AMC 8 · 2003 · #25
Grade 8 geometry-2dProblem
In the figure, the area of square WXYZ is 25 cm2. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In △ABC, AB=AC, and when △ABC is folded over side BC, point A coincides with O, the center of square WXYZ. What is the area of △ABC, in square centimeters?
Pick an answer.
AMC 8 2003 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: Square $WXYZ$ has area $25 \text{ cm}^2$, so its side length is $5$ cm. Four $1 \text{ cm}$-by-$1 \text{ cm}$ squares frame the figure (one at each corner of the surrounding rectangle), so the segment $\overline{BC}$ runs vertically just outside side $\overline{WZ}$. Triangle $ABC$ is isosceles with $AB = AC$, and folding it over $\overline{BC}$ lands $A$ exactly on $O$, the center of square $WXYZ$. Find the area of $\triangle ABC$.
Givens: Square $WXYZ$ has area $25 \text{ cm}^2$, so its side length is $5$ cm; $O$ is the center of square $WXYZ$; Four $1$-cm squares frame the figure; from the diagram, $\overline{BC}$ is parallel to side $\overline{WZ}$; From the diagram, $B$ sits $1$ cm below the top edge $\overline{WX}$ and $C$ sits $1$ cm above the bottom edge $\overline{ZY}$; From the diagram, the line containing $\overline{BC}$ lies $2$ cm to the left of side $\overline{WZ}$ (the small-square frame plus a $1$-cm gap); $\triangle ABC$ is isosceles with $AB = AC$; Folding $\triangle ABC$ over $\overline{BC}$ sends point $A$ onto $O$; Answer choices: (A) $\tfrac{15}{4}$, (B) $\tfrac{21}{4}$, (C) $\tfrac{27}{4}$, (D) $\tfrac{21}{2}$, (E) $\tfrac{27}{2}$
Unknowns: The area of $\triangle ABC$, in square centimeters
Understand
Restated: Square $WXYZ$ has area $25 \text{ cm}^2$, so its side length is $5$ cm. Four $1 \text{ cm}$-by-$1 \text{ cm}$ squares frame the figure (one at each corner of the surrounding rectangle), so the segment $\overline{BC}$ runs vertically just outside side $\overline{WZ}$. Triangle $ABC$ is isosceles with $AB = AC$, and folding it over $\overline{BC}$ lands $A$ exactly on $O$, the center of square $WXYZ$. Find the area of $\triangle ABC$.
Givens: Square $WXYZ$ has area $25 \text{ cm}^2$, so its side length is $5$ cm; $O$ is the center of square $WXYZ$; Four $1$-cm squares frame the figure; from the diagram, $\overline{BC}$ is parallel to side $\overline{WZ}$; From the diagram, $B$ sits $1$ cm below the top edge $\overline{WX}$ and $C$ sits $1$ cm above the bottom edge $\overline{ZY}$; From the diagram, the line containing $\overline{BC}$ lies $2$ cm to the left of side $\overline{WZ}$ (the small-square frame plus a $1$-cm gap); $\triangle ABC$ is isosceles with $AB = AC$; Folding $\triangle ABC$ over $\overline{BC}$ sends point $A$ onto $O$; Answer choices: (A) $\tfrac{15}{4}$, (B) $\tfrac{21}{4}$, (C) $\tfrac{27}{4}$, (D) $\tfrac{21}{2}$, (E) $\tfrac{27}{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The figure is the whole problem, so Tool #1 (Draw a Diagram) goes first: place coordinates on the picture so every length we need is a coordinate difference. Once axes are set, the area question splits cleanly into Tool #7 (Identify Subproblems): (a) find the base $BC$ by reading the vertical positions of $B$ and $C$ off the side of the square minus the two $1$-cm trims, and (b) find the height by using the fold. The fold is the key Grade 8 reflection idea — line $\overline{BC}$ is the perpendicular bisector of $\overline{AO}$, so the height from $A$ to $\overline{BC}$ equals the distance from $O$ to $\overline{BC}$, which we read off coordinates. Multiply, halve, done.
Execute — Answer: C
6.G.A.3 Step 1 - Set coordinates on the diagram.
- Place square $WXYZ$ with $Z = (0,0)$, $Y = (5,0)$, $X = (5,5)$, $W = (0,5)$, since the side length is $\sqrt{25} = 5$ cm.
- The center is $O = \left(\tfrac{5}{2},\tfrac{5}{2}\right)$.
- From the diagram, $\overline{BC}$ is the vertical segment to the left of the square: $B$ sits $1$ cm below the top edge (so $y_B = 5 - 1 = 4$) and $C$ sits $1$ cm above the bottom edge (so $y_C = 0 + 1 = 1$).
- The four $1$-cm corner squares and the visible gap between them and side $\overline{WZ}$ place the line of $\overline{BC}$ at $x = -2$.
💡 Grade 6 "draw polygons in the coordinate plane": once every important point has an $(x,y)$ address, the lengths become subtractions instead of measurements.
6.NS.C.8 Step 2 - Sub-problem A — find the base $BC$.
- Points $B$ and $C$ share the same $x$-coordinate, so segment $\overline{BC}$ is vertical and its length is the difference of $y$-coordinates.
- The top trim and the bottom trim each remove $1$ cm from the $5$-cm side, leaving $5 - 1 - 1 = 3$.
💡 Grade 6 "find distances between points with the same first coordinate" — for a vertical segment, the length is just $|y_B - y_C|$.
8.G.A.1 Step 3 - Sub-problem B — turn the fold into a height equation.
- Folding $\triangle ABC$ across $\overline{BC}$ is a reflection across the line through $\overline{BC}$.
- Since the fold sends $A$ to $O$, that line is the perpendicular bisector of $\overline{AO}$.
- A perpendicular bisector is the set of points equidistant from $A$ and $O$, so the distance from $A$ to line $\overline{BC}$ equals the distance from $O$ to line $\overline{BC}$.
- The first distance is the triangle's height $h$ (since $\overline{BC}$ is the base), so $h$ equals the distance from $O$ to line $\overline{BC}$ — a length we can read off coordinates.
💡 Grade 8 "reflections preserve distances and map lines to lines." Folding $A$ onto $O$ over $\overline{BC}$ is exactly that reflection, which forces $A$ and $O$ to be the same distance from the fold line.
6.NS.C.8 Step 4 - Compute the height from coordinates.
- The line containing $\overline{BC}$ is the vertical line $x = -2$.
- Point $O$ has $x$-coordinate $\tfrac{5}{2}$.
- The horizontal distance from $O$ to that line is the difference of $x$-coordinates: $\tfrac{5}{2} - (-2) = \tfrac{5}{2} + 2 = \tfrac{9}{2}$.
💡 Grade 6 "distance from a point to a vertical line" — when the line is $x = k$, the distance is $|x_O - k|$, here $\left|\tfrac{5}{2} - (-2)\right| = \tfrac{9}{2}$.
6.G.A.1 Step 5 - Combine the two sub-problems with the triangle-area formula.
- The base is $BC = 3$ and the height is $h = \tfrac{9}{2}$, so the area is half their product.
💡 Grade 6 "area of a triangle is half the base times the height" — once base and height are in hand, the area is one multiplication and a halving.
6.G.A.3 Set coordinates on the diagram. Place square $WXYZ$ with $Z = (0,0)$, $Y = (5,0) 6.NS.C.8 Sub-problem A — find the base $BC$. Points $B$ and $C$ share the same $x$-coordi 8.G.A.1 Sub-problem B — turn the fold into a height equation. Folding $\triangle ABC$ ac 6.NS.C.8 Compute the height from coordinates. The line containing $\overline{BC}$ is the 6.G.A.1 Combine the two sub-problems with the triangle-area formula. The base is $BC = 3 Review
Reasonableness: Sanity check the size. The fold lands $A$ on $O$ across $\overline{BC}$, so $A$ sits on the far side of $\overline{BC}$ at the same distance as $O$. With $O$ at $x = \tfrac{5}{2}$ and the fold line at $x = -2$, point $A$ must be at $x = -2 - \tfrac{9}{2} = -\tfrac{13}{2}$, well to the left of the small-square frame — which matches the diagram showing $A$ far outside the squares. A direct area check: base $3$ and height $\tfrac{9}{2}$ give $\tfrac{1}{2} \cdot 3 \cdot \tfrac{9}{2} = \tfrac{27}{4} = 6.75 \text{ cm}^2$, the same order of magnitude as a triangle with a $3$-cm base and a roughly $4.5$-cm height. Choices (A) $\tfrac{15}{4} = 3.75$ and (B) $\tfrac{21}{4} = 5.25$ would require a smaller height (under $4$), while (D) $\tfrac{21}{2} = 10.5$ and (E) $\tfrac{27}{2} = 13.5$ would require a height larger than the whole figure. Only (C) fits.
Alternative: Tool #17 (Visualize Spatial Relationships) gives the height without coordinates. Drop a perpendicular from $O$ to line $\overline{BC}$, hitting it at the midpoint $M$ of $\overline{BC}$ (because $OB = OC$ by the left-right symmetry of the framed figure). Walking from $O$ horizontally to $M$ passes through three pieces shown in the picture: the half-side from $O$ to $\overline{WZ}$ (length $\tfrac{5}{2}$), the $1$-cm gap between $\overline{WZ}$ and the small-square frame, and the $1$-cm small square itself — total $\tfrac{5}{2} + 1 + 1 = \tfrac{9}{2}$. With base $BC = 5 - 1 - 1 = 3$ from the same picture, the area is $\tfrac{1}{2} \cdot 3 \cdot \tfrac{9}{2} = \tfrac{27}{4}$, choice (C).
CCSS standards used (min grade 8)
6.G.A.1Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles (Computing the area of $\triangle ABC$ as $\tfrac{1}{2} \cdot BC \cdot h$ once the base $3$ and the height $\tfrac{9}{2}$ are known.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate (Placing $Z=(0,0)$, $W=(0,5)$, $X=(5,5)$, $Y=(5,0)$ and reading $B=(-2,4)$, $C=(-2,1)$, $O=\left(\tfrac{5}{2},\tfrac{5}{2}\right)$ off the diagram.)6.NS.C.8Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane; include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate (Computing the base $BC = |y_B - y_C| = 3$ and the height $\text{dist}(O,\overline{BC}) = \left|\tfrac{5}{2} - (-2)\right| = \tfrac{9}{2}$ as differences of coordinates.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations: lines are taken to lines, and line segments to line segments of the same length (Recognizing the fold over $\overline{BC}$ as a reflection that preserves distances, so the height from $A$ to $\overline{BC}$ equals the distance from $O$ to $\overline{BC}$.)
⭐ Folding $A$ onto $O$ across $\overline{BC}$ means $A$ and $O$ are mirror images across the fold line, so the triangle's height equals the distance from $O$ to $\overline{BC}$. Read base $3$ and height $\tfrac{9}{2}$ straight off the picture and the area is $\tfrac{1}{2} \cdot 3 \cdot \tfrac{9}{2} = \tfrac{27}{4}$.
⭐ Folding $A$ onto $O$ across $\overline{BC}$ means $A$ and $O$ are mirror images across the fold line, so the triangle's height equals the distance from $O$ to $\overline{BC}$. Read base $3$ and height $\tfrac{9}{2}$ straight off the picture and the area is $\tfrac{1}{2} \cdot 3 \cdot \tfrac{9}{2} = \tfrac{27}{4}$.
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