AMC 8 · 2003 · #6

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

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?

Pick an answer.

(A)

(B)

(C)

(D)

300

(E)

1800

AMC 8 2003 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.

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)

More like this