2024 AMC 10 A Problem 22

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To help you master 2024 AMC 10 A Problem 22, 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

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2024 AMC10 A Problem 22

Solution

Solution 1

Base-and-Height Area Method

Use the area formula for a right triangle to find its altitude to the hypotenuse. Then read the base and height of $△ABC$ by counting the repeated hypotenuse and altitude lengths in the tiling.

A triangle with base $b$ and perpendicular height $h$ has area $\frac{1}{2}bh$. The same small right triangle can be described using either its two legs or its hypotenuse and its altitude.

Step 1: Find the common hypotenuse of a small right triangle

The two legs have lengths $1$ and $\sqrt{3}$. Apply the Pythagorean Theorem to find the hypotenuse shared by the two right triangles in one kite.

$$1^2+(\sqrt{3}{)}^2=1+3=4$$

$$\text{hypotenuse}=\sqrt{4}=2$$

Step 2: Find the altitude to that hypotenuse

Let $h$ be the perpendicular distance from the right-angle vertex of one small right triangle to its hypotenuse. Compute the triangle’s area first from its legs and then from its hypotenuse and altitude.

$$[△]=\frac{1}{2}(1)(\sqrt{3})=\frac{\sqrt{3}}{2}$$

$$[△]=\frac{1}{2}(2)(h)=h$$

$$h=\frac{\sqrt{3}}{2}$$

Step 3: Read the base and height of the large triangle from the tiling

In the shown arrangement, $AB$ consists of three consecutive common hypotenuses. Also, the perpendicular distance from $C$ down to the line containing $AB$ passes through three altitude layers. These layers are perpendicular to the parallel hypotenuse segments, so they give the height of $△ABC$.

$$AB=3\cdot 2=6$$

$$\text{height from}C\text{to}AB=3\cdot \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$$

Step 4: Apply the triangle area formula

Use $AB$ as the base and the perpendicular distance from $C$ to the line containing $AB$ as the corresponding height.

$$[△ABC]=\frac{1}{2}\cdot 6\cdot \frac{3\sqrt{3}}{2}$$

$$[△ABC]=\frac{9\sqrt{3}}{2}$$

Correct answer: B. $\frac{9}{2}\sqrt{3}$
Therefore, the area is $\boxed{\frac{9}{2}\sqrt{3}}$, which is choice $\boxed{\text{B}}$. As a check, the small right triangle has area $\frac{\sqrt{3}}{2}$, and the large triangle’s area is $9$ times that amount: $9\cdot \frac{\sqrt{3}}{2}=\frac{9\sqrt{3}}{2}$.

Similar Problems in Real Exams

The following five AMC 10 A problems are closely related to AMC 10 A 2024 Question 22. These real AMC 10 A problems test the same key concept—Area. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2025 AMC 10 A Problem 15

2025 AMC 10 A Problem 10

2024 AMC 10 A Problem 14

2024 AMC 10 A Problem 16

2023 AMC 10 A Problem 15

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Area of a Triangle:

$b$ is any chosen base of the triangle, and $h$ is the perpendicular distance from the opposite vertex to the line containing that base. It is used first with the legs of a small right triangle and then with its hypotenuse and altitude. It is also used with base $AB$ for $△ABC$.

• Repeated-length counting in a tiling:

The pieces must lie end-to-end on the same line. For a height, the repeated layers must be perpendicular to the chosen base. The diagram shows three hypotenuse lengths along $AB$ and three equal altitude layers from $C$ to the line containing $AB$.

Key Takeaway: Remember This Strategy for Next Time

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