AMC 8 · 2002 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

pythagorean-theoremarea-trianglesperfect-squares identify-subproblemspattern-recognition ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Short solution 💡 2 insights 📊 Diagram

Problem

Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?

Pick an answer.

(A)

X+Z=W+Y

(B)

W+X=Z

(C)

3X+4Y=5Z

(D)

$X+W=\frac{1}{2}(Y+Z)$

(E)

X+Y=Z

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Toolkit + CCSS Solution

Understand

Restated: A $3\text{-}4\text{-}5$ right triangle has a right isosceles triangle built outward on each of its three sides. Call $W$ the area of the original right triangle, and $X$, $Y$, $Z$ the areas of the isosceles triangles on the sides of length $3$, $4$, $5$ respectively. Which of the five area equations is always true?

Givens: Inner triangle: right triangle with legs $3$ and $4$ and hypotenuse $5$, area $W$; Three outer right isosceles ($45\text{-}45\text{-}90$) triangles, each built on one side of the inner triangle with that shared side as one of its two equal legs; $X$ sits on the leg of length $3$, $Y$ on the leg of length $4$, $Z$ on the hypotenuse of length $5$; Answer choices: (A) $X+Z=W+Y$, (B) $W+X=Z$, (C) $3X+4Y=5Z$, (D) $X+W=\tfrac{1}{2}(Y+Z)$, (E) $X+Y=Z$

Unknowns: Which of the five candidate equations is true for these four areas

Understand

Plan

Primary tool: #10 Use a Related Problem

Secondary: #7 Break Into Subproblems

The setup — three shapes built on the sides of a right triangle, one per side — is the picture that appears in every proof of the Pythagorean theorem. Tool #10 (Use a Related Problem) tells us to lean on $3^2 + 4^2 = 5^2$ instead of grinding through five answer choices. Each outer triangle on a side of length $s$ has area $\tfrac{1}{2}s^2$, so the three outer areas are exactly half the three squared sides — multiply Pythagoras by $\tfrac{1}{2}$ and the equation $X+Y=Z$ pops out. Tool #7 (Break Into Subproblems) handles the bookkeeping: compute $W, X, Y, Z$ one at a time, then test the choices.

Execute — Answer: E

#7 Break Into Subproblems 6.G.A.1 Step 1

Compute $W$, the area of the inner $3\text{-}4\text{-}5$ right triangle.
The two legs $3$ and $4$ meet at a right angle, so they serve as base and height.

$$W = \tfrac{1}{2} \cdot 3 \cdot 4 = 6$$

💡 Grade 6 "area of a right triangle = $\tfrac{1}{2}$ base $\times$ height" applied to the legs.

#7 Break Into Subproblems 6.G.A.1 Step 2

Compute $X$, $Y$, $Z$.
Each outer triangle is right isosceles with its two equal legs along the shared side of the inner triangle, so the outer triangle on a side of length $s$ has area $\tfrac{1}{2}s^2$.

$$X = \tfrac{1}{2}\cdot 3^2 = 4.5,\quad Y = \tfrac{1}{2}\cdot 4^2 = 8,\quad Z = \tfrac{1}{2}\cdot 5^2 = 12.5$$

💡 A right isosceles triangle with legs $s$ has area $\tfrac{1}{2}s \cdot s = \tfrac{1}{2}s^2$ — half of the square on that side.

#10 Use a Related Problem 8.G.B.7 Step 3

Recognise the Pythagorean pattern.
The inner triangle is a $3\text{-}4\text{-}5$ right triangle, so $3^2 + 4^2 = 5^2$.
Multiply both sides by $\tfrac{1}{2}$ to match the outer-triangle areas.

$$3^2 + 4^2 = 5^2 \;\Rightarrow\; \tfrac{1}{2}\cdot 3^2 + \tfrac{1}{2}\cdot 4^2 = \tfrac{1}{2}\cdot 5^2 \;\Rightarrow\; X + Y = Z$$

💡 Grade 8 "apply the Pythagorean theorem" — three similar shapes on the sides of a right triangle always satisfy outer$_a$ + outer$_b$ = outer$_c$, because all three are the same fraction of the squares on the sides.

#7 Break Into Subproblems 6.EE.B.5 Step 4

Test against the answer choices using the computed areas $W=6,\,X=4.5,\,Y=8,\,Z=12.5$.
(A) $X+Z = 4.5+12.5 = 17$, $W+Y = 6+8 = 14$ — no.
(B) $W+X = 6+4.5 = 10.5 \ne 12.5 = Z$ — no.
(C) $3X+4Y = 13.5+32 = 45.5$, $5Z = 62.5$ — no.
(D) $X+W = 10.5$, $\tfrac{1}{2}(Y+Z) = \tfrac{1}{2}\cdot 20.5 = 10.25$ — no.
(E) $X+Y = 4.5+8 = 12.5 = Z$ — yes.

$$X+Y = 4.5 + 8 = 12.5 = Z \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 6 "check which value makes an equation true" — only (E) survives the numerical test, matching the Pythagorean argument.

[1] #7 6.G.A.1 Compute $W$, the area of the inner $3\text{-}4\text{-}5$ right triangle. The two

[2] #7 6.G.A.1 Compute $X$, $Y$, $Z$. Each outer triangle is right isosceles with its two equal

[3] #10 8.G.B.7 Recognise the Pythagorean pattern. The inner triangle is a $3\text{-}4\text{-}5$

[4] #7 6.EE.B.5 Test against the answer choices using the computed areas $W=6,\,X=4.5,\,Y=8,\,Z=

Review

Reasonableness: Two independent paths give the same answer. Direct computation: $W=6,\,X=4.5,\,Y=8,\,Z=12.5$, and only (E) checks out numerically. Structural argument: each outer area equals $\tfrac{1}{2}$ of the square on its side, so the Pythagorean equation $3^2+4^2=5^2$ scales by $\tfrac{1}{2}$ to $X+Y=Z$ for any right triangle, not just $3\text{-}4\text{-}5$. The structural view also explains why $W$ never appears in the correct equation — $W$ is the inner triangle, not built on any side.

Alternative: Tool #1 (Draw a Diagram) plus squares: replace each right isosceles triangle with the full square on that side (each square is exactly two copies of the isosceles triangle). Then the picture becomes the classical Pythagorean diagram — squares on the three sides of a right triangle — with areas $2X$ on leg $3$, $2Y$ on leg $4$, $2Z$ on hypotenuse $5$. Pythagoras says $2X + 2Y = 2Z$, and dividing by $2$ gives $X + Y = Z$, the same answer (E).

CCSS standards used (min grade 8)

6.G.A.1 Find the area of right triangles and other shapes by composing or decomposing (Computing $W = \tfrac{1}{2}\cdot 3 \cdot 4 = 6$ and $X,Y,Z = \tfrac{1}{2}s^2$ for sides $s = 3,4,5$.)
6.EE.B.5 Substitute specific values and check which values make an equation true (Plugging $W=6,\,X=4.5,\,Y=8,\,Z=12.5$ into each answer choice to confirm only (E) holds.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown lengths (Using $3^2 + 4^2 = 5^2$ and scaling by $\tfrac{1}{2}$ to derive $X + Y = Z$ without case-by-case answer-choice testing.)

⭐ Three triangles built on a $3\text{-}4\text{-}5$ right triangle is just the Pythagorean picture in disguise — each area is half of a square on a side, so $3^2+4^2=5^2$ becomes $X+Y=Z$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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