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15 Jun 2022, 07:00
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In the figure above, which shows rectangle ABCD tangent to a circle at each corner, AB is 2 units in length. Is rectangle ABCD a square?
(1) The length of minor arc AB is exactly half the length of arc ABC.
(2) The length of minor arc AD is \(\frac{π\sqrt{2}}{2}\)
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16 Jun 2022, 07:07
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(1)
As the length of minor arc AB is exactly half the length of arc ABC, that means that AB=BC. As AB and BC are not opposite sides, ABCD can only be a square.
SUFFICIENT
(2)
Let the centre of the circle = o
If ABCD is a square, then the radius of the circle must be \(\sqrt{2}\). If we create a triangle with sides oA, oB and AB, where AB is the hypotenuse and as oA and oB are both radii they will be equal to one another, we create a special 45-45-90 triangle with sides \(x, x\) & \(x\sqrt{2}\). The only way that the hypotenuse is \(2\) is if the other sides are each \(\sqrt{2}\)
If ABCD is a square then arcs AD=DC=CB=BA. If we multiply the length of the arc by 4 (which will give us the possible circumference) and solve for r, using the formula for the circumference of a circle, and r = \(\sqrt{2}\) then we know without a doubt that ABCD is a square.
Circumference: \(\frac{π\sqrt{2}}{2}*4\)
\(2π\sqrt{2}\)
\(2πr = 2π\sqrt{2}\)
\(r = \sqrt{2}\)
As r = \(\sqrt{2}\), we know ABCD is a square.
SUFFICIENT
Answer D
As the length of minor arc AB is exactly half the length of arc ABC, that means that AB=BC. As AB and BC are not opposite sides, ABCD can only be a square.
SUFFICIENT
(2)
Let the centre of the circle = o
If ABCD is a square, then the radius of the circle must be \(\sqrt{2}\). If we create a triangle with sides oA, oB and AB, where AB is the hypotenuse and as oA and oB are both radii they will be equal to one another, we create a special 45-45-90 triangle with sides \(x, x\) & \(x\sqrt{2}\). The only way that the hypotenuse is \(2\) is if the other sides are each \(\sqrt{2}\)
If ABCD is a square then arcs AD=DC=CB=BA. If we multiply the length of the arc by 4 (which will give us the possible circumference) and solve for r, using the formula for the circumference of a circle, and r = \(\sqrt{2}\) then we know without a doubt that ABCD is a square.
Circumference: \(\frac{π\sqrt{2}}{2}*4\)
\(2π\sqrt{2}\)
\(2πr = 2π\sqrt{2}\)
\(r = \sqrt{2}\)
As r = \(\sqrt{2}\), we know ABCD is a square.
SUFFICIENT
Answer D
