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The length and width of a rectangle are integer values. [#permalink]
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Updated on: 13 May 2013, 23:50
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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? (A) 12 (B) 24 (C) 36 (D) 48 (E) 60
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Originally posted by emmak on 06 May 2013, 00:12.
Last edited by Bunuel on 13 May 2013, 23:50, edited 1 time in total.
OA added.



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Re: The length and width of a rectangle are integer values. [#permalink]
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06 May 2013, 00:25



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Re: The length and width of a rectangle are integer values. [#permalink]
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06 May 2013, 01:08
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.
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Re: The length and width of a rectangle are integer values. [#permalink]
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06 May 2013, 01:59
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



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Mgmat challenge problem 5513 [#permalink]
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06 May 2013, 06:55
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



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Re: The length and width of a rectangle are integer values. [#permalink]
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06 May 2013, 07:13
emmak wrote: 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? (A) 12 (B) 24 (C) 36 (D) 48 (E) 60 OA will be added on Next monday It is one of those random questions in which no numbers are given. So often, the problem is 'where do we begin?' The only constraint is that length, width and radius must be integral. So how are the dimensions of a rectangle and the radius linked? The diagonal of the rectangle will be the circle's diameter. So the given constraint says that the length, width and the diagonal (i.e. the three sides of a right angled triangle) must be integral. The smallest pythagorean triplet is 3, 4, 5 but if the diameter is 5, radius will be 2.5 (non integer) Next triplet is 5, 12, 13 or 6, 8, 10 (multiply 3, 4, 5 by 2) We see that 6, 8, 10 gives us the answer.
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Re: The length and width of a rectangle are integer values. [#permalink]
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07 May 2013, 04:03
emmak wrote: 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? (A) 12 (B) 24 (C) 36 (D) 48 (E) 60 OA will be added on Next monday Option D for me as well.......
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Re: The length and width of a rectangle are integer values. [#permalink]
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07 May 2013, 04:03
VeritasPrepKarishma wrote: emmak wrote: 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? (A) 12 (B) 24 (C) 36 (D) 48 (E) 60 OA will be added on Next monday It is one of those random questions in which no numbers are given. So often, the problem is 'where do we begin?' The only constraint is that length, width and radius must be integral. So how are the dimensions of a rectangle and the radius linked? The diagonal of the rectangle will be the circle's diameter. So the given constraint says that the length, width and the diagonal (i.e. the three sides of a right angled triangle) must be integral. The smallest pythagorean triplet is 3, 4, 5 but if the diameter is 5, radius will be 2.5 (non integer) Next triplet is 5, 12, 13 or 6, 8, 10 (multiply 3, 4, 5 by 2) We see that 6, 8, 10 gives us the answer. Gud Explanation Karishma .................. !!
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The length and width of a rectangle are integer values. [#permalink]
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22 May 2017, 12:05
VeritasPrepKarishma wrote: emmak wrote: 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? (A) 12 (B) 24 (C) 36 (D) 48 (E) 60 OA will be added on Next monday It is one of those random questions in which no numbers are given. So often, the problem is 'where do we begin?' The only constraint is that length, width and radius must be integral. So how are the dimensions of a rectangle and the radius linked? The diagonal of the rectangle will be the circle's diameter. So the given constraint says that the length, width and the diagonal (i.e. the three sides of a right angled triangle) must be integral. The smallest pythagorean triplet is 3, 4, 5 but if the diameter is 5, radius will be 2.5 (non integer) Next triplet is 5, 12, 13 or 6, 8, 10 (multiply 3, 4, 5 by 2) We see that 6, 8, 10 gives us the answer. Hi VeritasPrepKarishmaReally 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
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Re: The length and width of a rectangle are integer values. [#permalink]
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22 May 2017, 21:22
Sash143 wrote: VeritasPrepKarishma wrote: emmak wrote: 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? (A) 12 (B) 24 (C) 36 (D) 48 (E) 60 OA will be added on Next monday It is one of those random questions in which no numbers are given. So often, the problem is 'where do we begin?' The only constraint is that length, width and radius must be integral. So how are the dimensions of a rectangle and the radius linked? The diagonal of the rectangle will be the circle's diameter. So the given constraint says that the length, width and the diagonal (i.e. the three sides of a right angled triangle) must be integral. The smallest pythagorean triplet is 3, 4, 5 but if the diameter is 5, radius will be 2.5 (non integer) Next triplet is 5, 12, 13 or 6, 8, 10 (multiply 3, 4, 5 by 2) We see that 6, 8, 10 gives us the answer. Hi VeritasPrepKarishmaReally 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.
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Re: The length and width of a rectangle are integer values. [#permalink]
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19 Aug 2017, 11:03
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
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The length and width of a rectangle are integer values. [#permalink]
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23 Sep 2017, 12:48
OFFICIAL SOLUTON FROM MANHATTANIf 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 nextsmallest 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 8by6 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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