AMC 8 · 2003 · #6

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition

pythagorean-theoremperfect-squaresarea-trianglesspatial-visualization identify-subproblemspattern-recognition ↑ Prerequisites: perfect-squaresarea-trianglesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Given the areas of the three squares in the figure, what is the area of the interior triangle?

$\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800$

Pick an answer.

(A)

(B)

(C)

(D)

300

(E)

1800

View mode:

Toolkit + CCSS Solution

Understand

Restated: A figure shows three squares meeting at the vertices of an interior triangle. The squares have areas $25$, $144$, and $169$. Find the area of the interior triangle.

Givens: Three squares are drawn on the sides of an interior triangle; The squares' areas are $25$, $144$, and $169$; Answer choices: (A) $13$, (B) $30$, (C) $60$, (D) $300$, (E) $1800$

Unknowns: The area of the interior triangle

Understand

Restated: A figure shows three squares meeting at the vertices of an interior triangle. The squares have areas $25$, $144$, and $169$. Find the area of the interior triangle.

Givens: Three squares are drawn on the sides of an interior triangle; The squares' areas are $25$, $144$, and $169$; Answer choices: (A) $13$, (B) $30$, (C) $60$, (D) $300$, (E) $1800$

Plan

Primary tool: #7 Break Into Subproblems

Secondary: #10 Use a Related Problem

The figure pairs each square with one side of the triangle, so Tool #7 (Break Into Subproblems) splits the problem into three clean parts: (1) turn each square's area into a side length, (2) check what kind of triangle has sides $5, 12, 13$, (3) compute its area. Tool #10 (Use a Related Problem) is the recognition that $5\text{-}12\text{-}13$ is a famous Pythagorean triple, so the converse of the Pythagorean theorem from a related problem makes step (3) easy — the triangle is right-angled, and its legs are the base and height.

Execute — Answer: B

#7 Break Into Subproblems 8.EE.A.2 Step 1

Turn each square's area into a side length.
A square with area $A$ has side $\sqrt{A}$.

$$\sqrt{25} = 5,\quad \sqrt{144} = 12,\quad \sqrt{169} = 13$$

💡 Grade 8 "use square root to solve $x^2 = p$" — each side is the square root of its square's area.

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

Read the side lengths of the triangle from the figure.
Each side of the interior triangle is shared with one of the squares, so the triangle has sides $5$, $12$, and $13$.

$$\text{triangle sides} = 5,\; 12,\; 13$$

💡 Grade 7 "draw geometric shapes with given conditions" — sides of the squares are exactly the sides of the triangle.

#10 Use a Related Problem 8.G.B.6 Step 3

Check whether the triangle is right-angled using the converse of the Pythagorean theorem.
If $a^2 + b^2 = c^2$, the triangle is right-angled with legs $a, b$ and hypotenuse $c$.

$$5^2 + 12^2 = 25 + 144 = 169 = 13^2 \;\Rightarrow\; \text{right triangle with legs } 5, 12$$

💡 Grade 8 "explain a proof of the converse of the Pythagorean theorem" — $5\text{-}12\text{-}13$ is the classic right-triangle pattern.

#7 Break Into Subproblems 6.G.A.1 Step 4

Compute the area.
In a right triangle the two legs serve as base and height, so the area is half their product.

$$\text{Area} = \tfrac{1}{2} \cdot 5 \cdot 12 = 30 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 6 "find area of right triangles" — once base and height are the legs, the formula gives the answer in one step.

[1] #7 8.EE.A.2 Turn each square's area into a side length. A square with area $A$ has side $\sq

[2] #7 7.G.A.2 Read the side lengths of the triangle from the figure. Each side of the interior

[3] #10 8.G.B.6 Check whether the triangle is right-angled using the converse of the Pythagorean

[4] #7 6.G.A.1 Compute the area. In a right triangle the two legs serve as base and height, so

Review

Reasonableness: Cross-check using the answer choices. (A) $13$ matches a side length, not an area — a trap. (E) $1800$ and (D) $300$ are far too big: the right triangle sits inside the $12 \times 12$ square, so its area cannot exceed $144$. (C) $60 = 5 \cdot 12$ forgets the factor of $\tfrac{1}{2}$. Only (B) $30$ is consistent with the bound and with $\tfrac{1}{2} \cdot 5 \cdot 12$, confirming the answer.

Alternative: Tool #11 (Find an Invariant): the area $25 + 144 = 169$ relation is the same equation as $5^2 + 12^2 = 13^2$, so the sum of the two smaller square areas equals the largest. That invariant immediately signals a right triangle without computing side lengths — then $\tfrac{1}{2} \cdot 5 \cdot 12 = 30$ finishes the problem.

CCSS standards used (min grade 8)

8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form $x^2 = p$ (Turning each square's area into a side length: $\sqrt{25}=5$, $\sqrt{144}=12$, $\sqrt{169}=13$.)
7.G.A.2 Draw geometric shapes with given conditions (Reading the figure to identify that the triangle's sides are exactly the sides of the three squares, giving $5, 12, 13$.)
8.G.B.6 Explain a proof of the Pythagorean Theorem and its converse (Using the converse of the Pythagorean theorem on $5^2 + 12^2 = 13^2$ to conclude the triangle is right-angled with legs $5$ and $12$.)
6.G.A.1 Find the area of right triangles by composing into rectangles or decomposing into triangles (Using $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ with the two legs as base and height to get area $30$.)

⭐ Each square's area gives you a side of the triangle. Spot the $5$-$12$-$13$ right triangle and the area is just $\tfrac{1}{2} \cdot 5 \cdot 12 = 30$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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