The length and width of a rectangle are integer values.

If we consider all pythagorean triplets with integral values, we have the following results:-
(3,4,5),(6,8,10)........
The diameter of the circle is 5 in the 1st case and 10 in the 2nd.
But the question says that radius of the circle is also an integer hence the 1st triplet will be ruled out.
We are left with the 2nd triplet where radius of the circle=5,
Hence the sides of the rectangle are 6 and 8 making the area =48.

Option D =48.

l^2+b^2 = 4r^2.

since all 3 l,b and r are integers, l and b cannot be 3,4 as 2r will then be 5 and r will not be an integer.
next l and b can be 6,8 and r will then be 5. hence area of the rectangle is 6*8 = 48

The length and width of a rectangle are integer values. What is the area of the smallest such rectangle that can be inscribed in a circle whose radius is also an integer?






see
L = int
B = int
r= int

let the rectangle in circle be ABCD

the diameter of circle will be diagonal of rectangle ABCD

B:L:2r(d) = 1:root3 :2

but they should be integers - smallest possible ints

so according to pytho - therom they will be 3:4:5

well and good - area will be 3 * 4 = 12

but chit is 5 i.e diameter and diagonal

if d= 5 then r=2.5 not an int

so 3:4:5 - multiply by 2

6:8:10

then area is 6 * 8 = 48

d= 10 then r = 5 an integer

so answer is 48

Option D for me as well.......

Gud Explanation Karishma .................. !!

Hi VeritasPrepKarishma

Really good explanation!
I have a doubt though: Can square be considered a rectangle here? If yes then can the answer be 36? If no why cant a square be a rectangle in this case?

Thanks in advance

A square is always a rectangle too. A square is a special rectangle. Here, a square will not satisfy our constraints. We need the radius of the circle to be an integer too. The diagonal of the rectangle/square will be the diameter of the circle (a right angle is inscribed in a semi circle). If the side of the square is an integer s, its diagonal will be \(\sqrt{2}s\) - not an integer. So the diameter (and hence the radius) of the circle will not be an integer. That is why we can't have a square here.

Clearly the Ans is D

In simple words, LxB is area and L^2+B^2= (2r)^2
IF LxB is 48
then L=8 and B=6
and sum of squares is 100 whose root is 10 and whose half is r which will be 5 , hence all integers. Ans is D

OFFICIAL SOLUTON FROM MANHATTAN

If a rectangle is inscribed in a circle, the rectangle’s diagonals are also diameters of the circle:

Call the length and the width of the rectangle L and W and the circle’s radius r. A right triangle can be constructed, with legs L and W and with hypotenuse 2r.
 
Use the Pythagorean theorem: a^2 + b^2 = c^2
L^2 + W^2 = (2r)^2, where L, W, and r are all integers.
 
The smallest three integers that satisfy the Pythagorean theorem are 3, 4, and 5, but those numbers don’t work here, because if 2r = 5, then r is not an integer. (Remember that the sides of the right triangle here are L, W, and 2r, not just L, W, and r.)
 
The next-smallest Pythagorean triplet is 6, 8, and 10. The desired situation can thus be constructed if L = 8, W = 6, and 2r = 10; in other words, an 8-by-6 rectangle can be inscribed in a circle of radius 5.  The area of such a rectangle is 8 x 6 = 48 square units.
 
The correct answer is D.

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