Middle School Math Quiz: Pythagorean Theorem Distance

2 questions · exam conditions

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Pythagorean Theorem DistanceQuestion 1 of 2

A square has vertices at $A(1, 3)$ , $B(5, 1)$ , $C(3, -3)$ , and $D(-1, -1)$ . Point $P$ is inside the square such that the sum of the distances from $P$ to all four vertices is minimized. If $P$ has coordinates $(a, b)$ , what is the value of $a + b$ ?

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Middle School Math Quiz

Middle School Math Quiz: Pythagorean Theorem Distance

Practice Pythagorean Theorem Distance in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Pythagorean Theorem Distance, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$1$
$2$ (correct answer)
$3$
$4$

Explanation: The point that minimizes the sum of distances to all vertices of a polygon is the centroid (center of mass). For a square, this is simply the intersection of the diagonals, which is the average of all vertex coordinates. Centroid:

P = \left(\frac{1+5+3+(-1)}{4}, \frac{3+1+(-3)+(-1)}{4}\right) = \left(\frac{8}{4}, \frac{0}{4}\right) = (2, 0)

. Therefore

a = 2

and

b = 0

, so

a + b = 2

. Choice A would result from calculation errors in finding the centroid. Choice C might come from using only three vertices. Choice D could result from adding coordinates incorrectly or using a different approach like the circumcenter.

Question 2

A rectangular garden has vertices at $P(2, 1)$ , $Q(8, 1)$ , $R(8, 5)$ , and $S(2, 5)$ . A water sprinkler is placed at point $T(5, 7)$ . What is the distance from the sprinkler to the closest corner of the garden?

$\sqrt{13}$ units (correct answer)
$2\sqrt{5}$ units
$\sqrt{17}$ units
$3\sqrt{2}$ units

Explanation: We need to find the distance from T(5,7) to each corner: to P(2,1):

\sqrt{(5-2)^2 + (7-1)^2} = \sqrt{9+36} = \sqrt{45} = 3\sqrt{5}

; to Q(8,1):

\sqrt{(5-8)^2 + (7-1)^2} = \sqrt{9+36} = 3\sqrt{5}

; to R(8,5):

\sqrt{(5-8)^2 + (7-5)^2} = \sqrt{9+4} = \sqrt{13}

; to S(2,5):

\sqrt{(5-2)^2 + (7-5)^2} = \sqrt{9+4} = \sqrt{13}

. The smallest distance is

\sqrt{13}

. Choice B equals

\sqrt{20}

which is larger than

\sqrt{13}

. Choice C equals

\sqrt{17}

which is also larger. Choice D equals

\sqrt{18}

which is larger than

\sqrt{13}