# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-9/2

Exercises: Try Out - Set 9: Real exam simulation to sharpen your skills and build confidence. - Problem 2

---

## Exercise 2

### Question

export const metadata = {
  title: "Problem 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph
    title="Geometry Visualization"
    description={
        <>
            Square <InlineMath math="ABCD" /> and a circle passing through points{" "}
            <InlineMath math="A" /> and <InlineMath math="D" />, tangent to side{" "}
            <InlineMath math="BC" />.
        </>
    }
/>

Square <InlineMath math="ABCD" /> has a side length of <InlineMath math="24 \text{ cm}" />. A circle passes through points <InlineMath math="A" /> and <InlineMath math="D" />, and is tangent to side <InlineMath math="BC" />. The area of the circle is...


### Choices

- [ ] $$144\pi\text{ cm}^2$$
- [x] $$225\pi\text{ cm}^2$$
- [ ] $$256\pi\text{ cm}^2$$
- [ ] $$336\pi\text{ cm}^2$$
- [ ] $$425\pi\text{ cm}^2$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Problem 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph
    mode="answer"
    title="Solution Visualization"
    description={
        <>
            Circle with center <InlineMath math="F" />, radius{" "}
            <InlineMath math="r" />, and auxiliary lines to calculate the radius.
        </>
    }
/>

Let <InlineMath math="F" /> be the center of the circle, so <InlineMath math="AF = FD = FG = r" /> (the radius of the circle).

Given the side length of the square <InlineMath math="AD = 24 \text{ cm}" />. Since <InlineMath math="E" /> is the midpoint of <InlineMath math="AD" /> (based on the symmetry of the circle passing through <InlineMath math="A" /> and <InlineMath math="D" />), then:

<MathContainer>
  <BlockMath math="AE = ED = \frac{1}{2} AD = 12 \text{ cm}" />
</MathContainer>

#### Finding the Radius of the Circle

We will find the radius of the circle (<InlineMath math="r" />) using the Pythagorean theorem on the right-angled triangle <InlineMath math="\triangle AEF" />.
Note that the length <InlineMath math="EF" /> is the distance from the center of the circle to side <InlineMath math="AD" />.
Since the circle is tangent to side <InlineMath math="BC" /> at point <InlineMath math="G" />, and the total distance from side <InlineMath math="AD" /> to side <InlineMath math="BC" /> is the side length of the square (<InlineMath math="24 \text{ cm}" />), then <InlineMath math="EF = AB - r = 24 - r" />.

Using the Pythagorean theorem on <InlineMath math="\triangle AEF" />:

<MathContainer>
  <BlockMath math="AF^2 = AE^2 + EF^2" />
  <BlockMath math="r^2 = 12^2 + (24 - r)^2" />
  <BlockMath math="r^2 = 144 + (576 - 48r + r^2)" />
  <BlockMath math="r^2 = r^2 - 48r + 720" />
  <BlockMath math="48r = 720" />
  <BlockMath math="r = \frac{720}{48} = 15 \text{ cm}" />
</MathContainer>

#### Calculating the Area of the Circle

After finding the radius <InlineMath math="r = 15 \text{ cm}" />, we can calculate the area:

<MathContainer>
  <BlockMath math="A_o = \pi r^2" />
  <BlockMath math="A_o = \pi (15)^2" />
  <BlockMath math="A_o = 225\pi \text{ cm}^2" />
</MathContainer>

Therefore, the area of the circle is <InlineMath math="225\pi \text{ cm}^2" />.


---