Sensim Math Original · sm-11
SM Original Grade 6 geometry-2dProblem
A landscape architect sketches a triangular flower bed on a coordinate grid (units in meters). The three corners are placed at P(3,1), Q(3,9), and R(x,5), with the apex R lying to the right of the vertical side PQ, so x>3. If the flower bed covers exactly 24 square meters, what is the value of x?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A triangular flower bed sits on a coordinate grid (1 unit = 1 meter) with corners $P(3,1)$, $Q(3,9)$, and $R(x,5)$, where $x > 3$ pins the apex to the right of side $PQ$. The bed has area $24$ square meters. Find $x$.
Givens: Vertex coordinates: $P(3,1)$, $Q(3,9)$, $R(x,5)$; Coordinate units are meters, so areas are in square meters; Condition: $x > 3$ (apex $R$ lies to the right of segment $PQ$); Area of the triangular flower bed $= 24$ square meters; Choices: (A) 6, (B) 7, (C) 8, (D) 9, (E) 10
Unknowns: The x-coordinate $x$ of vertex $R$
Understand
Restated: A triangular flower bed sits on a coordinate grid (1 unit = 1 meter) with corners $P(3,1)$, $Q(3,9)$, and $R(x,5)$, where $x > 3$ pins the apex to the right of side $PQ$. The bed has area $24$ square meters. Find $x$.
Givens: Vertex coordinates: $P(3,1)$, $Q(3,9)$, $R(x,5)$; Coordinate units are meters, so areas are in square meters; Condition: $x > 3$ (apex $R$ lies to the right of segment $PQ$); Area of the triangular flower bed $= 24$ square meters; Choices: (A) 6, (B) 7, (C) 8, (D) 9, (E) 10
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #6 Guess and Check
Coordinates are given but no figure is drawn, so the first move is #1 **Draw a Diagram** — plot the three corners on grid paper. The picture instantly shows that $P$ and $Q$ share the same $x$-value, so side $PQ$ is vertical and makes a tidy base. Next, #7 **Identify Subproblems** splits the work into two crisp questions: (i) how long is the vertical base, and (ii) how far is $R$ from the line containing it. Finally, because the problem is multiple choice, we lean on #6 **Guess and Check** with the answer choices rather than reaching for algebra (#13).
Execute — Answer: D
5.G.A.2 Step 1 - Plot the three corners on grid paper.
- $P(3,1)$ and $Q(3,9)$ both have $x = 3$, so they sit on the same vertical line.
- The third corner $R(x,5)$ has $y = 5$ — exactly halfway up between $P$ and $Q$ — and lies somewhere to the right of that vertical line (since $x > 3$).
- The picture makes side $PQ$ jump out as the natural base of the triangle.
💡 Plotting labelled points on a coordinate grid is the very first move taught in the Grade 5 coordinate-plane standard.
6.NS.C.8 Step 2 - Break the area question into smaller pieces.
- (i) How long is the base $PQ$?
- Since $P$ and $Q$ share the same $x$-coordinate, the segment is vertical and its length is just the difference of the $y$-coordinates: $|9 - 1| = 8$ meters.
💡 Finding the distance between two points that share a coordinate by subtracting the other coordinate is exactly the Grade 6 coordinate-distance standard.
6.NS.C.8 Step 3 - (ii) How tall is the triangle, measured from this vertical base?
- The 'height' is the perpendicular distance from $R(x,5)$ across to the vertical line $x = 3$ that holds $PQ$.
- The $y$-coordinate of $R$ doesn't matter for this distance — only the $x$-coordinates do: distance $= |x - 3|$.
- Since the problem says $x > 3$, we can drop the absolute value and write height $= x - 3$.
💡 The perpendicular distance from a point to a vertical line is again a coordinate-difference, the same Grade 6 standard — just along the x-axis this time.
6.G.A.1 Step 4 - Apply the triangle-area formula: $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$.
- Plugging in base $= 8$, height $= x - 3$, and area $= 24$ gives $24 = \tfrac{1}{2} \cdot 8 \cdot (x-3) = 4(x-3)$.
- The geometry has collapsed into a clean 'four times what equals twenty-four?' question.
💡 The triangle-area formula $\tfrac{1}{2}\cdot\text{base}\cdot\text{height}$ is a core Grade 6 geometry standard.
6.G.A.1 Step 5 - Instead of doing algebra, try #6 **Guess and Check** with the answer choices.
- (A) $x=6$: $4(6-3) = 12$, too small.
- (C) $x=8$: $4(8-3) = 20$, still short.
- (D) $x=9$: $4(9-3) = 4 \cdot 6 = 24$.
- **Exact match!** (E) $x=10$: $4(10-3) = 28$, too big.
- So $x = 9$, choice **(D)**.
💡 Plugging each candidate value into the area formula is verification inside the same Grade 6 triangle-area standard.
5.G.A.2 Plot the three corners on grid paper. $P(3,1)$ and $Q(3,9)$ both have $x = 3$, s 6.NS.C.8 Break the area question into smaller pieces. (i) How long is the base $PQ$? Sinc 6.NS.C.8 (ii) How tall is the triangle, measured from this vertical base? The 'height' is 6.G.A.1 Apply the triangle-area formula: $\text{Area} = \tfrac{1}{2} \times \text{base} 6.G.A.1 Instead of doing algebra, try #6 **Guess and Check** with the answer choices. (A Review
Reasonableness: Quick sanity check from the picture: for a triangle with base $8$ to have area $24$, the height must be $\tfrac{2 \times 24}{8} = 6$. So $R$ must sit exactly $6$ meters to the right of the line $x = 3$, giving $x = 3 + 6 = 9$. This also satisfies $x > 3$. As a trap-check, forgetting the $\tfrac{1}{2}$ would give $8(x-3) = 24 \Rightarrow x = 6$ — that is exactly distractor (A), so the test-writer planted it on purpose. The smallest choice (A) $x=6$ gives only height $3$ and area $12$ (too small), and the largest (E) $x=10$ gives height $7$ and area $28$ (too big), so $x = 9$ sitting in between is the right magnitude.
Alternative: We could also use #13 (Convert to Algebra): from $4(x-3) = 24$, divide both sides by $4$ to get $x - 3 = 6$, then add $3$ to get $x = 9$. Same answer, but for an elementary student plugging in the choices is more concrete and protects against the missing-$\tfrac{1}{2}$ slip.
CCSS standards used (min grade 6)
5.G.A.2Represent real-world and mathematical problems by graphing points (Plotting the three flower-bed corners $P(3,1)$, $Q(3,9)$, $R(x,5)$ on the coordinate grid to visualize the triangle.)6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Computing the vertical base length $|9-1|=8$ and the horizontal height $|x-3| = x-3$ as coordinate differences between points sharing a coordinate.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Applying $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ to set up $24 = 4(x-3)$ and confirming $x = 9$ by checking each answer choice.)
⭐ This AMC 8 problem only needs the Grade 6 triangle-area formula and coordinate-distance idea you already know!
⭐ This AMC 8 problem only needs the Grade 6 triangle-area formula and coordinate-distance idea you already know!
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