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Square PQRS is tilted 90 degrees anticlockwise direction around the point P, so that points Q, R, S reach the points Q', R', S' respectively. What is the distance covered by the point R if the length of PQ is 2.

Since point P is the hinge point we can take point Q either on left side or on right side of P. As square PQRS is rotated anticlockwise with center at point P, radius would be equal to diagonal PR where PR s equal to √2 * side of square.

PR = 2√2

Here R would cover a distance along the length of perimeter of circle with radius 2√2.
Since only 90 deg is rotated, point R would cover only one - fourth the length of circle's perimeter.

Perimeter of circle with radius PR = 2 * π * PR
= 2π2√2
= 4π√2

Distance covered by point R = \(\frac{1}{4} * 4π√2\)
= π√2

Answer C.
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Please, Find The Attached

90º Rotation anticlockwise direction around the point P is inscribed in a circle.
Attachments

Problem_solving.jpg
Problem_solving.jpg [ 146.71 KiB | Viewed 6668 times ]

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Quote:
Square PQRS is tilted 90 degrees anti-clockwise direction around the point P, so that points Q, R, S reach the points Q', R', S' respectively. What is the distance covered by the point R if the length of PQ is 2.

A. (π√2)/3
B. (π√2)/2
C. π√2
D. π√3
E. 2π√3

"AROUND POINT P" means that P is the origin;
therefore, the diagonal of the square PQRS is the radius, not the diameter!

square PQRS side = 2
square PQRS diagonal = circle's radius r = side√2 = (2)√2
circle's circumference = 2πr = 2π(2)√2 = 4π√2
distance covered by R to R' (90 deg…360/90…1/4) = (1/4)•4π√2 = π√2

Answer (C)
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If we treat it as though it is an a square inscribed within a circle, then we know that the radius of the circle within which a square is inscribed is half the diagonal of the square. So we already know that PQ=2, and the diagonal of the square = length of side x √2
hence diagonal = 2*√2
Radius of the circle = half of the diagonal = √2

Now the distance covered by the square when it is rotated 90degrees = circumference of the circle * 90/360 = 2*π*√2 *90/36
= 2*π*√2 * 1/4 = π*√2 / 2.

The answer is therefore B.
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We can calculate the r =2sqrt2 and then we can calculate the area of the arc in the sector

The answer should be C.
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Dillesh4096
—> RR’ = θ/360*2*π*r = 90/360*2*π*\(\sqrt{2}\) = π*\(\sqrt{2}\)

your calculation is wrong:
90/360*2*π*\(\sqrt{2}\)…
1/4*2*π*\(\sqrt{2}\)…
1/2*π*\(\sqrt{2}\)…
π*\(\sqrt{2}\)/2…
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key to this question in my case was to correctly draw the rotation.
Rotation is about Center point P - I initially didn’t read the question correctly and rotated the whole square rather than just Q, R,S
Which resulted in diagonal of the square = radius of the circle rather than my initial view of diagonal being the diameter which was screwing up the equation.



Thanks,
Lasya
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