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

Let the re\(2lw = \frac{P^2}{4}-4R^2 lw = \frac{P^2}{8} - 2R^2\)tangle have length 'l' and width 'w'.Since all the vertices of the rectangle lie on a circle of radius 'R',the diagonal of the rectangle is equal to the diameter of the circle, which is 2R.
Using the Pythagorean theorem, we can relate the length, width, and diagonal of the rectangle:
\(l^2 + w^2 = (2R)^2 l^2 + w^2 = 4R^2\)
Now, the perimeter of the rectangle is given by:
P=2l+2w
We can rewrite this in terms of either 'l' or 'w' using the relation above:
P=2l+2wP=2(l+w)l+w=\(\frac{P}{2}\)
Now, we can use the sum of squares identity to factor the left-hand side of the equation:
\(l^2 + w^2 + 2lw = \frac{P^2}{4}\)
Substitute the relation between \(l^2 + w^2 and 4R^2\):
\(4R^2 + 2lw = \frac{P^2}{4}\)
Now, solve for 'lw':
\(2lw = \frac{P^2}{4} - 4R^2 lw = \frac{P^2}{8} - 2R^2\)
The area of the rectangle (A) is given by:
\(A=l\times{wA}=(\frac{P^2}{8}-2R^2)\)
So, the correct answer is option 2:
\(\frac{P^2}{8}-2R^2\)

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): \(\frac {p^2}{8} - 2r^2\)