Math 1 Quiz: Multi Step Geometric Problems

5 questions · exam conditions

0:00

Multi Step Geometric ProblemsQuestion 1 of 5

A regular hexagonal prism has a base edge length of 4 and height of 6. A diagonal is drawn from one vertex of the top base to the opposite vertex of the bottom base. What is the length of this diagonal?

10

2\sqrt{31}

\sqrt{148}

2\sqrt{37}

← Back to quizzes

Math 1 Quiz

Math 1 Quiz: Multi Step Geometric Problems

Practice Multi Step Geometric Problems in Math 1 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 Multi Step Geometric Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

$10$ (correct answer)
$2\sqrt{31}$
$\sqrt{148}$
$2\sqrt{37}$

Explanation: In a regular hexagon with edge length 4, the distance from one vertex to the opposite vertex (across the center) is 2 × 4 = 8. This can be seen because a regular hexagon can be divided into 6 equilateral triangles, and the distance across opposite vertices is twice the edge length. Now consider the hexagonal prism: if we take a vertex on the top base and the opposite vertex on the bottom base, we have a right triangle where one leg is the distance between opposite vertices of the hexagon (8), the other leg is the height of the prism (6), and the hypotenuse is the desired diagonal. Using the Pythagorean theorem: diagonal = √(8² + 6²) = √(64 + 36) = √100 = 10.

Question 2

A cylindrical water tank has a radius of 6 feet and a height of 10 feet. A smaller cylindrical tank with radius 3 feet is placed inside the larger tank such that both tanks share the same central axis. If water is poured into the space between the two tanks until the water level reaches 8 feet high, what is the volume of water in the tank? (Use $\pi ≈ 3.14$ )

678.24 cubic feet (correct answer)
904.32 cubic feet
1130.4 cubic feet
1356.48 cubic feet

Explanation: The volume of water equals the volume of the annular space up to 8 feet high. This is the volume of the larger cylinder minus the volume of the smaller cylinder, both with height 8 feet. Volume = π(6²)(8) - π(3²)(8) = π(36)(8) - π(9)(8) = 288π - 72π = 216π ≈ 216(3.14) = 678.24 cubic feet. Choice B uses the full 10-foot height instead of 8 feet. Choice C calculates only the larger cylinder's volume to 8 feet. Choice D uses the full height and doesn't subtract the inner cylinder.

Question 3

A spherical balloon with radius 9 inches is placed inside a cubic box. The balloon touches all six faces of the cube. A second, smaller spherical balloon with radius 3 inches is then placed in the remaining space inside the cube. What is the maximum possible distance between the centers of the two balloons?

6 inches
$6\sqrt{2}$ inches
$6\sqrt{3}$ inches (correct answer)
$9\sqrt{2}$ inches

Explanation: Since the large balloon touches all six faces of the cube, the cube has side length 18 inches (diameter of the large balloon), and the large balloon's center is at the cube's center. The small balloon has radius 3 inches, so its center must be at least 3 inches from any face of the cube. The maximum distance occurs when the small balloon is positioned at a corner region of the cube. The small balloon's center will be at coordinates (3,3,3) relative to a corner of the cube, which places it at coordinates (3,3,3) if we put the corner at the origin and the cube center at (9,9,9). The distance from the cube center (9,9,9) to the small balloon center (3,3,3) is √[(9-3)² + (9-3)² + (9-3)²] = √(6² + 6² + 6²) = √(108) = 6√3 inches.

Question 4

A regular octagon is inscribed in a circle with radius 10 units. Four alternating vertices of the octagon are connected to form a square. What is the area of the region inside the circle but outside this inscribed square?

$100\pi - 200$ square units (correct answer)
$100\pi - 100\sqrt{2}$ square units
$100\pi - 100$ square units
$100\pi - 50\sqrt{2}$ square units

Explanation: The circle has area π(10)² = 100π square units. For a regular octagon inscribed in a circle of radius 10, the vertices are at angles 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°. Taking alternating vertices (0°, 90°, 180°, 270°) forms a square. These vertices are at coordinates (10,0), (0,10), (-10,0), (0,-10), forming a square with vertices on the circle. The side length of this square is the distance between adjacent vertices: √[(10-0)² + (0-10)²] = √(100+100) = 10√2. The area of the square is (10√2)² = 200 square units. The area inside the circle but outside the square is 100π - 200. Choice B incorrectly calculates the square's side length, Choice C uses an incorrect square area, Choice D uses half the correct square area.

Question 5

A regular tetrahedron has edge length 6. A sphere is inscribed in the tetrahedron such that it touches all four faces. What is the radius of the inscribed sphere?

$\frac{3\sqrt{6}}{2}$
$\frac{\sqrt{6}}{2}$ (correct answer)
$\frac{3\sqrt{6}}{4}$
$\frac{\sqrt{6}}{4}$

Explanation: For a regular tetrahedron with edge length a, the radius of the inscribed sphere is r = a√6/12. With edge length 6, r = 6√6/12 = √6/2. To verify: the volume of a regular tetrahedron with edge length a is V = a³√2/12. With a = 6, V = 216√2/12 = 18√2. The surface area is 4 times the area of an equilateral triangle with side 6: SA = 4 × (√3/4) × 6² = 36√3. Using V = (1/3) × SA × r: 18√2 = (1/3) × 36√3 × r, so 18√2 = 12√3 × r, giving r = 18√2/(12√3) = 3√2/(2√3) = 3√2√3/(2×3) = √6/2.