AMC 10 · 2022 · #16
Grade 8 geometry-2dProblem
The diagram below shows a rectangle with side lengths 4 and 8 and a square with side length 5. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
Pick an answer.
AMC 10 2022 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 $4 \times 8$ rectangle contains a square of side $5$. Three corners of the square sit on three different sides of the rectangle (bottom, right, top). Find the area of the region that lies inside both shapes.
Givens: Rectangle has side lengths $4$ and $8$; Square has side length $5$; Three vertices of the square lie on three different sides of the rectangle (one on bottom, one on right, one on top); Choices: (A) $15\tfrac{1}{8}$, (B) $15\tfrac{3}{8}$, (C) $15\tfrac{1}{2}$, (D) $15\tfrac{5}{8}$, (E) $15\tfrac{7}{8}$
Unknowns: Area of the region inside both the square and the rectangle
Understand
Restated: A $4 \times 8$ rectangle contains a square of side $5$. Three corners of the square sit on three different sides of the rectangle (bottom, right, top). Find the area of the region that lies inside both shapes.
Givens: Rectangle has side lengths $4$ and $8$; Square has side length $5$; Three vertices of the square lie on three different sides of the rectangle (one on bottom, one on right, one on top); Choices: (A) $15\tfrac{1}{8}$, (B) $15\tfrac{3}{8}$, (C) $15\tfrac{1}{2}$, (D) $15\tfrac{5}{8}$, (E) $15\tfrac{7}{8}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #16 Change Focus / Complement
Tool #1 (Diagram): drop the picture on coordinates so the rectangle is $[0,8] \times [0,4]$. Tool #7 (Subproblems): the two right triangles formed by the square's slanted sides and the bottom/right edges of the rectangle are both 3-4-5 — that fixes every vertex coordinate. Tool #16 (Complement): instead of computing the overlap directly (a pentagon), compute the whole square ($25$) and subtract the small triangle that pokes above the rectangle's top edge.
Execute — Answer: D
5.G.A.1 Step 1 - Place the rectangle at $(0,0)$, $(8,0)$, $(8,4)$, $(0,4)$.
- Label the three square vertices that lie on rectangle sides as $A$ (top), $B$ (bottom), $C$ (right).
- From the picture, $A$-$B$-$C$ are consecutive corners with the right angle at $B$, so $AB = BC = 5$.
💡 Coordinates turn a picture problem into an arithmetic problem.
8.G.B.7 Step 2 - Drop a perpendicular from $A$ to the bottom edge; call the foot $A'$.
- Triangle $AA'B$ has the vertical leg $AA' = 4$ (rectangle height) and hypotenuse $AB = 5$.
- So the horizontal leg is $\sqrt{5^2 - 4^2} = 3$ — the classic 3-4-5 right triangle.
💡 Vertical leg $4$, hypotenuse $5$ — the missing leg has to be $3$.
5.G.A.2 Step 3 - By symmetry the second triangle at $B$ (formed with the right edge $x = 8$) is also 3-4-5 with legs $4$ (horizontal) and $3$ (vertical).
- So $B$ sits $4$ to the left of the corner $(8,0)$: $B = (4, 0)$.
- Then $A$ is $3$ to the left of $B$: $A = (1, 4)$.
- And $C$ is $3$ above $(8,0)$: $C = (8, 3)$.
💡 Two identical 3-4-5 triangles flank the square's right-angle corner $B$.
5.G.A.2 Step 4 - Find the fourth square vertex $D$.
- Since $ABCD$ is a square, $D = A + (C - B) = (1,4) + (4,3) = (5, 7)$.
- The rectangle's top edge is $y = 4$, and $D$ has $y = 7$, so $D$ sticks $3$ units above the rectangle.
💡 Add the side vector $\vec{BC}$ to $A$ to get the opposite corner $D$.
8.EE.B.6 Step 5 - The square's side $CD$ crosses the top edge $y = 4$ somewhere.
- Slope of $CD$ from $C(8,3)$ to $D(5,7)$ is $\tfrac{7-3}{5-8} = -\tfrac{4}{3}$.
- Setting $y = 4$: $4 - 3 = -\tfrac{4}{3}(x - 8)$, giving $x - 8 = -\tfrac{3}{4}$, so $x = \tfrac{29}{4}$.
- Call this crossing point $G = (\tfrac{29}{4},\ 4)$.
💡 Find where the slanted side crosses the rectangle's ceiling.
6.G.A.1 Step 6 - The piece of the square that sticks above the rectangle is the triangle $A$-$G$-$D$.
- Its base lies on $y = 4$ from $A(1,4)$ to $G(\tfrac{29}{4}, 4)$, so base $= \tfrac{29}{4} - 1 = \tfrac{25}{4}$.
- Its height is the vertical drop from $D(5,7)$ down to $y = 4$, which is $3$.
- Triangle area $= \tfrac{1}{2} \cdot \tfrac{25}{4} \cdot 3 = \tfrac{75}{8}$.
💡 The leaked-out piece is one clean triangle: base on the ceiling, apex at $D$.
5.NF.A.1 Step 7 - Use complement (Tool #16): overlap = square area $-$ leaked triangle $= 25 - \tfrac{75}{8} = \tfrac{200 - 75}{8} = \tfrac{125}{8} = 15\tfrac{5}{8}$.
- Match to choices: $\textbf{(D)}$.
💡 Whole square minus the spilled-out triangle equals the overlap.
5.G.A.1 Place the rectangle at $(0,0)$, $(8,0)$, $(8,4)$, $(0,4)$. Label the three squar 8.G.B.7 Drop a perpendicular from $A$ to the bottom edge; call the foot $A'$. Triangle $ 5.G.A.2 By symmetry the second triangle at $B$ (formed with the right edge $x = 8$) is a 5.G.A.2 Find the fourth square vertex $D$. Since $ABCD$ is a square, $D = A + (C - B) = 8.EE.B.6 The square's side $CD$ crosses the top edge $y = 4$ somewhere. Slope of $CD$ fro 6.G.A.1 The piece of the square that sticks above the rectangle is the triangle $A$-$G$- 5.NF.A.1 Use complement (Tool #16): overlap = square area $-$ leaked triangle $= 25 - \tf Review
Reasonableness: The overlap must be less than the square ($25$) and less than the rectangle ($32$). $15\tfrac{5}{8}$ sits comfortably below both. The spilled triangle has area $\tfrac{75}{8} \approx 9.4$, which feels right for a triangle of base $\approx 6.25$ and height $3$. Also, the answer choices are tightly bunched between $15\tfrac{1}{8}$ and $15\tfrac{7}{8}$ — only careful arithmetic on the eighth ($75/8$) picks the right one.
Alternative: Tool #2 (Systematic List) on the overlap polygon directly: vertices in order are $B(4,0)$, $(8,0)$, $C(8,3)$, $G(\tfrac{29}{4}, 4)$, $A(1,4)$, $(1,0)$? No — re-trace: the overlap is bounded by parts of both shapes. Use the Shoelace formula on the overlap pentagon $B(4,0)$, $C(8,3)$, $G(\tfrac{29}{4}, 4)$, $A(1,4)$, $(1,0)$... this is messier than the complement route, but it gives the same $\tfrac{125}{8}$.
CCSS standards used (min grade 8)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Placing the rectangle at $(0,0)$–$(8,4)$ so vertices become numerical.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the missing leg of length $3$ in the 3-4-5 triangle from leg $4$ and hypotenuse $5$.)5.G.A.2Represent real-world and mathematical problems by graphing points (Pinning the four square vertices to coordinates $A(1,4)$, $B(4,0)$, $C(8,3)$, $D(5,7)$.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Slope of $CD$ and using it to find where the side crosses $y = 4$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Area of the spilled-out triangle as $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$.)5.NF.A.1Add and subtract fractions with unlike denominators (Subtracting $\tfrac{75}{8}$ from $25 = \tfrac{200}{8}$ to land on $\tfrac{125}{8}$.)
⭐ Two hidden 3-4-5 right triangles fix every corner of the square. The square's top corner $(5, 7)$ pokes above the rectangle, so chop off that little triangle — area $\tfrac{75}{8}$ — and the leftover is $25 - \tfrac{75}{8} = 15\tfrac{5}{8}$, choice $\textbf{(D)}$.
⭐ Two hidden 3-4-5 right triangles fix every corner of the square. The square's top corner $(5, 7)$ pokes above the rectangle, so chop off that little triangle — area $\tfrac{75}{8}$ — and the leftover is $25 - \tfrac{75}{8} = 15\tfrac{5}{8}$, choice $\textbf{(D)}$.
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