All the vertices of a rectangle lie on a circle of radius $R$. If the perimeter of the rectangle is $P$, then the area of the rectangle is

When a rectangle is inscribed in a circle, the diagonal of the rectangle equals the diameter of the circle. This is because the diagonal connects two points on the circle and passes through the center, making it the longest possible chord.

We define our variables:

Circle radius = $R$

Rectangle length = $l$

Rectangle breadth = $b$

Rectangle perimeter = $P$

Rectangle area = $A$ (what we want to find)

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Since the diagonal of the rectangle equals the diameter of the circle:

Diagonal of rectangle = Diameter of circle

Using the Pythagorean theorem for the rectangle:

$\sqrt{l^2 + b^2} = 2R$

$l^2 + b^2 = 4R^2 \quad \text{...(1)}$

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We're given that the perimeter is $P$:

$P = 2(l + b)$

Therefore:

$l + b = \tfrac{P}{2} \quad \text{...(2)}$

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From equation (2):

$(l + b)^2 = \left(\tfrac{P}{2}\right)^2$

Expanding using $(a + b)^2 = a^2 + 2ab + b^2$:

$l^2 + 2lb + b^2 = \dfrac{P^2}{4}$

We can rewrite this as:

$(l^2 + b^2) + 2lb = \dfrac{P^2}{4}$

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From equation (1), we know that $l^2 + b^2 = 4R^2$:

$4R^2 + 2lb = \dfrac{P^2}{4}$

Since the area $A = lb$, we can write $2lb = 2A$:

$4R^2 + 2A = \dfrac{P^2}{4}$

$2A = \dfrac{P^2}{4} - 4R^2$

$A = \dfrac{P^2}{8} - 2R^2$

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The area of the rectangle is: $A = \dfrac{P^2}{8} - 2R^2$

This formula makes sense because:

If $R$ is very small, the rectangle becomes very thin, so area decreases

If $P$ is large, we have more "material" to work with, so area increases

The formula gives us area in terms of the two given quantities: $P$ and $R$