AMC 10 · 2007 · #23

Grade 8 geometry-3d

similar-figuressurface-areaquadratic-equations convert-to-algebra ↑ Prerequisites: similar-figures

📏 Medium solution 💡 3 insights

Problem

A pyramid with a square base is cut by a plane that is parallel to its base and $2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

Pick an answer.

(A)

(B)

$2+\sqrt{2}$

(C)

$1+2\sqrt{2}$

(D)

(E)

$4+2\sqrt{2}$

AMC 10 2007 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: A square-based pyramid is sliced by a plane parallel to the base, 2 units above the base. The top piece is a smaller pyramid whose total surface area is exactly half the total surface area of the original pyramid. Find the height of the original pyramid.

Givens: The original solid is a pyramid with a square base.; The cutting plane is parallel to the base and sits 2 units above it.; The small pyramid cut from the top has surface area equal to half the original pyramid's surface area.

Unknowns: The altitude (height) of the original pyramid.

Understand

Plan

Primary tool: #17 Visualize Spatial Relationships

Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities

A plane parallel to the base cuts off a top pyramid that is a shrunk copy of the whole pyramid, so the two solids are similar. Seeing that similarity is the whole game: it links the surface areas to the heights. Then naming the height with a variable turns the area condition into a single equation to solve.

Execute — Answer: E

#17 Visualize Spatial Relationships 8.G.A.4 Step 1

See the two similar pyramids

A cut parallel to the base produces a top pyramid that is a scaled-down copy of the original.
Every length in the small pyramid is the same fraction of the matching length in the big one, so the two pyramids are similar.

$$\text{small pyramid} \sim \text{original pyramid}$$

💡 Slicing parallel to the base shrinks the whole shape uniformly, keeping it the same shape.

#17 Visualize Spatial Relationships 8.G.A.4 Step 2

Areas scale as height squared

For similar solids, every surface area scales with the square of the linear scale factor.
If the small pyramid's height is a fraction k of the original's height, then its surface area is k^2 times the original's surface area.
The problem says that fraction of area is 1/2.

$$\left(\frac{\text{small height}}{\text{original height}}\right)^2 = \frac{1}{2}$$

💡 Area is a two-dimensional measure, so it grows with the square of the scale factor.

#4 Introduce a Variable 6.RP.A.3 Step 3

Name the height and build the equation

Let h be the altitude of the original pyramid.
The cut is 2 units above the base, so the small top pyramid has altitude h - 2.
Substitute these two heights into the area ratio.

$$\left(\frac{h-2}{h}\right)^2 = \frac{1}{2}$$

💡 The small pyramid keeps only the top portion, so its height is the full height minus the 2-unit stub near the base.

#4 Introduce a Variable 8.EE.A.2 Step 4

Take the square root

Take the positive square root of both sides.
Since h - 2 and h are both positive lengths, the ratio is positive, so we keep the positive root.

$$\frac{h-2}{h} = \frac{1}{\sqrt{2}}$$

💡 Undo the square by rooting both sides; lengths are positive so only the plus sign survives.

#4 Introduce a Variable 8.EE.C.7 Step 5

Solve for the height

Multiply out and collect the h terms, then divide.
Rationalizing the denominator turns the answer into the form of the choices: h = 4 + 2\sqrt{2}.
This matches choice (E).

$$h - 2 = \frac{h}{\sqrt{2}} \;\Rightarrow\; h\left(1 - \frac{1}{\sqrt{2}}\right) = 2 \;\Rightarrow\; h = \frac{2\sqrt{2}}{\sqrt{2}-1} = 2\sqrt{2}(\sqrt{2}+1) = 4 + 2\sqrt{2}$$

💡 Once the square root is gone, it is an ordinary linear equation in h.

[1] #17 8.G.A.4 A cut parallel to the base produces a top pyramid that is a scaled-down copy of

[2] #17 8.G.A.4 For similar solids, every surface area scales with the square of the linear scal

[3] #4 6.RP.A.3 Let h be the altitude of the original pyramid. The cut is 2 units above the base

[4] #4 8.EE.A.2 Take the positive square root of both sides. Since h - 2 and h are both positive

[5] #4 8.EE.C.7 Multiply out and collect the h terms, then divide. Rationalizing the denominator

Review

Reasonableness: The height 4 + 2\sqrt{2} \approx 6.83, so the small pyramid's height is about 4.83. Their ratio is about 0.707, which is 1/\sqrt{2}, and squaring gives 0.5 exactly. The area of the top piece is indeed half the original, so the answer checks out.

Alternative: Eliminate possibilities: the height must be more than 2 for the cut to make sense, and the ratio (h-2)/h has to equal 1/\sqrt{2} \approx 0.707. Testing the choices, only 4 + 2\sqrt{2} gives (h-2)/h close to 0.707; the smaller choices make that ratio too small. So (E) is forced without full algebra.

CCSS standards used (min grade 8)

8.G.A.4 Understand similarity and that similar figures scale by a common factor (Recognizing the top pyramid is similar to the original and that surface areas scale as the square of the height ratio.)
6.RP.A.3 Use ratio reasoning to solve real-world problems (Setting the ratio of the small pyramid's area to the original's area equal to 1/2.)
8.EE.A.2 Use square-root symbols and evaluate square roots (Taking the square root of both sides of the area equation, producing 1/\sqrt{2}.)
8.EE.C.7 Solve linear equations in one variable (Solving h(1 - 1/\sqrt{2}) = 2 and rationalizing to get h = 4 + 2\sqrt{2}.)

⭐ When one solid is a shrunk copy of another, its areas shrink by the square of the height ratio, so a half-area copy has height 1 over root 2 of the original.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

See the two similar pyramids

Areas scale as height squared

Name the height and build the equation

Take the square root

Solve for the height

Review

CCSS standards used (min grade 8)

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