AMC 10 · 2022 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

pythagorean-theoremsimilar-trianglescoordinate-geometryarea-trianglesarea-rectangles identify-subproblemsarea-differencecomplementary-counting ↑ Prerequisites: pythagorean-theoremsimilar-triangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

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?

$\textbf{(A) }15\dfrac{1}{8} \qquad \textbf{(B) }15\dfrac{3}{8} \qquad \textbf{(C) }15\dfrac{1}{2} \qquad \textbf{(D) }15\dfrac{5}{8} \qquad \textbf{(E) }15\dfrac{7}{8}$

Pick an answer.

(A)

$15\dfrac{1}{8}$

(B)

$15\dfrac{3}{8}$

(C)

$15\dfrac{1}{2}$

(D)

$15\dfrac{5}{8}$

(E)

$15\dfrac{7}{8}$

View mode:

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

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

#1 Draw a Diagram 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 $AB = BC = 5$.

$$AB = BC = 5,\ \angle ABC = 90°$$

💡 Coordinates turn a picture problem into an arithmetic problem.

#7 Identify Subproblems 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.

$$A'B = \sqrt{25 - 16} = 3$$

💡 Vertical leg $4$, hypotenuse $5$ — the missing leg has to be $3$.

#7 Identify Subproblems 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 $(8,0)$: $C = (8, 3)$.

$$A = (1, 4),\ B = (4, 0),\ C = (8, 3)$$

💡 Two identical 3-4-5 triangles flank the square's right-angle corner $B$.

#1 Draw a Diagram 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.

$$D = (1+4,\ 4+3) = (5, 7)$$

💡 Add the side vector $\vec{BC}$ to $A$ to get the opposite corner $D$.

#7 Identify Subproblems 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 $G = (\tfrac{29}{4},\ 4)$.

$$G = \left(\tfrac{29}{4},\ 4\right)$$

💡 Find where the slanted side crosses the rectangle's ceiling.

#7 Identify Subproblems 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 $= \tfrac{1}{2} \cdot \tfrac{25}{4} \cdot 3 = \tfrac{75}{8}$.

$$\text{outside area} = \tfrac{1}{2} \cdot \tfrac{25}{4} \cdot 3 = \tfrac{75}{8}$$

💡 The leaked-out piece is one clean triangle: base on the ceiling, apex at $D$.

#16 Change Focus / Complement 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: $\textbf{(D)}$.

$$25 - \tfrac{75}{8} = \tfrac{125}{8} = 15\tfrac{5}{8}$$

💡 Whole square minus the spilled-out triangle equals the overlap.

[1] #1 5.G.A.1 Place the rectangle at $(0,0)$, $(8,0)$, $(8,4)$, $(0,4)$. Label the three squar

[2] #7 8.G.B.7 Drop a perpendicular from $A$ to the bottom edge; call the foot $A'$. Triangle $

[3] #7 5.G.A.2 By symmetry the second triangle at $B$ (formed with the right edge $x = 8$) is a

[4] #1 5.G.A.2 Find the fourth square vertex $D$. Since $ABCD$ is a square, $D = A + (C - B) =

[5] #7 8.EE.B.6 The square's side $CD$ crosses the top edge $y = 4$ somewhere. Slope of $CD$ fro

[6] #7 6.G.A.1 The piece of the square that sticks above the rectangle is the triangle $A$-$G$-

[7] #16 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.1 Use 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.7 Apply 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.2 Represent 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.6 Use 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.1 Find 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.1 Add 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 $\textbf{(D)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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