# Find the Area of a Rectangle Bounded by a Semicircle | Geometry Problem Solution

• slayer16
In summary, the rectangle bounded by the x-axis and the semicircle y=√(36-x^2) has a base of length 2x and a height of √(36-x^2). Therefore, the area A can be expressed as A=2x√(36-x^2).
slayer16

## Homework Statement

A rectangle is bounded by the x-axis and the semicircle y=√(36-x^2) (see figure). Write the area A of the rectangle as a function of x.

I will try to explain the figure as best as possible. The figure is basically a semicircle on the Cartesian plane with a domain of [-6,6]. There is a point on the semicircle of (X,Y) where y is the height of the rectangle and x is the maximum width of the rectangle. The value of x is around five on the graph, while the value for y is around 3.5 on the graph (note: they are just estimates to help visualize the figure better).

A=(w)(h)

## The Attempt at a Solution

I substitute the (x,y) for (w,h). I then realized that the height stays constant, therefore I can write the height as √(36-w^2), where w is a constant. Now, I do not know how to write the width because the x has a limited domain.

Couldn't you use a definite integral to evaluate this?

jegues said:
Couldn't you use a definite integral to evaluate this?

Not really. The question is found in a precalculus textbook.

slayer16 said:

## Homework Statement

A rectangle is bounded by the x-axis and the semicircle y=√(36-x^2) (see figure). Write the area A of the rectangle as a function of x.

I will try to explain the figure as best as possible. The figure is basically a semicircle on the Cartesian plane with a domain of [-6,6]. There is a point on the semicircle of (X,Y) where y is the height of the rectangle and x is the maximum width of the rectangle. The value of x is around five on the graph, while the value for y is around 3.5 on the graph (note: they are just estimates to help visualize the figure better).

A=(w)(h)

## The Attempt at a Solution

I substitute the (x,y) for (w,h). I then realized that the height stays constant, therefore I can write the height as √(36-w^2), where w is a constant. Now, I do not know how to write the width because the x has a limited domain.

The rectangle with a corner at (x,y) on your circle in the first quadrant has a base of length 2x and its height is y = sqrt(36-x2). Multiply those together and you will have the area as a function of x. Apparently you are not asking for the dimensions of such a rectangle with largest area, just the expression for A as a function of x.

LCKurtz said:
The rectangle with a corner at (x,y) on your circle in the first quadrant has a base of length 2x and its height is y = sqrt(36-x2). Multiply those together and you will have the area as a function of x. Apparently you are not asking for the dimensions of such a rectangle with largest area, just the expression for A as a function of x.

Thanks LCKurtz

## 1. What is the formula for finding the area of a rectangle bounded by a semicircle?

The formula for finding the area of a rectangle bounded by a semicircle is A = l * w + (π/2) * (r^2), where l is the length of the rectangle, w is the width of the rectangle, and r is the radius of the semicircle.

## 2. How do you know if a given shape can be divided into a rectangle bounded by a semicircle?

A shape can be divided into a rectangle bounded by a semicircle if it has a straight edge that is the same length as the diameter of the semicircle. This straight edge will be one side of the rectangle and the semicircle will be another side. The other two sides of the rectangle will be perpendicular to the straight edge and tangent to the semicircle.

## 3. Can a rectangle bounded by a semicircle have any dimensions or are there limitations?

A rectangle bounded by a semicircle can have any dimensions as long as it follows the formula for finding the area. However, the rectangle must have a straight edge that is the same length as the diameter of the semicircle in order for it to be divided into a rectangle bounded by a semicircle.

## 4. How is the area of a rectangle bounded by a semicircle different from a regular rectangle?

The area of a rectangle bounded by a semicircle is different from a regular rectangle because it also includes the area of the semicircle. This means that the area formula for a rectangle bounded by a semicircle will have an additional term, (π/2) * (r^2), which represents the area of the semicircle.

## 5. What are some real-life applications of finding the area of a rectangle bounded by a semicircle?

Finding the area of a rectangle bounded by a semicircle can be useful in a variety of real-life situations. For example, it can be used in landscaping to determine the amount of space taken up by a semicircular flower bed or patio. It can also be used in construction to calculate the amount of material needed for a curved wall or window. Additionally, it can be used in engineering for designing curved structures or for calculating the area of a cross-section of a curved pipe or duct.

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