ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Pythagorean Theorem
16 questions · exam conditions
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Pythagorean TheoremQuestion 1 of 16

A rectangular box has a length of 12 cm, a width of 4 cm, and a height of 3 cm. What is the length of the space diagonal connecting two opposite corners of the box?

5
13
19
169
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Pythagorean Theorem

Practice Pythagorean Theorem in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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

A rectangular box has a length of 12 cm, a width of 4 cm, and a height of 3 cm. What is the length of the space diagonal connecting two opposite corners of the box?

  1. 5
  2. 13 (correct answer)
  3. 19
  4. 169
Explanation: The space diagonal DD can be found using the formula D2=l2+w2+h2D^2 = l^2 + w^2 + h^2. Substituting the given values: D2=122+42+32=144+16+9=169D^2 = 12^2 + 4^2 + 3^2 = 144 + 16 + 9 = 169. Taking the square root of both sides gives D=169=13D = \sqrt{169} = 13 cm. Alternatively, one could first find the diagonal of the base (dbase2=122+42=160d_{base}^2 = 12^2+4^2=160) and then use that with the height (D2=160+32=169D^2 = 160 + 3^2 = 169).

Question 2

A 25-foot ladder is leaning against a vertical wall, with its base 7 feet from the wall. If the base is then pulled 8 feet further away from the wall, how many feet does the top of the ladder slide down the wall?

  1. 4 (correct answer)
  2. 8
  3. 20
  4. 24
Explanation: This is a two-step problem. Initially, the ladder forms a right triangle with hypotenuse 25 and base 7. The initial height h1h_1 is 252−72=625−49=576=24\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 feet. When the base is pulled 8 feet further, the new base is 7+8=157 + 8 = 15 feet. The new height h2h_2 is 252−152=625−225=400=20\sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20 feet. The distance the ladder slides down is the difference in heights: h1−h2=24−20=4h_1 - h_2 = 24 - 20 = 4 feet.

Question 3

A 12-meter tall pole is supported by a wire that runs from the top of the pole to a point on the ground 5 meters from the base. From that same ground point, the wire continues to an anchor point 4 meters farther away. What is the total length of wire needed?

  1. 13
  2. 15
  3. 17 (correct answer)
  4. 21
Explanation: The wire has two segments. The first segment goes from the top of the pole to the ground point, forming a right triangle with legs of 12 m (pole height) and 5 m (ground distance). Using the Pythagorean theorem: 122+52=144+25=169=13\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 meters. The second segment runs along the ground for 4 meters. Total wire length: 13+4=1713 + 4 = 17 meters.

Question 4

In triangle ABC, angle C is a right angle. If the altitude from C to the hypotenuse AB has length 6, and one of the segments of the hypotenuse created by this altitude is 4, what is the length of the other segment?

  1. 8
  2. 9 (correct answer)
  3. 12
  4. 10
Explanation: When an altitude is drawn from the right angle to the hypotenuse in a right triangle, it creates the relationship: altitude² = segment₁ × segment₂. Given altitude = 6 and one segment = 4: 6² = 4 × segment₂, so 36 = 4 × segment₂, therefore segment₂ = 9. Choice A (8) might result from incorrectly using 6² - 4² = 36 - 16 = 20, then guessing. Choice C (12) might come from incorrectly thinking altitude × segment = 6 × 4 = 24, then dividing by 2. Choice D (10) might result from using the Pythagorean theorem incorrectly on the segments themselves.

Question 5

A rectangular box has dimensions where the length is twice the width, and the height equals the width. If the space diagonal of the box (from one corner to the opposite corner) measures 6√6 units, what is the width of the box?

  1. 6 units (correct answer)
  2. 4 units
  3. 3 units
  4. 2 units
Explanation: Let width = w, then length = 2w and height = w. The space diagonal of a rectangular box uses the 3D Pythagorean theorem: d² = length² + width² + height². So (6√6)² = (2w)² + w² + w² = 4w² + w² + w² = 6w². Therefore 216 = 6w², so w² = 36, and w = 6. Choice B (4) results from solving 4w² = 216 instead of 6w². Choice C (3) comes from forgetting to take the square root: w² = 36 but choosing w = 3. Choice D (2) might result from incorrectly calculating (6√6)² as 72 instead of 216.

Question 6

A right triangle has one leg of length 5 and a hypotenuse of length 10. What is the length of the other leg?

  1. 5\sqrt{5}
  2. 5
  3. 555\sqrt{5}
  4. 535\sqrt{3} (correct answer)
Explanation: Let the legs be aa and bb, and the hypotenuse be cc. We have a=5a=5 and c=10c=10. Using the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2, we get 52+b2=1025^2 + b^2 = 10^2, which is 25+b2=10025 + b^2 = 100. Subtracting 25 gives b2=75b^2 = 75. The length of the other leg is b=75=25×3=53b = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}.

Question 7

Which of the following sets of side lengths forms a right triangle?

  1. 6, 8, 9
  2. 7, 11, 13
  3. 10, 15, 20
  4. 9, 40, 41 (correct answer)
Explanation: The converse of the Pythagorean theorem states that if a2+b2=c2a^2 + b^2 = c^2, where c is the longest side, the triangle is a right triangle. We test each option: A) 62+82=36+64=100≠92=816^2+8^2=36+64=100 \neq 9^2=81. B) 72+112=49+121=170≠132=1697^2+11^2=49+121=170 \neq 13^2=169. C) 92+402=81+1600=16819^2+40^2=81+1600=1681, and 412=168141^2=1681. Since they are equal, this forms a right triangle. D) 102+152=100+225=325≠202=40010^2+15^2=100+225=325 \neq 20^2=400.

Question 8

A right triangle has a hypotenuse of length 26 and one leg of length 10. What is the perimeter of the triangle?

  1. 36
  2. 52
  3. 60 (correct answer)
  4. 120
Explanation: Let the legs be a and b, and the hypotenuse be c. We are given a=10a=10 and c=26c=26. According to the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2. So, 102+b2=26210^2 + b^2 = 26^2, which simplifies to 100+b2=676100 + b^2 = 676. Subtracting 100 from both sides gives b2=576b^2 = 576. Taking the square root, we find b=24b = 24. The perimeter is the sum of the side lengths: P=a+b+c=10+24+26=60P = a + b + c = 10 + 24 + 26 = 60.

Question 9

What is the distance between the points (−3,4)(-3, 4) and (5,−2)(5, -2) on the coordinate plane?

  1. 6
  2. 8
  3. 10 (correct answer)
  4. 14
Explanation: The distance formula is an application of the Pythagorean theorem: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. The change in x is 5−(−3)=85 - (-3) = 8 and the change in y is −2−4=−6-2 - 4 = -6. These are the legs of the right triangle. The distance (hypotenuse) is d=82+(−6)2=64+36=100=10d = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.

Question 10

The lengths of the legs of a right triangle are pp and qq. If p2+q2=100p^2 + q^2 = 100, what is the length of the hypotenuse?

  1. 10 (correct answer)
  2. 20
  3. 50
  4. 100
Explanation: According to the Pythagorean theorem, if the legs of a right triangle are pp and qq and the hypotenuse is cc, then p2+q2=c2p^2 + q^2 = c^2. The problem states that p2+q2=100p^2 + q^2 = 100. Therefore, c2=100c^2 = 100. To find the length of the hypotenuse cc, we take the square root of both sides: c=100=10c = \sqrt{100} = 10.

Question 11

A circle has a radius of 9 cm. From a point P, which is 41 cm from the center of the circle, a line is drawn tangent to the circle at point T. What is the length of the tangent segment PT?

  1. 32
  2. 40 (correct answer)
  3. 50
  4. 1600
Explanation: The radius to the point of tangency (T) is perpendicular to the tangent line (PT). This forms a right triangle with vertices at the center of the circle (C), point P, and point T. The hypotenuse is the distance from the center to P (CP = 41 cm). One leg is the radius (CT = 9 cm). The other leg is the tangent segment PT. Using the Pythagorean theorem: (PT)2+92=412⇒(PT)2+81=1681⇒(PT)2=1600⇒PT=40(PT)^2 + 9^2 = 41^2 \Rightarrow (PT)^2 + 81 = 1681 \Rightarrow (PT)^2 = 1600 \Rightarrow PT = 40 cm.

Question 12

An isosceles triangle has two congruent sides of length 10 inches and a base of 12 inches. What is the area of the triangle?

  1. 24
  2. 48 (correct answer)
  3. 60
  4. 96
Explanation: To find the area, first find the height. The altitude to the base of an isosceles triangle bisects the base, creating two right triangles. Each has a hypotenuse of 10 and a leg of 12/2=612/2 = 6. The height hh is the other leg. h2+62=102⇒h2+36=100⇒h2=64⇒h=8h^2 + 6^2 = 10^2 \Rightarrow h^2 + 36 = 100 \Rightarrow h^2 = 64 \Rightarrow h=8. The area is A=12×base×height=12×12×8=48A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 8 = 48 square inches.

Question 13

The lengths of the legs of a right triangle are represented by xx and x−7x-7. The length of the hypotenuse is x+2x+2. What is the length of the hypotenuse?

  1. 8
  2. 15
  3. 17 (correct answer)
  4. 40
Explanation: By the Pythagorean theorem, x2+(x−7)2=(x+2)2x^2 + (x-7)^2 = (x+2)^2. Expanding the terms gives x2+(x2−14x+49)=(x2+4x+4)x^2 + (x^2 - 14x + 49) = (x^2 + 4x + 4). Combining like terms, we get 2x2−14x+49=x2+4x+42x^2 - 14x + 49 = x^2 + 4x + 4. Rearranging into a standard quadratic form gives x2−18x+45=0x^2 - 18x + 45 = 0. Factoring the quadratic yields (x−3)(x−15)=0(x-3)(x-15) = 0, so x=3x=3 or x=15x=15. If x=3x=3, one leg would be 3−7=−43-7=-4, which is impossible. So, x=15x=15. The hypotenuse is x+2=15+2=17x+2 = 15+2 = 17.

Question 14

Two friends start at the same point. One walks east at 3 miles per hour, and the other walks north at 4 miles per hour. After 2 hours, how many miles apart are they?

  1. 5
  2. 7
  3. 10 (correct answer)
  4. 14
Explanation: First, calculate the distance each friend has traveled. The friend walking east travels 3 mph×2 hours=63 \text{ mph} \times 2 \text{ hours} = 6 miles. The friend walking north travels 4 mph×2 hours=84 \text{ mph} \times 2 \text{ hours} = 8 miles. Their paths form the legs of a right triangle. The distance between them is the hypotenuse, dd. d2=62+82=36+64=100d^2 = 6^2 + 8^2 = 36 + 64 = 100. So, d=100=10d = \sqrt{100} = 10 miles.

Question 15

In right triangle PQR with right angle at Q, the median from Q to the hypotenuse PR has length 5. What is the length of the hypotenuse PR?

  1. 8
  2. 10 (correct answer)
  3. 12
  4. 15
Explanation: In a right triangle, the median from the right angle to the hypotenuse has a special property: it equals half the length of the hypotenuse. This is because the median to the hypotenuse in a right triangle is the radius of the circumcircle, and the hypotenuse is the diameter. Given that the median length is 5, the hypotenuse length is 2 × 5 = 10. Choice A (8) might result from incorrectly thinking the median is 5/8 of the hypotenuse. Choice C (12) could come from incorrectly adding: 5 + 7 = 12, where 7 is some mistaken calculation. Choice D (15) might result from incorrectly thinking the median is 1/3 of the hypotenuse.

Question 16

A right circular cone has a height of 12 units and a base radius of 5 units. What is the slant height of the cone?

  1. 11 units
  2. 13 units (correct answer)
  3. 17 units
  4. 15 units
Explanation: The slant height, height, and radius of a cone form a right triangle where the slant height is the hypotenuse. Using the Pythagorean theorem: slant height² = height² + radius² = 12² + 5² = 144 + 25 = 169. Therefore, slant height = √169 = 13. This uses the well-known 5-12-13 Pythagorean triple. Choice A (11) might result from incorrectly calculating √(144 - 25) = √119 ≈ 10.9 ≈ 11. Choice C (17) could come from adding 12 + 5 = 17. Choice D (15) might result from calculation errors or confusing with other Pythagorean triples.