← Sensim Lab Sensim Math

AMC 8 · 2009 · #7

Grade 6 geometry-2d
Archetype Coordinate Plane Figure Computation
coordinate-geometryarea-triangles coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?

Pick an answer.

(A)
2
(B)
3
(C)
4.5
(D)
6
(E)
9
View mode:

Toolkit + CCSS Solution

Understand

Restated: A triangular plot of land has corners $A$, $C$, and $D$. $A$ lies $3$ miles west of point $B$ on Main Street (east-west). $C$ lies $3$ miles north of $B$ on the railroad (north-south). $D$ lies another $3$ miles north of $C$ on the railroad. Find the area of triangle $ACD$ in square miles.

Givens: $AB = 3$ miles (Main Street runs east-west through $A$ and $B$); $BC = 3$ miles (railroad runs north-south through $B$, $C$, $D$); $CD = 3$ miles (on the railroad, $D$ is north of $C$); Main Street $\perp$ railroad, so $\angle ABC = \angle ABD = 90^\circ$; Answer choices: (A) $2$, (B) $3$, (C) $4.5$, (D) $6$, (E) $9$ (square miles)

Unknowns: The area of the triangular plot $ACD$, in square miles

Understand

Restated: A triangular plot of land has corners $A$, $C$, and $D$. $A$ lies $3$ miles west of point $B$ on Main Street (east-west). $C$ lies $3$ miles north of $B$ on the railroad (north-south). $D$ lies another $3$ miles north of $C$ on the railroad. Find the area of triangle $ACD$ in square miles.

Givens: $AB = 3$ miles (Main Street runs east-west through $A$ and $B$); $BC = 3$ miles (railroad runs north-south through $B$, $C$, $D$); $CD = 3$ miles (on the railroad, $D$ is north of $C$); Main Street $\perp$ railroad, so $\angle ABC = \angle ABD = 90^\circ$; Answer choices: (A) $2$, (B) $3$, (C) $4.5$, (D) $6$, (E) $9$ (square miles)

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement

Two perpendicular roads make this a coordinate-plane problem in disguise. Tool #1 (Draw a Diagram) says: put $B$ at the origin, Main Street on the $x$-axis, and the railroad on the $y$-axis. Then $A$, $C$, $D$ all have clean integer coordinates. With $C$ and $D$ both on the $y$-axis, side $CD$ is vertical, so the perpendicular distance from $A$ to that side is just $AB$ — Tool #7 (Identify Subproblems) splits the area question into two easy subproblems: "how long is $CD$?" and "how far is $A$ from the railroad?" Tool #16 (Count the Complement) gives a check: $\triangle ACD = \triangle ABD - \triangle ABC$.

Execute — Answer: C

#1 Draw a Diagram 5.G.A.1 Step 1
  • Set up coordinates.
  • Put $B$ at the origin, Main Street along the $x$-axis, and the railroad along the $y$-axis.
  • Then $A$ is $3$ miles west of $B$, so $A = (-3, 0)$.
  • $C$ is $3$ miles north of $B$, so $C = (0, 3)$.
  • $D$ is $3$ miles north of $C$, so $D = (0, 6)$.
$$A = (-3, 0), \; C = (0, 3), \; D = (0, 6)$$

💡 Putting the intersection $B$ at the origin turns the road map into a coordinate plane — a Grade 5 graphing skill.

#7 Identify Subproblems 6.NS.C.8 Step 2
  • Pick $CD$ as the base of $\triangle ACD$.
  • Both $C$ and $D$ sit on the railroad (the $y$-axis), so $CD$ is vertical.
  • Its length is the difference of $y$-coordinates.
$$CD = 6 - 3 = 3 \text{ miles}$$

💡 Two points on the same vertical line have a distance equal to the gap in their $y$-coordinates — Grade 6 coordinate-distance reasoning.

#7 Identify Subproblems 6.NS.C.6 Step 3
  • Find the height of $\triangle ACD$ from vertex $A$ to side $CD$.
  • Since $CD$ lies on the $y$-axis (the line $x = 0$), the perpendicular distance from $A = (-3, 0)$ to that line is just the absolute value of $A$'s $x$-coordinate.
  • This is exactly the length $AB$.
$$h = |-3| = 3 \text{ miles}$$

💡 Distance from a point to a vertical line is the absolute value of the $x$-difference — Grade 6 absolute-value-on-the-number-line idea.

#1 Draw a Diagram 6.G.A.1 Step 4

Apply the triangle area formula with base $CD = 3$ and height $h = 3$.

$$\text{Area} = \tfrac{1}{2} \cdot \text{base} \cdot \text{height} = \tfrac{1}{2} \cdot 3 \cdot 3 = 4.5 \text{ sq mi} \;\Rightarrow\; \textbf{(C)}$$

💡 Half of base times height is the standard Grade 6 triangle-area formula.

[1] #1 5.G.A.1 Set up coordinates. Put $B$ at the origin, Main Street along the $x$-axis, and t
[2] #7 6.NS.C.8 Pick $CD$ as the base of $\triangle ACD$. Both $C$ and $D$ sit on the railroad (
[3] #7 6.NS.C.6 Find the height of $\triangle ACD$ from vertex $A$ to side $CD$. Since $CD$ lies
[4] #1 6.G.A.1 Apply the triangle area formula with base $CD = 3$ and height $h = 3$.

Review

Reasonableness: Sanity-check the size. Triangle $ABD$ (corners at $(-3,0)$, $(0,0)$, $(0,6)$) is a right triangle with legs $3$ and $6$, so its area is $\tfrac{1}{2} \cdot 3 \cdot 6 = 9$ sq mi — the largest answer choice (E). The plot $ACD$ chops off the bottom half ($\triangle ABC$ has legs $3$ and $3$, area $4.5$), so $\triangle ACD$ should be the leftover $9 - 4.5 = 4.5$ sq mi. That matches (C), and it sits comfortably between $3$ and $6$ — exactly where we'd expect the upper half of the big triangle to fall.

Alternative: Tool #16 (Count the Complement). Instead of computing $\triangle ACD$ directly, compute the big right triangle $\triangle ABD$ and subtract the small right triangle $\triangle ABC$. $\triangle ABD$ has legs $AB = 3$ and $BD = 6$, area $\tfrac{1}{2}\cdot 3 \cdot 6 = 9$. $\triangle ABC$ has legs $AB = 3$ and $BC = 3$, area $\tfrac{1}{2}\cdot 3 \cdot 3 = 4.5$. So $\triangle ACD = 9 - 4.5 = 4.5$ sq mi — same answer (C).

CCSS standards used (min grade 6)

  • 5.G.A.1 Use a pair of perpendicular number lines (axes) to define a coordinate system (Modeling the perpendicular Main Street and railroad as $x$- and $y$-axes, with $B$ at the origin, to assign coordinates to $A$, $C$, and $D$.)
  • 6.NS.C.6 Understand a rational number as a point on the number line; extend to the coordinate plane (Using $|-3| = 3$ to read the perpendicular distance from $A = (-3, 0)$ to the $y$-axis (the line $x = 0$).)
  • 6.NS.C.8 Solve real-world problems by graphing points in the coordinate plane (Computing $CD = 6 - 3 = 3$ as the gap between $C = (0,3)$ and $D = (0,6)$ on a vertical line.)
  • 6.G.A.1 Find the area of triangles by composing into rectangles or decomposing into triangles (Applying $\text{Area} = \tfrac{1}{2} \cdot \text{base} \cdot \text{height} = \tfrac{1}{2}\cdot 3 \cdot 3 = 4.5$ sq mi to get the area of $\triangle ACD$.)

⭐ This AMC 8 problem only needs Grade 6 coordinate-plane area: pick a side as the base, measure the perpendicular distance from the opposite vertex, and use $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$.

⭐ This AMC 8 problem only needs Grade 6 coordinate-plane area: pick a side as the base, measure the perpendicular distance from the opposite vertex, and use $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$.

More like this

Same archetype — closest grade level first.