15 Aug, 2024

· Mathematics

What is the quarter circle's area

The perimeter of a quarter circle is 3.57 cm.

Short Answer

Some answer Some answer Some answer

Long Explanation

Explanation

Given the perimeter of a quarter circle is 3.57 cm, we need to find its area.

First, let's break down the components of the given perimeter. The perimeter $P$ of a quarter circle includes:

One-fourth of the circle's circumference
The two radii

If $r$ is the radius of the circle, the total perimeter of the quarter circle can be represented as:

P = \frac{1}{4} \times 2\pi r + 2r

Given:

P = 3.57 \, \text{cm}

We can write the equation as:

\frac{1}{2} \pi r + 2r = 3.57

Now, let's solve for the radius $r$ .

First, we'll combine the terms involving $r$ :

r \left( \frac{\pi}{2} + 2 \right) = 3.57

Solving for $r$ :

r = \frac{3.57}{\frac{\pi}{2} + 2}

r = \frac{3.57}{1.57 + 2}

r \approx \frac{3.57}{3.57} = 1 \, \text{cm}

So, the radius $r$ is approximately $1 \, \text{cm}$ .

Next, we compute the area $A$ of the quarter circle. The area of a full circle is $\pi r^2$ , so the area of a quarter circle is:

A = \frac{1}{4} \pi r^2

Using the radius we found:

A = \frac{1}{4} \pi (1)^2

A = \frac{1}{4} \pi

Plugging in the value of $\pi \approx 3.14159$ :

A \approx \frac{3.14159}{4}

A \approx 0.7854 \, \text{cm}^2

Therefore, the area of the quarter circle is approximately 0.7854 cm $^2$ .

Verified By

Emily Rosen

Mathematics Content Writer at Math AI

Emily Rosen is a recent graduate with a Master's in Mathematics from the University of Otago. She has been tutoring math students and working as a part-time contract writer for the past three years. She is passionate about helping students overcome their fear of math through easily digestible content.

mathematics

Concept

The Specific And Relevant Concepts Used To Answer This Question Are: Perimeter Calculation For Quarter Circles

Explanation

Calculating the perimeter of a quarter circle involves understanding the definitions and formulas related to the geometry of circles. Below are the specific and relevant concepts for this calculation:

Basics of Circle Geometry

Radius ( $r$ ): The distance from the center of the circle to any point on its circumference.
Circumference ( $C$ ): The total distance around the circle, given by the formula: $C = 2\pi r$

Quarter Circle Perimeter

A quarter circle is one-fourth of a full circle. To find its perimeter, we need to consider both the arc length and the straight edges:

Arc Length: The arc length of a quarter circle is one-fourth of the circumference of the full circle:
$\text{Arc Length} = \frac{C}{4} = \frac{2\pi r}{4} = \frac{\pi r}{2}$
Straight Edges: The quarter circle has two straight edges, each equal to the radius:
$\text{Straight Edges Total} = r + r = 2r$
Total Perimeter: Combining the arc length and the straight edges, the total perimeter ( $P$ ) of the quarter circle is given by:
$P = \text{Arc Length} + \text{Straight Edges Total} = \frac{\pi r}{2} + 2r$

Summary

The perimeter of a quarter circle can be calculated using the formula:

P = \frac{\pi r}{2} + 2r

where $r$ is the radius of the quarter circle.

Key Points

Remember to include both the curved arc and the straight edges in the perimeter calculation.
The formula is derived from basic circle geometry principles.

Concept

Solving Algebraic Equations For The Variable

Explanation

Solving algebraic equations for the variable involves finding the value(s) of the variable that make the equation true. This is a fundamental skill in algebra and is used to determine unknown quantities.

Steps to Solve Algebraic Equations

Simplify Both Sides of the Equation:
- Combine like terms
- Use the distributive property
Isolate the Variable:
- Use inverse operations to move terms involving the variable to one side of the equation and constants to the other side.
Solve the Equation:
- Simplify the expression to find the value of the variable.
Verify the Solution:
- Substitute the solution back into the original equation to ensure it satisfies the equation.

Examples

Example 1: Simple Linear Equation

Solve the equation:

3x + 5 = 14

Step 1: Subtract 5 from both sides

3x + 5 - 5 = 14 - 5

3x = 9

Step 2: Divide both sides by 3

x = \frac{9}{3}

x = 3

Verification: Substitute $x = 3$ back into the original equation:

3(3) + 5 = 14

9 + 5 = 14

14 = 14 (True)

Example 2: Quadratic Equation

Solve the equation:

x^2 - 5x + 6 = 0

Step 1: Factor the quadratic equation

(x - 2)(x - 3) = 0

Step 2: Set each factor equal to zero and solve

\begin{align*} x - 2 &= 0 \\ x &= 2 \end{align*}

\begin{align*} x - 3 &= 0 \\ x &= 3 \end{align*}

Verification: Substitute $x = 2$ and $x = 3$ back into the original equation: For $x = 2$ :

(2)^2 - 5(2) + 6 = 0

4 - 10 + 6 = 0

0 = 0 (True)

For $x = 3$ :

(3)^2 - 5(3) + 6 = 0

9 - 15 + 6 = 0

0 = 0 (True)

Important Points

Inverse Operations: To isolate the variable, use operations that undo each other, e.g., addition and subtraction or multiplication and division.
Verification: Always check your solutions by substituting them back into the original equation to ensure they satisfy the equation.
Factoring: For quadratic equations, factoring can simplify finding solutions but other methods like completing the square or using the quadratic formula may also be used.