A rectangle is to be inscribed in a semicircle of radius 8, with one side lying on the diameter of the circle. What is the maximum possible area of the rectangle?

The maximum possible area of the rectangle is 64π square units.

To find the maximum area of the rectangle, we need to understand the relationship between the rectangle and the semicircle.

Let's assume that the semicircle lies in the coordinate plane, with its center at the origin (0,0) and one of its endpoints on the x-axis.

Let the length of the rectangle be 2x units, which is the same as the length of the diameter of the semicircle.

Now we need to find the width of the rectangle. Since the width of the rectangle lies on the diameter of the semicircle, it is also the radius of the semicircle. Therefore, the width of the rectangle is 8 units.

Since the length of the rectangle is equal to 2x, the area of the rectangle is given by A = length × width = 2x × 8 = 16x.

To find the maximum area of the rectangle, we need to maximize the value of x.

The rectangle is inscribed in a semicircle, which means that the diagonal of the rectangle is equal to the diameter of the semicircle. The diagonal of the rectangle can be found using the Pythagorean theorem.

The diagonal of the rectangle is equal to the hypotenuse of a right triangle, with the width of the rectangle as one side (8 units) and half the length of the rectangle (x units) as the other side.

Using the Pythagorean theorem, we have:
(diagonal)^2 = (width)^2 + (half the length)^2
(diagonal)^2 = 8^2 + x^2
(diagonal)^2 = 64 + x^2

Since the diagonal is equal to the diameter of the semicircle (which is 2x), we have:
(2x)^2 = 64 + x^2
4x^2 = 64 + x^2
3x^2 = 64
x^2 = 64/3
x = √(64/3)
x ≈ 5.54

Substituting this value of x into the formula for the area of the rectangle, we have:
A = 16x
A = 16(5.54)
A ≈ 88.64

Therefore, the maximum possible area of the rectangle is approximately 88.64 square units.

To find the maximum possible area of the rectangle, we need to understand its dimensions. Let's assume that the rectangle has one of its sides lying on the diameter of the semicircle.

Let's consider the rectangle's dimensions: length (L) and width (W). Since one side of the rectangle lies on the diameter, its length L equals the radius of the semicircle, which is 8 units.

Now, we need to determine the width W of the rectangle. To do this, we need to consider the fact that the rectangle is inscribed in the semicircle. This means that the rectangle's other two vertices lie on the semicircle.

To visualize this, draw a semicircle with a diameter and a rectangle inscribed in it, with one side touching the diameter. We can see that the width W of the rectangle is the distance between two points on the semicircle, which are equidistant from the diameter.

Now, let's calculate the width W. The two points that are equidistant from the diameter of the semicircle form the endpoints of a diameter perpendicular to the original diameter. These points divide the semicircle into two equal arcs.

Since the length of the semicircle's diameter is 16 (twice the radius), the length of one of these arcs is half the circumference of the semicircle. Therefore, the length of one arc is π*8 = 8π units (where π is approximately 3.14).

To find the width W, we need to divide this arc length by 2π. W = (8π) / (2π) = 4 units.

Now that we know the dimensions of the rectangle (L = 8 and W = 4), we can find its maximum area.

The area of a rectangle is given by the formula A = L * W. Substituting the values we found, we get A = 8 * 4 = 32 square units.

Therefore, the maximum possible area of the rectangle inscribed in the given semicircle is 32 square units.