AMC 8 · 2002 · #16
Grade 8 geometry-2dProblem
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.
AMC 8 2002 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.
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
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) $X+Y=Z$
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
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.
💡 Grade 6 "area of a right triangle = $\tfrac{1}{2}$ base $\times$ height" applied to the legs.
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 $\tfrac{1}{2}s^2$.
💡 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.
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.
💡 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.
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.
💡 Grade 6 "check which value makes an equation true" — only (E) survives the numerical test, matching the Pythagorean argument.
6.G.A.1 Compute $W$, the area of the inner $3\text{-}4\text{-}5$ right triangle. The two 6.G.A.1 Compute $X$, $Y$, $Z$. Each outer triangle is right isosceles with its two equal 8.G.B.7 Recognise the Pythagorean pattern. The inner triangle is a $3\text{-}4\text{-}5$ 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.1Find 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.5Substitute 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.7Apply 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 $X+Y=Z$.
⭐ 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 $X+Y=Z$.
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