AMC 10 · 2009 · #14

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

area-rectanglesratio-proportionsystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: area-rectangles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

Pick an answer.

(A)

(B)

$\sqrt {10}$

(C)

$2 + \sqrt2$

(D)

$2\sqrt3$

(E)

AMC 10 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: Four identical rectangles are set in a pinwheel around a small square, and together they fill a larger square. The big square's area is 4 times the small square's area. Find the ratio of a rectangle's longer side to its shorter side.

Givens: Four congruent rectangles arranged in a pinwheel form an outer square with a smaller square hole in the middle.; Area of the outer square is 4 times the area of the inner square.

Unknowns: The ratio (longer side) : (shorter side) of each rectangle.

Understand

Givens: Four congruent rectangles arranged in a pinwheel form an outer square with a smaller square hole in the middle.; Area of the outer square is 4 times the area of the inner square.

Plan

Primary tool: #4 Introduce a Variable

Secondary: #1 Draw a Diagram, #13 Convert to Algebra

The figure hides two squares whose sides are built from the rectangle's two side lengths. Name those lengths, read the two square sides off the picture, and the area fact becomes a single equation to solve.

Execute — Answer: A

#1 Draw a Diagram 6.G.A.1 Step 1

Read the two squares off the figure

Trace one straight edge of the big square.
It runs along one full long side of a rectangle and then along one short side of the next rectangle, so the outer square's side is (long side) + (short side).
The inner hole's side is the leftover: the long side of a rectangle minus the short side that overlaps it, so the inner square's side is (long side) - (short side).

$$\text{outer side} = \ell + s, \qquad \text{inner side} = \ell - s$$

💡 Walking along the big square's edge, you cross one long side and one short side end to end.

#4 Introduce a Variable 6.EE.B.6 Step 2

Turn the area fact into an equation

Let the shorter side be s and the longer side be \ell.
A square's area is its side length squared, so the outer square has area (\ell + s)^2 and the inner square has area (\ell - s)^2.
The outer area is 4 times the inner area.

$$(\ell + s)^2 = 4(\ell - s)^2$$

💡 Both regions are squares, so 'area = side squared' converts the picture into one clean equation.

#13 Convert to Algebra 8.EE.A.2 Step 3

Take the square root of both sides

Both sides are perfect squares of positive lengths, and 4 = 2^2.
Taking the positive square root of each side keeps the equation true and removes the exponents.

$$\ell + s = 2(\ell - s)$$

💡 Areas in ratio 4 : 1 mean side lengths in ratio 2 : 1.

#13 Convert to Algebra 8.EE.C.7 Step 4

Solve for the side ratio

Expand the right side and collect terms.
This gives 3s = \ell, so the longer side is 3 times the shorter side.
The ratio of longer to shorter is 3, which is choice (A).

$$\ell + s = 2\ell - 2s \;\Rightarrow\; 3s = \ell \;\Rightarrow\; \frac{\ell}{s} = 3$$

💡 Once the exponents are gone, it is a one-line linear equation in the two lengths.

[1] #1 6.G.A.1 Trace one straight edge of the big square. It runs along one full long side of a

[2] #4 6.EE.B.6 Let the shorter side be s and the longer side be \ell. A square's area is its si

[3] #13 8.EE.A.2 Both sides are perfect squares of positive lengths, and 4 = 2^2. Taking the posi

[4] #13 8.EE.C.7 Expand the right side and collect terms. This gives 3s = \ell, so the longer sid

Review

Reasonableness: Test with s = 1, \ell = 3. Outer side = 1 + 3 = 4 and inner side = 3 - 1 = 2, giving areas 16 and 4. Indeed 16 = 4 \times 4, so the area condition holds and the ratio 3 is confirmed. It also matches the sample rectangle in the figure.

Alternative: Set the inner square's side to 1 first. Then the outer square's side must be 2 (since its area is 4 times as big). The rectangle's short and long sides satisfy \ell - s = 1 and \ell + s = 2, giving \ell = 3/2, s = 1/2, and again \ell / s = 3.

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Reading the outer and inner square side lengths from the composed pinwheel figure.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the rectangle's sides s and \ell and writing the two square areas as expressions.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Taking the square root of the area equation to relate the side lengths.)
8.EE.C.7 Solve linear equations in one variable (Solving the resulting linear equation to get the side ratio.)

⭐ If one square has 4 times the area of another, its side is twice as long, so turn the area clue into a side clue first.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Read the two squares off the figure

Turn the area fact into an equation

Take the square root of both sides

Solve for the side ratio

Review

CCSS standards used (min grade 8)

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