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Let's solve the problem in the provided image using the quadratic equation.

### Problem Statement
We have two plane figures:
1. Half a circle of radius \( x \) units.
2. A quarter of a circle of radius \( x + 2 \) units.

The areas of these two figures are equal.

We need to:
1. Construct a quadratic equation in terms of \( x \).
2. Solve for \( x \) correct to the first decimal place using \( \sqrt{2} \approx 1.41 \).

### Step-by-Step Solution

1. **Area Calculation:**
   - **Area of the half-circle:**
     \[
     \text{Area} = \frac{1}{2} \pi x^2
     \]

   - **Area of the quarter-circle:**
     \[
     \text{Area} = \frac{1}{4} \pi (x + 2)^2
     \]

2. **Equating the areas:**
   \[
   \frac{1}{2} \pi x^2 = \frac{1}{4} \pi (x + 2)^2
   \]

   Simplify by multiplying both sides by 4:
   \[
   2 \pi x^2 = \pi (x + 2)^2
   \]

   Cancel \( \pi \) from both sides:
   \[
   2 x^2 = (x + 2)^2
   \]

   Expanding the right side:
   \[
   2 x^2 = x^2 + 4x + 4
   \]

   Rearrange to form a quadratic equation:
   \[
   2 x^2 - x^2 - 4x - 4 = 0
   \]
   \[
   x^2 - 4x - 4 = 0
   \]

3. **Solving the quadratic equation:**
   We solve the quadratic equation \( x^2 - 4x - 4 = 0 \) using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For our equation \( x^2 - 4x - 4 = 0 \):
   \[
   a = 1, \quad b = -4, \quad c = -4
   \]
   \[
   x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
   \]
   \[
   x = \frac{4 \pm \sqrt{16 + 16}}{2}
   \]
   \[
   x = \frac{4 \pm \sqrt{32}}{2}
   \]
   \[
   x = \frac{4 \pm 4\sqrt{2}}{2}
   \]
   \[
   x = 2 \pm 2\sqrt{2}
   \]
   Since \( \sqrt{2} \approx 1.41 \):
   \[
   x = 2 + 2(1.41)
   \]
   \[


Explore how to solve a geometry problem involving the areas of a half circle and a quarter circle with equal areas. This problem utilizes geometric concepts and equations, including the area formula of circles. Suitable for Grades 10-12 students.

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