AMC 8 · 2017 · #18

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

pythagorean-theoremarea-triangles area-differenceidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ , $BC=4$ , $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$ ?

$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A non-convex quadrilateral $ABCD$ has a right angle at $C$ (so $\angle BCD = 90^\circ$) with legs $BC = 4$ and $CD = 3$. The other two sides are $AB = 12$ and $AD = 13$. The picture shows that vertex $C$ pokes inward, so the quadrilateral looks like the big triangle $\triangle ABD$ with the small triangle $\triangle BCD$ scooped out. Find the area of $ABCD$.

Givens: $\angle BCD = 90^\circ$; $BC = 4$, $CD = 3$; $AB = 12$, $AD = 13$; Quadrilateral $ABCD$ is non-convex with $C$ on the inside of $\triangle ABD$; Answer choices: (A) $12$, (B) $24$, (C) $26$, (D) $30$, (E) $36$

Unknowns: The area of quadrilateral $ABCD$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

The shape is an awkward non-convex quadrilateral, but the diagonal $BD$ splits it into two right triangles we already understand. Tool #7 (Identify Subproblems) breaks the area question into three pieces: (a) find area of $\triangle BCD$, (b) find $BD$ so we can study $\triangle ABD$, (c) find area of $\triangle ABD$, then subtract. Tool #1 (Draw a Diagram) is the natural companion — sketching the figure and drawing diagonal $BD$ makes the 3-4-5 and 5-12-13 right triangles jump out, which is the whole shortcut. Tool #3 (Eliminate Possibilities) gives a quick sanity check at the end against the five answer choices.

Execute — Answer: B

#1 Draw a Diagram 6.G.A.1 Step 1

Sketch the quadrilateral and draw the diagonal $BD$.
This single extra line decomposes $ABCD$ into the small inner triangle $\triangle BCD$ (with the marked right angle at $C$) and the larger outer triangle $\triangle ABD$.
Because $C$ sits inside $\triangle ABD$, the quadrilateral's area equals area($\triangle ABD$) $-$ area($\triangle BCD$).

$$\text{Area}(ABCD) = \text{Area}(\triangle ABD) - \text{Area}(\triangle BCD)$$

💡 Cutting a hard shape with one diagonal turns it into shapes whose areas we already know how to compute.

#7 Identify Subproblems 6.G.A.1 Step 2

Compute the area of $\triangle BCD$.
The right angle is at $C$, so the two legs $BC$ and $CD$ are the base and height.
Multiply and halve.

$$\text{Area}(\triangle BCD) = \tfrac{1}{2} \cdot BC \cdot CD = \tfrac{1}{2} \cdot 4 \cdot 3 = 6$$

💡 For a right triangle the two legs are automatically a base and a perpendicular height.

#7 Identify Subproblems 8.G.B.7 Step 3

Find the diagonal $BD$ with the Pythagorean theorem inside $\triangle BCD$.
The legs are $3$ and $4$, so the hypotenuse $BD$ is the familiar $3$-$4$-$5$ triple.

$$BD = \sqrt{BC^2 + CD^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$$

💡 Pythagoras turns the two leg lengths into the hypotenuse length whenever the angle between them is $90^\circ$.

#7 Identify Subproblems 8.G.B.6 Step 4

Check whether $\triangle ABD$ is also a right triangle.
Its sides are $AB = 12$, $BD = 5$, and $AD = 13$.
Test the converse of the Pythagorean theorem: if $12^2 + 5^2 = 13^2$ then the angle opposite the longest side ($\angle ABD$) is $90^\circ$.

$$12^2 + 5^2 = 144 + 25 = 169 = 13^2 \;\checkmark$$

💡 The converse of Pythagoras lets us upgrade a side-length match into the existence of a right angle — here giving us the bonus $5$-$12$-$13$ right triangle.

#7 Identify Subproblems 6.G.A.1 Step 5

Compute the area of $\triangle ABD$.
Now that we know the right angle is at $B$, the legs $AB = 12$ and $BD = 5$ play the roles of base and height.

$$\text{Area}(\triangle ABD) = \tfrac{1}{2} \cdot AB \cdot BD = \tfrac{1}{2} \cdot 12 \cdot 5 = 30$$

💡 Same right-triangle area trick as before: two perpendicular legs, half their product.

#3 Eliminate Possibilities 6.G.A.1 Step 6

Subtract to get the quadrilateral's area, and compare with the answer choices to eliminate the four wrong ones.

$$\text{Area}(ABCD) = 30 - 6 = 24 \;\Rightarrow\; \textbf{(B)}$$

💡 $24$ matches choice (B); $12$, $26$, $30$, and $36$ are eliminated, with $30$ being the trap that forgets to remove the scooped-out piece.

[1] #1 6.G.A.1 Sketch the quadrilateral and draw the diagonal $BD$. This single extra line deco

[2] #7 6.G.A.1 Compute the area of $\triangle BCD$. The right angle is at $C$, so the two legs

[3] #7 8.G.B.7 Find the diagonal $BD$ with the Pythagorean theorem inside $\triangle BCD$. The

[4] #7 8.G.B.6 Check whether $\triangle ABD$ is also a right triangle. Its sides are $AB = 12$,

[5] #7 6.G.A.1 Compute the area of $\triangle ABD$. Now that we know the right angle is at $B$,

[6] #3 6.G.A.1 Subtract to get the quadrilateral's area, and compare with the answer choices to

Review

Reasonableness: A quick reality check: $\triangle ABD$ alone has area $30$ and $\triangle BCD$ has area $6$, so the answer must sit strictly between $30 - 6 = 24$ (carve out the dent) and $30 + 6 = 36$ (if the dent were instead an outward bump). The non-convex picture forces the subtraction, giving $24$. The trap answer (D) $= 30$ corresponds to forgetting that $C$ is on the inside; (E) $= 36$ would be the convex case. So $24$ is the only number consistent with both the arithmetic and the picture.

Alternative: Tool #1 (Draw a Diagram) plus coordinates: place $C = (0,0)$, $B = (4,0)$, $D = (0,3)$ using the right angle at $C$. Then $A$ is the point with $AB = 12$ and $AD = 13$; solving gives $A = (-2.4, -3.6) \cdot (-5)\ldots$ or, more cleanly, apply the Shoelace formula to the vertices $A$, $B$, $C$, $D$ in order. Both routes recover area $= 24$, but they require more setup than the subproblem decomposition above.

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing or decomposing (Decomposing the non-convex quadrilateral $ABCD$ into $\triangle ABD$ minus $\triangle BCD$, and computing each right-triangle area as $\tfrac{1}{2} \cdot \text{leg}_1 \cdot \text{leg}_2$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using $BC = 4$ and $CD = 3$ with the right angle at $C$ to compute the diagonal $BD = 5$.)
8.G.B.6 Explain a proof of the Pythagorean theorem and its converse (Using the converse to verify that $\triangle ABD$ with sides $5$, $12$, $13$ is a right triangle (because $5^2 + 12^2 = 13^2$), so its legs can serve as base and height.)

⭐ This AMC 8 problem only needs Grade 8 Pythagorean theorem (and its converse) you already know — once you spot the 3-4-5 and 5-12-13 right triangles, it's just one subtraction!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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