Question:medium

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

Updated On: Jan 15, 2026
  • \(\frac{P^2}{2}−2PR\)
  • \(\frac{P^2}{8}−2R^2\)
  • \(\frac{P^2}{16}−R^2\)
  • \(\frac{P^2}{8}−\frac{R^2}{2}\)
Show Solution

The Correct Option is B

Solution and Explanation

To determine the area of a rectangle inscribed within a circle, where all vertices lie on the circle, we must consider specific geometric properties. Let the rectangle's side lengths be \(a\) and \(b\). The diagonals of such a rectangle are equivalent to the circle's diameter, which is \(2R\). By applying the Pythagorean theorem, we establish the relationship: \(a^2 + b^2 = (2R)^2 = 4R^2\). The perimeter \(P\) of the rectangle is defined as \(P = 2(a + b)\).
From the perimeter equation, we can derive:
\[a + b = \frac{P}{2}\]
The objective is to find the area \(A\) of the rectangle, represented by the formula:
\[A = a \times b\]
Squaring the sum of the sides expression, \(a + b\), and utilizing the algebraic identity \((a + b)^2 = a^2 + b^2 + 2ab\), we proceed:
\[\left(\frac{P}{2}\right)^2 = 4R^2 + 2ab\]
Rearranging to solve for \(ab\):
\[2ab = \frac{P^2}{4} - 4R^2\]
\[ab = \frac{P^2}{8} - 2R^2\]
Therefore, the area of the rectangle is given by:
\[A = \frac{P^2}{8} - 2R^2\]
This calculation confirms that the correct area is \(\frac{P^2}{8} - 2R^2\).
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