Find the area of the largest rectangle that can be inscribed in a semicircle of radius "r"?
place your sketch on the x-y grid with the centre at (0,0)
let the radius of the semicircle be r, where r is a constant
so the base of the rectangle is 2x, let its height be y
then x^2 + y^2 = r^2
y =(r^2 - x^2)^(1/2)
area = 2xy
= 2x(r^2-x^2)^(1/2)
d(area)/dx = (2x)(1/2)(r^2 - x^2)^(-1/2) (-2x) + 2(r^2-x^2)^(1/2)
= 0 for a max/min of area
2(r^2 - x^2)^(1/2) = 2x^2/(r^2-x^2)^(1/2)
x^2 = r^2 - x^2
2x^2 = r^2
x^2 = r^2/2
x = r/√2 = (√2/2)r
then largest area
= 2x√(r^2 - x^2)
= (4√2/2)r (r^2 - r^2/2^(1/2)
= 2√2r(r/√2) = 2r^2
or, in a real simple way
suppose we look at the whole circle, the largest "rectangle we can fit inside the circle is a square, where x = y
then the sides of the square are 2x and 2y,
and the area is 4xy, but x=y and x^2 + x^2 = r^2 ---> x^2 = r^2/2
so the largest area = 4x^2
= 4(r^2/2) = 2r^2
To find the area of the largest rectangle that can be inscribed in a semicircle of radius "r," follow these steps:
Step 1: Draw a diagram of the semicircle and the rectangle inscribed within it. The rectangle should have its longer side parallel to the diameter of the semicircle.
Step 2: Since the rectangle is inscribed within the semicircle, the longer side of the rectangle will be equal to the diameter of the semicircle, which is 2r.
Step 3: To find the area of the rectangle, multiply the length by the width. In this case, the length is 2r, and the width is perpendicular to the diameter, which means its length is r.
Step 4: Calculate the area by multiplying the length and the width. Therefore, the area of the rectangle inscribed in the semicircle is (2r) * (r) = 2r^2.
Hence, the area of the largest rectangle that can be inscribed in a semicircle of radius "r" is 2r^2.
To find the area of the largest rectangle that can be inscribed in a semicircle of radius "r", we will need to apply some geometry concepts and optimization techniques.
Step 1: Understand the problem
In this problem, we are given a semicircle with a radius of "r". We need to find the dimensions of the largest rectangle that can be inscribed within this semicircle, and then calculate its area.
Step 2: Identify the shape and its properties
To solve this problem, we need to understand the geometric properties of the rectangle inscribed in a semicircle. By visualizing the problem, we can observe that the rectangle will have its longer side aligned with the diameter of the semicircle. The shorter side of the rectangle will be parallel to the curved edge of the semicircle.
Step 3: Determine the dimensions of the rectangle
Let's assume that the longer side of the rectangle is "2x" units (twice the length of the shorter side). To determine the value of "x", we will use Pythagoras' theorem.
In a circle, the diameter is twice the radius, so the diameter of the semicircle will be "2r" units. Since the longer side of the rectangle coincides with the diameter, we have:
(x^2) + (2x)^2 = (2r)^2
Simplifying this equation, we get:
x^2 + 4x^2 = 4r^2
5x^2 = 4r^2
x^2 = (4r^2) / 5
x = sqrt((4r^2) / 5)
x = (2r / sqrt(5))
Now we have the value of "x". The dimensions of the rectangle will be "2x" and "x".
Step 4: Calculate the area of the rectangle
The area of a rectangle is given by the product of its length and width. In this case, the length is "2x" and the width is "x". Thus, the area of the rectangle is:
Area = (2x) * (x)
Area = 2x^2
Area = 2 * ((2r / sqrt(5))^2)
Area = 2 * ((4r^2) / 5)
Area = (8r^2) / 5
Therefore, the area of the largest rectangle that can be inscribed in a semicircle of radius "r" is equal to (8r^2) / 5.