Question

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

• A. \(\frac{P^2}{2}−2PR\)
• B. \(\frac{P^2}{8}−2R^2\)
• C. \(\frac{P^2}{16}−R^2\)
• D. \(\frac{P^2}{8}−\frac{R^2}{2}\)

Answer 1

The Correct Option is B

To find the area of a rectangle inscribed in a circle (i.e., with all vertices on the circle), we need to consider the properties of such a rectangle. If we denote the rectangle's side lengths as \(a\) and \(b\), the diagonals are equal to the circle's diameter, which is \(2R\). Thus, using the Pythagorean theorem: \(a^2 + b^2 = (2R)^2 = 4R^2\). Additionally, the perimeter \(P\) of the rectangle is given by: \(P = 2(a + b)\).

Using the perimeter expression, we have:

\[a + b = \frac{P}{2}\]

We are asked to find the area \(A\) of the rectangle, which can be expressed as:

\[A = a \times b\]

We can square the sum expression \(a + b\) and use the identity:

\[(a + b)^2 = a^2 + b^2 + 2ab\]

Substituting the known values:

\[\left(\frac{P}{2}\right)^2 = 4R^2 + 2ab\]

Solve for \(ab\):

\[2ab = \frac{P^2}{4} - 4R^2\]

\[ab = \frac{P^2}{8} - 2R^2\]

Thus, the area of the rectangle is:

\[A = \frac{P^2}{8} - 2R^2\]

This confirms the correct answer is: \(\frac{P^2}{8} - 2R^2\)

Answer 2

Apply Pythagoras Theorem,
\(4r^2 = l^2+b^2\)
Area \(= lb\)
Perimeter, \(P = 2(l+b)\)
\(\frac p2 = l+b\)
Squaring both side,
\(\frac {p^2}{4} = l^2+b^2+2lb\)

\(\frac {p^2}{4} = 4r^2+2lb\)

\(\frac {p^2}{8} - 2r^2 = lb\)
Then, area of the rectangle = \(\frac {p^2}{8} - 2r^2\)

So, the correct option is (B): \(\frac {p^2}{8} - 2r^2\)