← Sensim Lab Sensim Math

AMC 8 · 2004 · #25

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-rectanglesarea-circlespythagorean-theoremspatial-visualizationreflection-symmetry area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-circlespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Two 4×44 \times 4 squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

Pick an answer.

(A)
$16-4\pi$
(B)
$16-2\pi$
(C)
$28-4\pi$
(D)
$28-2\pi$
(E)
$32-2\pi$
View mode:

Toolkit + CCSS Solution

Understand

Restated: Two $4 \times 4$ squares are placed so they meet at right angles and each one cuts through the midpoints of the other's two intersected sides. A circle is drawn whose diameter is the segment between the two points where the square boundaries cross. Find the area of the shaded region — the union of the two squares with the circle removed.

Givens: Each square has side length $4$; The squares cross at right angles and bisect each other's intersected sides; The circle's diameter is the segment between the two intersection points of the square boundaries; Answer choices: (A) $16 - 4\pi$, (B) $16 - 2\pi$, (C) $28 - 4\pi$, (D) $28 - 2\pi$, (E) $32 - 2\pi$

Unknowns: The area of the shaded region (union of the two squares minus the circle)

Understand

Restated: Two $4 \times 4$ squares are placed so they meet at right angles and each one cuts through the midpoints of the other's two intersected sides. A circle is drawn whose diameter is the segment between the two points where the square boundaries cross. Find the area of the shaded region — the union of the two squares with the circle removed.

Givens: Each square has side length $4$; The squares cross at right angles and bisect each other's intersected sides; The circle's diameter is the segment between the two intersection points of the square boundaries; Answer choices: (A) $16 - 4\pi$, (B) $16 - 2\pi$, (C) $28 - 4\pi$, (D) $28 - 2\pi$, (E) $32 - 2\pi$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

The shaded region is a compound shape, so Tool #7 (Identify Subproblems) splits it into three clean pieces: (a) the area of the union of the two squares, which itself splits into "two squares minus their overlap"; (b) the radius of the circle, which comes from the geometry of the overlap square; and (c) the area of the circle. Tool #1 (Draw a Diagram) is the unlock: marking the overlap as a $2 \times 2$ square and labeling its diagonal makes (a) and (b) visible at a glance. Each subproblem is then a single formula away.

Execute — Answer: D

#1 Draw a Diagram 3.MD.C.7 Step 1
  • Sketch the figure and identify the overlap.
  • Each square has side $4$.
  • "Bisecting their intersecting sides" means each square's edge enters the other at the midpoint of a side, so the region common to both squares is a square of side $4/2 = 2$.
$$\text{overlap side} = \tfrac{4}{2} = 2, \quad \text{overlap area} = 2 \times 2 = 4$$

💡 A Grade 3 area-by-multiplication fact: a $2 \times 2$ square has area $4$. The diagram makes the side length obvious from the word "bisect."

#7 Identify Subproblems 6.G.A.1 Step 2
  • Subproblem A — area of the union of the two squares.
  • Use inclusion-exclusion: add the two square areas and subtract the overlap so the central $2 \times 2$ region is counted once.
$$\text{union area} = 4^2 + 4^2 - 2^2 = 16 + 16 - 4 = 28$$

💡 Grade 6 "area by composing/decomposing": the cross-shape is two squares glued at the central overlap, so add and subtract that overlap once.

#7 Identify Subproblems 8.G.B.7 Step 3
  • Subproblem B — radius of the circle.
  • The two boundary-intersection points are diagonally opposite corners of the $2 \times 2$ overlap square, so the diameter is that diagonal.
  • The diagonal of a $2 \times 2$ square is the hypotenuse of a right triangle with legs $2$ and $2$.
$$d = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}, \quad r = \tfrac{d}{2} = \sqrt{2}$$

💡 Grade 8 Pythagorean theorem on a $2$-$2$-? right triangle gives $2\sqrt{2}$; halving it gives the radius $\sqrt{2}$.

#7 Identify Subproblems 7.G.B.4 Step 4
  • Subproblem C — area of the circle.
  • Apply the circle area formula with $r = \sqrt{2}$.
$$\text{circle area} = \pi r^2 = \pi (\sqrt{2})^2 = 2\pi$$

💡 Grade 7 circle area: squaring $\sqrt{2}$ neatly gives $2$, so the circle contributes exactly $2\pi$.

#7 Identify Subproblems 6.G.A.1 Step 5
  • Combine.
  • Shaded area $=$ (union of squares) $-$ (circle).
$$\text{shaded} = 28 - 2\pi \;\Rightarrow\; \textbf{(D)}$$

💡 Same Grade 6 composing/decomposing move: the final region is the cross minus the disk, so subtract their areas.

[1] #1 3.MD.C.7 Sketch the figure and identify the overlap. Each square has side $4$. "Bisecting
[2] #7 6.G.A.1 Subproblem A — area of the union of the two squares. Use inclusion-exclusion: ad
[3] #7 8.G.B.7 Subproblem B — radius of the circle. The two boundary-intersection points are di
[4] #7 7.G.B.4 Subproblem C — area of the circle. Apply the circle area formula with $r = \sqrt
[5] #7 6.G.A.1 Combine. Shaded area $=$ (union of squares) $-$ (circle).

Review

Reasonableness: Check magnitudes. Each square covers $16$, so the union is between $16$ (if they coincided) and $32$ (if they were disjoint); our $28$ sits right in that band. The circle is inscribed in the $2 \times 2$ overlap region's diagonal — its area $2\pi \approx 6.28$ is comfortably less than the overlap's $4 \cdot \pi/2 \approx$ a small fraction of the cross, leaving shaded $\approx 28 - 6.28 \approx 21.7 > 0$. Eliminations: (A) $16 - 4\pi$ and (B) $16 - 2\pi$ start from a single square, ignoring the second. (C) $28 - 4\pi$ would need radius $2$, which is the overlap's side, not the half-diagonal. (E) $32 - 2\pi$ forgets to subtract the overlap once. Only (D) is consistent.

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: the union of two $4 \times 4$ squares that share a $2 \times 2$ overlap has area $16 + 16 - 4 = 28$, so any answer starting with $16$ or $32$ is wrong — that cuts the list to (C) $28 - 4\pi$ and (D) $28 - 2\pi$. To pick between them, note that the circle's diameter is the diagonal of the $2 \times 2$ overlap, not its side, so the radius is $\sqrt{2}$, not $2$; area $\pi (\sqrt{2})^2 = 2\pi$ selects (D).

CCSS standards used (min grade 8)

  • 3.MD.C.7 Relate area to the operations of multiplication and addition (Computing the overlap's area as $2 \times 2 = 4$ and each square's area as $4 \times 4 = 16$.)
  • 6.G.A.1 Find the area of polygons by composing into rectangles or decomposing into triangles and other shapes (Inclusion-exclusion on the cross shape: $16 + 16 - 4 = 28$, then subtracting the disk to obtain the shaded region.)
  • 7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Applying $\pi r^2$ with $r = \sqrt{2}$ to get a circle area of $2\pi$.)
  • 8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Finding the diagonal of the $2 \times 2$ overlap square, $\sqrt{2^2 + 2^2} = 2\sqrt{2}$, which is the circle's diameter.)

⭐ Break a tangled picture into pieces: the cross is two $4 \times 4$ squares minus their $2 \times 2$ overlap ($28$), the circle's radius is half the overlap's diagonal ($\sqrt{2}$), and subtracting the circle ($2\pi$) leaves $28 - 2\pi$.

⭐ Break a tangled picture into pieces: the cross is two $4 \times 4$ squares minus their $2 \times 2$ overlap ($28$), the circle's radius is half the overlap's diagonal ($\sqrt{2}$), and subtracting the circle ($2\pi$) leaves $28 - 2\pi$.

More like this

Same archetype — closest grade level first.