AMC 8 · 2012 · #25

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

pythagorean-theoremarea-trianglesarea-rectanglesspatial-visualization area-differenceidentify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A square with area $4$ is inscribed in a square with area $5$ , with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$ , and the other of length $b$ . What is the value of $ab$ ?

Pick an answer.

(A)

$hspace{.05in}\frac{1}5$

(B)

$hspace{.05in}\frac{2}5$

(C)

$hspace{.05in}\frac{1}2$

(D)

$hspace{.05in}1$

(E)

$hspace{.05in}4$

AMC 8 2012 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: An outer square has area $5$ and an inner (tilted) square has area $4$. Each vertex of the inner square sits on a side of the outer square, splitting that side into two pieces of lengths $a$ and $b$. Find the product $ab$.

Givens: Outer square area $= 5$, so its side length is $\sqrt{5}$; Inner square area $= 4$, so its side length is $2$; Each side of the outer square is divided into a segment of length $a$ and a segment of length $b$, with $a + b = \sqrt{5}$; Answer choices: (A) $\tfrac{1}{5}$, (B) $\tfrac{2}{5}$, (C) $\tfrac{1}{2}$, (D) $1$, (E) $4$

Unknowns: The value of the product $ab$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The figure is a classic "square inside a square" picture. Tool #1 (Draw a Diagram) makes the four corner right triangles visible — each one has legs of length $a$ and $b$. Tool #7 (Identify Subproblems) then turns the question into a single accounting equation: the four corner triangles fill exactly the gap between the two squares, so their total area equals $5 - 4 = 1$. This avoids algebra entirely (no Pythagorean setup, no expanding $(a+b)^2$). For a multiple-choice geometry problem, comparing areas is the simplest path.

Execute — Answer: C

#1 Draw a Diagram 5.NF.B.7 Step 1

Draw the picture.
The outer square has area $5$, so its side is $\sqrt{5}$.
The tilted inner square sits inside it with one vertex on each side.
Each side of the outer square is cut into a short piece $a$ and a long piece $b$, with $a + b = \sqrt{5}$.

$$a + b = \sqrt{5}$$

💡 Drawing the figure exposes the four corner triangles — the key objects we will measure.

#1 Draw a Diagram 4.G.A.2 Step 2

Look at one corner of the outer square.
It is cut off by an edge of the inner square.
The two legs of this corner are an $a$-piece on one side of the outer square and a $b$-piece on the adjacent side.
So each corner is a right triangle with legs $a$ and $b$.

$$\text{legs} = a \text{ and } b, \quad \text{right angle at the corner}$$

💡 Recognizing that the corner pieces are right triangles is a Grade 4 shape-classification move.

#7 Identify Subproblems 6.G.A.1 Step 3

Find the area of the four corner triangles two different ways.
First way: the four triangles fill exactly the region between the inner and outer squares, so their total area is the difference of the two square areas.

$$\text{Total triangle area} = 5 - 4 = 1$$

💡 Splitting the outer square into "inner square + 4 triangles" is the Tool #7 decomposition — a Grade 6 "compose and decompose figures" move.

#7 Identify Subproblems 6.G.A.1 Step 4

Second way: each right triangle has legs $a$ and $b$, so its area is $\tfrac{1}{2}ab$.
There are four such triangles, all congruent, so their total area is $4 \times \tfrac{1}{2}ab = 2ab$.

$$4 \times \tfrac{1}{2}ab = 2ab$$

💡 Triangle area $= \tfrac{1}{2} \times \text{base} \times \text{height}$ for a right triangle uses the two legs directly — Grade 6 area work.

#7 Identify Subproblems 6.EE.B.7 Step 5

Set the two expressions for the same total area equal, then solve for $ab$.

$$2ab = 1 \;\Rightarrow\; ab = \tfrac{1}{2} \;\Rightarrow\; \textbf{(C)}$$

💡 Solving the one-step equation $2ab = 1$ for $ab$ is basic Grade 6 algebra — divide both sides by $2$.

[1] #1 5.NF.B.7 Draw the picture. The outer square has area $5$, so its side is $\sqrt{5}$. The

[2] #1 4.G.A.2 Look at one corner of the outer square. It is cut off by an edge of the inner sq

[3] #7 6.G.A.1 Find the area of the four corner triangles two different ways. First way: the fo

[4] #7 6.G.A.1 Second way: each right triangle has legs $a$ and $b$, so its area is $\tfrac{1}{

[5] #7 6.EE.B.7 Set the two expressions for the same total area equal, then solve for $ab$.

Review

Reasonableness: Check the magnitude with concrete numbers. We need $a + b = \sqrt{5} \approx 2.236$ and $ab = \tfrac{1}{2}$. Solving the quadratic $t^2 - \sqrt{5}\,t + \tfrac{1}{2} = 0$ gives $t = \tfrac{\sqrt{5} \pm \sqrt{3}}{2}$, so $a \approx 0.252$ and $b \approx 1.984$. Both are positive, both fit inside a side of length $\sqrt{5}$, and the right triangle with these legs has hypotenuse $\sqrt{a^2+b^2} = \sqrt{(a+b)^2 - 2ab} = \sqrt{5 - 1} = 2$ — exactly the inner square's side length. Everything checks.

Alternative: Tool #13 (Convert to Algebra) with the Pythagorean theorem: each corner triangle has hypotenuse $2$ (a side of the inner square), so $a^2 + b^2 = 4$. Squaring $a + b = \sqrt{5}$ gives $a^2 + 2ab + b^2 = 5$. Subtracting: $2ab = 1$, so $ab = \tfrac{1}{2}$. Same answer (C), but heavier machinery — the area-subtraction route is one move shorter.

CCSS standards used (min grade 6)

5.NF.B.7 Apply and extend previous understandings of division of fractions (Reading $a + b = \sqrt{5}$ as a length split into two pieces — a Grade 5 "partition a quantity" idea applied to the side of the outer square.)
4.G.A.2 Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size (Recognizing that each corner cut off the outer square is a right triangle (the outer square's corner is a right angle).)
6.G.A.1 Find the area of right triangles, other triangles, and polygons by composing into rectangles or decomposing into triangles (Decomposing the outer square into the inner square plus four corner right triangles, and computing each triangle's area as $\tfrac{1}{2}ab$.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Solving the one-step equation $2ab = 1$ to get $ab = \tfrac{1}{2}$.)

⭐ This AMC 8 problem only needs Grade 6 area reasoning — split the big square into the small square plus four corner triangles, and the rest is a one-step equation.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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