Frances stands 15 meters directly south of a locked gate in a fence that runs east-west. Immediately behind the fence is a box of chocolates, located x meters east of the locked gate. An unlocked gate lies 9 meters east of the box, and another unlocked gate lies 8 meters west of the locked gate. Frances can reach the box by walking toward an unlocked gate, passing through it, and walking toward the box. It happens that the total distance Frances would travel is the same via either unlocked gate. What is the value of x?
Try it yourself first — the explanation is most useful after you’ve attempted it.
View mode:
Toolkit + CCSS Solution
Understand
Restated: Frances stands 15 m due south of a locked gate on an east-west fence. A box sits on the fence line $x$ m east of the locked gate. One open gate is 9 m east of the box; another open gate is 8 m west of the locked gate. To reach the box Frances walks straight to an open gate, through it, then straight to the box. Both open gates give the same total walking distance. Find $x$.
Givens: Frances is 15 m directly south of the locked gate, and the fence runs east-west, so the line from Frances to the fence is perpendicular to the fence.; The box is $x$ m east of the locked gate.; The east open gate is 9 m east of the box; the west open gate is 8 m west of the locked gate.; The two routes (via the east gate and via the west gate) cover the same total distance.
Unknowns: The distance $x$ from the locked gate to the box.
Understand
Restated: Frances stands 15 m due south of a locked gate on an east-west fence. A box sits on the fence line $x$ m east of the locked gate. One open gate is 9 m east of the box; another open gate is 8 m west of the locked gate. To reach the box Frances walks straight to an open gate, through it, then straight to the box. Both open gates give the same total walking distance. Find $x$.
Givens: Frances is 15 m directly south of the locked gate, and the fence runs east-west, so the line from Frances to the fence is perpendicular to the fence.; The box is $x$ m east of the locked gate.; The east open gate is 9 m east of the box; the west open gate is 8 m west of the locked gate.; The two routes (via the east gate and via the west gate) cover the same total distance.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #13 Convert to Algebra
The scene is all about positions and paths, so drawing it on a grid is the first move: it reveals that Frances, being straight south of the fence, sees every gate at the top of a right triangle with a 15 m vertical leg. That turns each diagonal walk into a Pythagorean length. Then name the box distance $x$, write each route's total length, set the two equal, and solve the resulting equation.
Execute — Answer: C
#1 Draw a Diagram 6.NS.C.8Step 1
Put the scene on a grid
Place the locked gate at the origin, with east as positive $x$ and north as positive $y$.
Then Frances is at $(0,-15)$, the box is at $(x,0)$, the west open gate is at $(-8,0)$, and the east open gate, being 9 m east of the box, is at $(x+9,0)$.
Since Frances is straight south of the fence, the segment from Frances up to any point on the fence makes a right angle with the fence.
💡 Pinning every object to a coordinate makes the hidden right angles at the fence pop into view.
#7 Identify Subproblems 8.G.B.7Step 2
Measure the west route
Going west, Frances first walks from $(0,-15)$ to the west gate at $(-8,0)$.
That diagonal is the hypotenuse of a right triangle with legs 8 (east-west) and 15 (north-south), so by the Pythagorean theorem it is $\sqrt{8^2+15^2}=\sqrt{289}=17$.
From the west gate she then walks east along the fence to the box, a distance of $x+8$.
💡 Isolating the square root before squaring keeps the algebra clean, and the $x^2$ terms conveniently vanish.
[1]
#1 6.NS.C.8Place the locked gate at the origin, with east as positive $x$ and north as posi
[2]
#7 8.G.B.7Going west, Frances first walks from $(0,-15)$ to the west gate at $(-8,0)$. Tha
[3]
#4 6.EE.B.6Going east, Frances walks from $(0,-15)$ to the east gate at $(x+9,0)$. This dia
[4]
#13 8.EE.C.7The two totals are equal, so $x+25=\sqrt{(x+9)^2+225}+9$. Move the 9 over to get
Review
Reasonableness: Plug $x=\tfrac{25}{7}$ back in. West route: $x+25=\tfrac{25}{7}+25=\tfrac{200}{7}\approx 28.57$. East route: $x+9=\tfrac{88}{7}$, so the diagonal is $\sqrt{(88/7)^2+225}=\sqrt{7744/49+11025/49}=\sqrt{18769/49}=\tfrac{137}{7}$, and adding 9 gives $\tfrac{137}{7}+\tfrac{63}{7}=\tfrac{200}{7}\approx 28.57$. Both routes match, and $3\tfrac{4}{7}$ sits in the middle of the answer choices, so (C) is consistent.
Alternative: Skip coordinates and reason with the two right triangles directly: the west leg 8 and height 15 make the 8-15-17 triangle, so the west trip is $17+x+8$; the east trip is $\sqrt{(x+9)^2+15^2}+9$. Setting them equal gives the same equation $14x=50$, hence $x=\tfrac{25}{7}$.
CCSS standards used (min grade 8)
6.NS.C.8 Solve real-world problems by graphing points in all four quadrants (Placing Frances, the box, and both open gates as coordinates so the right angles at the fence become visible.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding each diagonal walk-to-gate distance as the hypotenuse of a right triangle with vertical leg 15.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $x$ stand for the box distance and writing each route's total length as an expression in $x$.)
8.EE.C.7 Solve linear equations in one variable (Setting the two route lengths equal, squaring away the radical, and solving $14x=50$ for $x$.)
⭐ Drop everything onto a grid, use the right triangles to measure the diagonal walks, set the two equal-length routes equal, and solve for the unknown.
⭐ Drop everything onto a grid, use the right triangles to measure the diagonal walks, set the two equal-length routes equal, and solve for the unknown.
