2017 AMC 8 Problem 18

Improve Your AMC 8 Score

Practice with AMC 8 problems by topic and difficulty level.

Get Free AMC 8 Practice

Improve Your AMC 8 Score

Practice with AMC 8 problems by topic and difficulty level.

Get Free AMC 8 Practice

One Page to Learn. One Point to Earn.

In This Page:

To help you master 2017 AMC 8 Problem 18, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

Want more practice? Scroll down to download Think Academy’s AMC 8 practice packs and keep building your AMC 8 skills.

2017 AMC8 Problem 18

Solution

Solution 1

Pythagorean Decomposition Method

By drawing the diagonal BD, we can split the complicated area problem into two simpler triangle area problems. Because the shape is non-convex (shaped like a dart), its area is the difference between the larger triangle and the smaller inward-pointing triangle.

Converse of the Pythagorean Theorem

Step 1: Identify the missing diagonal

The shape is a non-convex quadrilateral. To calculate its area, it is helpful to divide it into simpler shapes, such as triangles, by drawing a diagonal. Connect point B to point D. This splits the total area into two triangles: a larger triangle ABD and a smaller triangle BCD that represents the ‘bite’ taken out of the larger triangle.

Area (A BC D) = Area (△ A B D) - Area (△ BC D)

Step 2: Analyze the smaller triangle BCD

The problem states that angle BCD is a right angle, meaning triangle BCD is a right triangle. We are given the lengths of its two legs. We can calculate its area directly.

Area (△ BC D) = \frac{1}{2} \cdot BC \cdot C D

Area (△ BC D) = \frac{1}{2} \cdot 4 \cdot 3 = 6

Step 3: Find the length of the diagonal BD

We can use the Pythagorean Theorem on right triangle BCD to find the length of the diagonal BD, which is the hypotenuse.

B D^{2} = B C^{2} + C D^{2}

B D^{2} = 4^{2} + 3^{2} = 16 + 9 = 25

B D = 5

Step 4: Analyze the larger triangle ABD

Now we look at triangle ABD. We know all three of its side lengths: AB=12, AD=13, and we just found BD=5. Let’s check if the squares of these sides satisfy the Pythagorean relationship.

5^{2} + 1 2^{2} = 25 + 144 = 169

1 3^{2} = 169

B D^{2} + A B^{2} = A D^{2}

Step 5: Calculate the area of triangle ABD

Because the side lengths satisfy the Pythagorean Theorem, the Converse of the Pythagorean Theorem tells us that triangle ABD is a right triangle with the right angle at B (opposite the longest side AD). Therefore, we can easily find its area using its perpendicular legs of 5 and 12.

Area (△ A B D) = \frac{1}{2} \cdot A B \cdot B D

Area (△ A B D) = \frac{1}{2} \cdot 12 \cdot 5 = 30

Step 6: Compute the final area of quadrilateral ABCD

As established in the first step, the area of the non-convex quadrilateral is the area of the larger right triangle minus the area of the smaller right triangle.

Area (A BC D) = 30 - 6 = 24

Correct answer: B. $24$
24 (Answer choice B). We can verify that subtracting is correct because the word ‘non-convex’ implies a dart-like shape, meaning triangle BCD is a ‘bite’ taken out of the larger triangle ABD. If we had mistakenly added the areas, we would get 36, which is the area of a convex kite-like shape with these same sides. 36 is explicitly offered as a trap in the answer choices, confirming our subtraction logic is correct.

Solution 2

Cartesian Coordinates and Shoelace Method

We can place the entire figure onto the xy-coordinate plane to find the exact coordinates of every vertex. Once we have the coordinates, we can use the Shoelace Formula to calculate the area of the polygon algebraically without needing a visual decomposition.

Shoelace Formula

Step 1: Set up a coordinate system

First, we note that the distance from B to D is the hypotenuse of right triangle BCD, so BD = 5. Because the sides of triangle ABD are 5, 12, and 13, it is a right triangle with its right angle at B. This makes the figure perfectly suited for the Cartesian coordinate plane. We can place the right angle at the origin, with the legs lying along the x and y axes.

B D = 3^{2} + 4^{2} = 5

5^{2} + 1 2^{2} = 1 3^{2} ⟹ ∠ A B D = 9 0^{\circ}

B = (0, 0), A = (12, 0), D = (0, 5)

Step 2: Create equations for point C

Let point C have coordinates (x, y). We know the distances from C to B and D. Using the distance formula, we can set up two equations based on these lengths.

BC = 4 ⟹ x^{2} + y^{2} = 16

C D = 3 ⟹ x^{2} + (y - 5)^{2} = 9

Step 3: Solve for the y-coordinate of C

We can expand the second equation. Notice that it contains the expression x^2 + y^2, which we know equals 16 from the first equation. We can substitute 16 into the expanded equation to quickly solve for y.

x^{2} + y^{2} - 10 y + 25 = 9

(x^{2} + y^{2}) - 10 y + 25 = 9

16 - 10 y + 25 = 9

41 - 10 y = 9

10 y = 32

y = 3.2

Step 4: Solve for the x-coordinate of C

Substitute the value of y back into the first equation. We choose the positive square root because point C must lie in the first quadrant to be inside the right triangle ABD, which creates the non-convex ‘dart’ shape.

x^{2} + (3.2)^{2} = 16

x^{2} + 10.24 = 16

x^{2} = 5.76

x = 2.4

Step 5: Apply the Shoelace Formula

Now that we have the coordinates of all four vertices, we can list them in order around the perimeter of the quadrilateral (A, B, C, D) and use the Shoelace Formula to calculate the area. First, calculate the sum of the down-right diagonal products.

(x_{A} \cdot y_{B}) + (x_{B} \cdot y_{C}) + (x_{C} \cdot y_{D}) + (x_{D} \cdot y_{A})

(12 \cdot 0) + (0 \cdot 3.2) + (2.4 \cdot 5) + (0 \cdot 0)

0 + 0 + 12 + 0 = 12

Step 6: Complete the Shoelace calculation

Next, calculate the sum of the down-left diagonal products. Then, find half of the absolute difference between the two sums to get the final area.

(y_{A} \cdot x_{B}) + (y_{B} \cdot x_{C}) + (y_{C} \cdot x_{D}) + (y_{D} \cdot x_{A})

(0 \cdot 0) + (0 \cdot 2.4) + (3.2 \cdot 0) + (5 \cdot 12)

0 + 0 + 0 + 60 = 60

Area = \frac{1}{2} ∣12 - 60∣ = \frac{1}{2} ∣ - 48∣ = 24

Correct answer: B. $24$
24. The Shoelace formula calculation perfectly matches our geometric deduction, proving that the geometric decomposition was mathematically rigorous.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Pythagorean Theorem:

a^{2} + b^{2} = c^{2}

a and b are the lengths of the legs of a right triangle, and c is the length of the hypotenuse. It is used to find the missing diagonal length BD in the right triangle BCD.

Converse of the Pythagorean Theorem:

If the square of one side of a triangle equals the sum of the squares of the other two sides, the triangle is a right triangle.

Applies to any triangle where the side lengths are known. It proves that triangle ABD is a right triangle because its side lengths 5, 12, and 13 satisfy 5^2 + 12^2 = 13^2.

Key Takeaway: Remember This Strategy for Next Time

A quadrilateral with missing side lengths or angles, but enough given segments to form common triangles.

Try: Draw a diagonal to divide the quadrilateral into two triangles.

Why: Triangles are much easier to solve than quadrilaterals, and diagonal lines often reveal hidden right triangles or familiar Pythagorean triples.

The phrase 'non-convex quadrilateral' alongside a diagram that looks like a dart.

Try: Calculate the area by finding a larger enclosing triangle and subtracting the smaller 'cutout' triangle.

Why: A non-convex polygon has an inward-pointing vertex, meaning a portion of the enclosing bounding shape has been removed.

A geometry problem with multiple right angles or known Pythagorean side lengths.

Try: Consider placing the figure on a Cartesian coordinate plane.

Why: Right angles align perfectly with the x and y axes, allowing you to easily find vertex coordinates and use powerful tools like the Shoelace formula.

Unlock More Targeted AMC 8 Practice Resource

Build Stronger Skills. Practice with Purpose. Prepare with Confidence.

Get a complete set of module-focused AMC 8 practice workbooks designed to help students strengthen core concepts, improve problem-solving strategies, and feel more confident on test day.

With 250 practice problems organized across six essential AMC 8 modules, students can focus on one skill area at a time and learn from clear, step-by-step solutions.

AMC8 Practice Resource Pack Promo Cropped Under 2MB

Download the Free AMC 8 Practice Resource Pack

Inside, You’ll Find:

6 module-focused practice workbooks: Algebra, Geometry, Number Theory, Combinatorics, and Probability & Statistics.
250 AMC 8-style practice problems covering essential concepts and common competition question types.
Complete answer keys and step-by-step solutions to help students understand the reasoning behind each answer.
Organized, print-ready materials for independent study, guided practice, or extra preparation at home.

Get Instant Access: Fill out the form below to receive our exclusive resource pack.

*Please enter a valid email address to access the materials and receive future prep updates.

About Think Academy

Think Academy US Online, part of TAL Education Group (NYSE: TAL), provides structured math and ELA programs that help students build strong academic foundations and grow with confidence. With expert teachers, proven methods, and innovative AI tools, we nurture learning ability, curiosity, and healthy study habits that support long-term growth and a lifelong love of learning.

Check our programs:

Think Kids: Ages 3-5
Math
- Lower Elementary Math — Grades 1–2
- Upper Elementary Math — Grades 3–5
- Middle School Math — Grade 6–8
- High School Math — Grade 9-12
ELA
- Elementary ELA — Grades 2–5

Want more insights on Math & ELA learning and parenting? Fill out the form on this page and subscribe to our newsletter for weekly tips and the latest resources. Join 100,000+ families in our Parent Community!

Categories: Math CompetitionsTags: AMC 8, AMC 8 2017

Get expert PreK–12 insights in your inbox every week

To support your child’s learning, confidence, and growth:
✓ Competition & Test Prep
✓ Printable Worksheets
✓ Webinar Invites

Join 100,000+ subscribers in our parent community.

2017 AMC 8 Problem 18

In This Page:

2017 AMC8 Problem 18

Solution

Solution 1

Pythagorean Decomposition Method

Step 1: Identify the missing diagonal

Step 2: Analyze the smaller triangle BCD

Step 3: Find the length of the diagonal BD

Step 4: Analyze the larger triangle ABD

Step 5: Calculate the area of triangle ABD

Step 6: Compute the final area of quadrilateral ABCD

Solution 2

Cartesian Coordinates and Shoelace Method

Step 1: Set up a coordinate system

Step 2: Create equations for point C

Step 3: Solve for the y-coordinate of C

Step 4: Solve for the x-coordinate of C

Step 5: Apply the Shoelace Formula

Step 6: Complete the Shoelace calculation

Similar Problems in Real Exams

2026 AMC 8 Problem 13

2019 AMC 8 Problem 2

2018 AMC 8 Problem 22

2014 AMC 8 Problem 6

2013 AMC 8 Problem 24

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

Key Takeaway: Remember This Strategy for Next Time

Unlock More Targeted AMC 8 Practice Resource

Build Stronger Skills. Practice with Purpose. Prepare with Confidence.

Download the Free AMC 8 Practice Resource Pack

AMC 8 Practice Resource Pack Free Download

About Think Academy

Leave a Comment & Share Your Thoughts!

Post Content

Get expert PreK–12 insights in your inbox every week

Share This Story, Choose Your Platform!

Related Posts

Download Free AMC 8 Practice Resource Pack

Download Free AMC 8 Practice Resource Pack

Get expert PreK-12 insights 100,000+ families rely on.

AMC 8 Practice Resource Pack
Free Download