Question

A rectangle is to be inscribed in a semicircle of radius 2 . Determine the maximum area that the rectangle can have and the dimensions of the rectangle. 2. A rectangular poster will contain 50in

Answer 1

The maximum area of the rectangle inscribed in the semicircle is 4 square units, with dimensions of 2 units by 2 units.

Step-by-step explanation:

To find the maximum area of the rectangle inscribed in the semicircle, it helps to visualize the situation. The rectangle's width will be the same as the diameter of the semicircle to maximize its area. Using the Pythagorean theorem for a right-angled triangle within the semicircle, where the rectangle's width is the diameter (2r), and the rectangle's length is the height, the length (l) can be found.

`(r^2 = l^2 + (2r)^2)`

`(4 = l^2 + 4r^2)`

`(l^2 = 4 - 4r^2)`

`(l = sqrt{4 - 4r^2})`

Substituting (r = 2) into the equation yields
`(l = sqrt{4 - 4(2)^2} = sqrt{4 - 16} = sqrt{-12})`. However, considering the rectangle's length can't be negative, the maximum area occurs when the rectangle's width is the semicircle's diameter (2 units) and the rectangle's length is also 2 units. Thus, the maximum area is (2 times 2 = 4) square units.

This maximum area is achieved when the rectangle is a square, fitting perfectly within the semicircle, maximizing its coverage without extending beyond the semicircular boundary.

Learning about how the rectangle fits snugly into the semicircle, forming a square, reveals the relationship between geometric shapes and their maximum utilization within given constraints, which is a common problem-solving approach in mathematics. This concept helps in optimizing space utilization and understanding geometric relationships in various practical scenarios.

Answer 2

The maximum area of the inscribed rectangle is
`\(4\, \text{in}^2\)`, and its dimensions are
`\(4\, \text{in}\) by \(2\, \text{in}\)`.

Step-by-step explanation:

In a semicircle, the rectangle with the maximum area is always a square. This is because, in a square, the diagonal is also the diameter of the semicircle. Let
`\(s\)` be the side length of the square. The area
`\(A\)` of the square can be expressed as
`\(A = s^2\).` In this case, the radius of the semicircle is given as 2 inches. The diameter of the semicircle is twice the radius, which is
`\(2 * 2 = 4\)` inches. Therefore, the side length of the square (and consequently the rectangle inscribed in the semicircle) is
`\(s = 4/√(2) = 2√(2)\).` Thus, the maximum area
`\(A\) is \(4\, \text{in}^2\) as \(A = (2√(2))^2\).`

This result can also be derived by considering the rectangle's width as the diameter of the semicircle. The width
`\(w\)`is equal to
`\(4\)`inches, and the height \(h\) is equal to
`\(2\)`inches, resulting in an area of
`\(A = w * h = 4 * 2 = 8\, \text{in}^2\)`. However, this is not the maximum area, as the rectangle with a square shape (where
`\(w = h\))` yields a larger area. Therefore, the maximum area of
`\(4\, \text{in}^2\)` is achieved when the rectangle is a square.

To summarize, the maximum area of the inscribed rectangle is
`\(4\,`
`\text{in}^2\),`and this occurs when the rectangle is a square with side lengths of
`\(2√(2)\)`inches.