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Re: A square ABCD and a rectangle PQRS lie in the xy-plane. Coordinates of [#permalink]
Carcass wrote:
OE


For Square ABCD, Coordinates of A and C are given as (3, 5) and (0, 2) respectively.

So, we can find the length of diagonal for square ABCD. AC = √(0 − 3)^2 + (2 − 5)^2 = 3√2
If we have a length of diagonal AC, we can find out the area of the square.

Area of square = 1/2 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑠)

Area of Square ABCD = 1/2 × 3√2 × 3√2 = 9 Square units

For rectangle PQRS, as coordinates of P and R are given as (3, - 4) and (4,6), we can find out the length of diagonal PR. PR = √(4 − 3)^2 + (6 + 4)^2 = √101

Area of the rectangle is maximum, when it is a square.

So maximum area of Rectangle PQRS = 1/2× √101 × √101 = 101/2 square units.

Minimum area of Rectangle PQRS could go close to Zero as well.
Area of rectangle PQRS varies from a ‘value close to zero’ to 101/2 square units.

As it varies, we cannot compare the area of Rectangle PQRS to the area of Square
ABCD.
Ans. (D)





Wait, since we have the opposite vertices, we can calculate the individual sides for both the square and the rectangle. There are no other possibilities. What you have described is elegant, but this would be true if we only had the lengths of the diagnols. Answer should be B.
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Re: A square ABCD and a rectangle PQRS lie in the xy-plane. Coordinates of [#permalink]
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