AMC 10 · 2002 · #24
Grade 8 geometry-2dRiders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius 20 feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point 10 vertical feet above the bottom?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A Ferris wheel is a circle of radius $20$ feet that spins at one full turn per minute. A rider starts at the very bottom. Find how many seconds pass before the rider is $10$ feet higher than the bottom.
Givens: The wheel is a circle with radius $20$ feet; It turns at a steady rate of one full revolution per minute (so one turn takes $60$ seconds); The rider begins at the lowest point of the wheel; Answer choices: (A) $5$, (B) $6$, (C) $7.5$, (D) $10$, (E) $15$
Unknowns: The number of seconds until the rider is $10$ vertical feet above the bottom
Understand
Restated: A Ferris wheel is a circle of radius $20$ feet that spins at one full turn per minute. A rider starts at the very bottom. Find how many seconds pass before the rider is $10$ feet higher than the bottom.
Givens: The wheel is a circle with radius $20$ feet; It turns at a steady rate of one full revolution per minute (so one turn takes $60$ seconds); The rider begins at the lowest point of the wheel; Answer choices: (A) $5$, (B) $6$, (C) $7.5$, (D) $10$, (E) $15$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #8 Analyze the Units
The question mixes a spinning circle with a straight-up height, so Tool #1 (Draw a Diagram) is the anchor: sketch the wheel as a circle, mark the center, the bottom point, and the rider. That picture turns 'height above the bottom' into a length inside a right triangle. Tool #7 (Identify Subproblems) splits the work into two clean pieces — first find the angle the wheel has turned through, then convert that angle into time. Tool #4 (Introduce a Variable) names the rider's sideways distance so the Pythagorean theorem can pin the point down. Finally Tool #8 (Analyze the Units) carries the angle over to seconds, since a constant one-turn-per-minute rate means an equal share of the angle is an equal share of the $60$ seconds.
Execute — Answer: D
7.G.B.4 Step 1 Draw the wheel and place the points
- Sketch the wheel as a circle.
- Its center $O$ sits $20$ feet above the ground-level bottom point, because the radius is $20$.
- Call the bottom point $B$ (height $0$) and the rider's target point $P$ (height $10$).
- Both $OB$ and $OP$ are radii, so $OB = OP = 20$.
- Set up simple coordinates with $B$ at $(0,0)$; then the center is $O=(0,20)$, and $P$ sits somewhere on the circle at height $10$, so $P=(x,10)$ for some sideways distance $x$.
💡 The center of the wheel is exactly one radius above the bottom, which turns 'height' into lengths you can measure.
8.G.B.7 Step 2 Find where the rider is
- Point $P=(x,10)$ lies on the circle, so its distance from the center $O=(0,20)$ must equal the radius $20$.
- The horizontal gap is $x$ and the vertical gap is $20-10=10$.
- The Pythagorean theorem says $x^2 + 10^2 = 20^2$, so $x^2 = 400 - 100 = 300$ and $x = \sqrt{300} = 10\sqrt{3}$.
- That locates the rider exactly: $P = (10\sqrt{3},\,10)$.
💡 Any point on the wheel is one radius from the center, and the Pythagorean theorem cashes that fact into an exact position.
8.G.A.5 Step 3 Measure the angle turned
- Now measure the straight chord from the bottom $B=(0,0)$ to the rider $P=(10\sqrt{3},10)$.
- Its length is $BP = \sqrt{(10\sqrt{3})^2 + 10^2} = \sqrt{300 + 100} = \sqrt{400} = 20$.
- So all three sides of triangle $OBP$ equal $20$: $OB = OP = BP = 20$.
- A triangle with three equal sides is equilateral, and every angle in an equilateral triangle is $60^\circ$.
- The angle at the center, $\angle BOP$, is exactly the angle the wheel has turned through, so the wheel has spun $60^\circ$.
💡 When a chord is as long as the radius, the slice it cuts is an equilateral triangle, so the center angle is $60^\circ$.
7.RP.A.2 Step 4 Turn the angle into seconds
- A full turn is $360^\circ$ and takes $60$ seconds at the steady rate.
- The rider has swept $60^\circ$, which is $\frac{60}{360} = \frac16$ of a full turn.
- Because the speed is constant, the rider used $\frac16$ of the time as well: $\frac16 \times 60 = 10$ seconds.
- The answer is (D).
💡 At a steady spin, the same fraction of the circle you turn is the same fraction of the minute you spend.
7.G.B.4 Sketch the wheel as a circle. Its center $O$ sits $20$ feet above the ground-lev 8.G.B.7 Point $P=(x,10)$ lies on the circle, so its distance from the center $O=(0,20)$ 8.G.A.5 Now measure the straight chord from the bottom $B=(0,0)$ to the rider $P=(10\sqr 7.RP.A.2 A full turn is $360^\circ$ and takes $60$ seconds at the steady rate. The rider Review
Reasonableness: Check the size of the answer. Half the wheel (from bottom all the way to the top, a rise of $40$ feet) is $180^\circ$ and would take $30$ seconds. A rise of only $10$ feet is far less than halfway up, so the time should be much less than $30$ seconds, and $10$ seconds fits. The traps make sense too: $15$ (E) is a quarter turn, which would lift the rider to the center height of $20$ feet, not $10$; $5$ (A) and $6$ (B) are too small for a $60^\circ$ sweep. Only (D) $10$ matches the $60^\circ$ the geometry forces.
Alternative: Skip coordinates and use the right triangle directly. Draw the radius $OP$ and drop a perpendicular from the center's height down to the rider's level: the rider at height $10$ is $10$ feet below the center (height $20$). That makes a right triangle whose hypotenuse is the radius $OP = 20$ and whose vertical leg is $10$ — exactly half the hypotenuse. A right triangle with a leg equal to half the hypotenuse is a $30$-$60$-$90$ triangle, so the radius makes a $60^\circ$ angle with the downward vertical. That is the same $60^\circ$ turn, giving $\frac16$ of a minute, or $10$ seconds.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Modeling the wheel as a circle whose center is one radius (20 ft) above the bottom, so every rim point sits 20 ft from the center.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Solving $x^2 + 10^2 = 20^2$ to locate the rider at height 10, and computing the chord length $BP = \sqrt{300+100} = 20$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Recognizing triangle $OBP$ as equilateral (all sides 20), so its central angle $\angle BOP$ must be $60^\circ$.)7.RP.A.2Recognize and represent proportional relationships between quantities (Converting the $60^\circ$ sweep ($\frac16$ of a full turn) into $\frac16$ of the 60-second period, giving 10 seconds.)
⭐ Turn 'how high' into an angle at the center of the wheel, then trade that angle for time — the same fraction of the circle is the same fraction of the minute.
⭐ Turn 'how high' into an angle at the center of the wheel, then trade that angle for time — the same fraction of the circle is the same fraction of the minute.
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