Let the re
2lw=4P2−4R2lw=8P2−2R2tangle 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:
l2+w2=(2R)2l2+w2=4R2Now, 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=
2PNow, we can use the sum of squares identity to factor the left-hand side of the equation:
l2+w2+2lw=4P2Substitute the relation between
l2+w2and4R2:
4R2+2lw=4P2Now, solve for 'lw':
2lw=4P2−4R2lw=8P2−2R2The area of the rectangle (A) is given by:
A=l×wA=(8P2−2R2)So, the correct answer is option 2:
8P2−2R2