4$. Using the Pythagorean Theorem on this right triangle: $d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.
Two points on a coordinate plane are (-2, 5) and (4, 1). What is the distance between the two points?
$2\sqrt{13}$
$\sqrt{10}$
52
$\sqrt{40}$
Explanation
The legs are the horizontal and vertical changes: $\Delta x=|4-(-2)|=6$, $\Delta y=|1-5|=4$. So $d=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$.
Two points on a coordinate plane are (7, -3) and (1, -3). What is the distance between the two points?
$\sqrt{6}$
36
0
6
Explanation
The horizontal leg is $\Delta x=|1-7|=6$ and the vertical leg is $\Delta y=|-3-(-3)|=0$. Then $d=\sqrt{6^2+0^2}=\sqrt{36}=6$.
Two points on a coordinate plane are (2, -1) and (-1, 5). What is the distance between the two points?
45
$3\sqrt{5}$
3
$\sqrt{15}$
Explanation
Use horizontal and vertical legs: $\Delta x=|-1-2|=3$, $\Delta y=|5-(-1)|=6$. Then $d=\sqrt{3^2+6^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}$.
Two points are at (0, 0) and (3, 4). What is the distance between the two points?
7
5
25
1
Explanation
The horizontal change is 3 and the vertical change is 4, so the right triangle's legs are 3 and 4. By the Pythagorean Theorem, $d=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. (7 comes from adding the legs; 25 comes from skipping the square root.) The legs are the horizontal and vertical distances on the grid.
Points are at (2, -1) and (-4, 5). What is the distance between the two points?
12
6
72
$6\sqrt{2}$
Explanation
The horizontal change is $|2-(-4)|=6$ and the vertical change is $|-1-5|=6$. Using the legs (horizontal and vertical distances), $d=\sqrt{6^2+6^2}=\sqrt{72}=6\sqrt{2}$. (12 comes from adding the legs; 6 is just one leg; 72 skips the square root.)
Find the distance between the points (-2, 3) and (4, 6).
$3\sqrt{5}$
9
45
$\sqrt{39}$
Explanation
The legs of the right triangle are the horizontal change $|-2-4|=6$ and vertical change $|3-6|=3$. Then $d=\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$. (9 adds the leg lengths; 45 skips the square root; $\sqrt{39}$ mis-squares or mis-adds.)
What is the distance between (1, -3) and (5, 2)?
9
41
$\sqrt{41}$
$3\sqrt{5}$
Explanation
The horizontal change is $|1-5|=4$ and the vertical change is $|-3-2|=5$, so the legs are 4 and 5. By the Pythagorean Theorem, $d=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}$. (41 skips the square root; 9 adds the legs; $3\sqrt{5}$ comes from mis-adding the squares to 45.) The horizontal and vertical differences form the right triangle's legs.