amanvermagmat
Rectangle R has an area A square units. Is the rectangle R a square?
(1) A = 36 square units, and sides of rectangle R are in integer units.
(2) For all the given rectangles with area A, R has minimum perimeter.
OA: B
Rectangle R has an area A square units. Is the rectangle R a square?
Statement 1: A = 36 square units, and sides of rectangle R are in integer units.
Possible combinations of dimensions of side are (1,36);(2,18);(3,12);(4,9);(6,6)
Only one combination results into square.
So statement 1 is insufficient for answering :Is the rectangle R a square?
Statement 2 :For all the given rectangles with area A, R has minimum perimeter.
Let the length and width of rectangle be x and y.
Perimeter(P)=\(2*(x+y)\) ; Area(A)= \(x*y\)
taking arithmetic mean and geometric mean of both sides x and y , we get
A.M = \(\frac{x+y}{2}\) ; G.M = \(\sqrt{x*y}\)\(\)
as we know \(arithmetic mean\geq{geometric mean}\)
\(\frac{x+y}{2}\geq{\sqrt{x*y}}\)
Multiplying both side by 4 , we get
\(2*(x+y)\geq{4*\sqrt{x*y}}\)
\(P\geq{4*\sqrt{A}}\)
So Lowest value of Perimter of rectangle is \(4*\sqrt{A}\)
It occurs when Arithmetic mean and Geometric mean of the sides of rectangle are equal.
A.M= G.M occurs only when both the term are equal. In this case, it will occurs when x=y.
As it is given in question stem that R is rectangle, and statement 2 is implying that length = width for rectangle R.
So Rectangle R is a square.
Statement 2 is alone sufficient