Problem Solving

Not sure about the source of the question. Can someone help with this question?

When a square with a center c is rotated through an angle 45 degrees, it has a new centre c'. If
the length of the square side is 5. Find the distance between c and c'?

ramyaravindran wrote:Not sure about the source of the question. Can someone help with this question?

When a square with a center c is rotated through an angle 45 degrees, it has a new centre c'. If
the length of the square side is 5. Find the distance between c and c'?

I think c and c' will be same.The distance between them will be 0.
Just visualize the question thinking the rotation is taking through " c ",taking it as an axis.

was there a picture illustrating this? i'm having a hard time visualizing the set up... you don't have to post it, but I'm just curious.

money9111 wrote:was there a picture illustrating this? i'm having a hard time visualizing the set up... you don't have to post it, but I'm just curious.

i think the answer is 5/sqrt(2). to visualize take a square page and rotate it from any corner to 45 degrees, since it is given that new centre is c'. so diagonal will be 5root2. and the new centre will be halfway on diagonal. so distancy will be half the diagonal which is 5/sqrt(2).

what are the options in this question? plz post them. so that more details can be obtained.

No options were given. I only got the question.

ramyaravindran wrote:Not sure about the source of the question. Can someone help with this question?

When a square with a center c is rotated through an angle 45 degrees, it has a new centre c'. If
the length of the square side is 5. Find the distance between c and c'?

If a square is rotated about its centre, then its centre will be same.
The centre willl change if it is rotated about some other point. Well if u can check ur question source again and find put the point about which this square is rotated. I hope it must be getting rotated abut its corner.

so please check it.

Can this be solved using arcs?

Centre C moves along a circle of radius 2.5*root2, through angle pi/4, to become C'

Therefore our answer is the length of the arc described = [spoiler](pi*2.5*root2)/4[/spoiler]

(assuming we a rotating the square about the corner, which seems the only sensible question to ask)

There is an error in the construction of the question, if the rotation is made at the center of the square c = c' and the difference is zero. If the rotation is made at any vertex then it makes some change. This is not a question to work through conventional methods it chould be done by approximation.

In general, since each side of a square is of length 5 units, the diagonal is of length 5√2.

The point C is center on the diagonal, when the square is rotated 450 the point C is moved to the point C' which is obviously on the side of the square as shown in the figure .

So CB = 5√2/2 and C'B = 5√2/2 , CA = 5/2 and AC' = (5√2/2) - 5/2.

By Pythagoras theorem CC' = (5/2)( √(4-2√2)).

But by plugging method we can understand that the value of CC' lies between 5/2 and 5/√2.

I would argue that the arc method I proposed lets you calculate the distance moved by C usuing conventional methods?

florencejennifer wrote:There is an error in the construction of the question, if the rotation is made at the center of the square c = c' and the difference is zero. If the rotation is made at any vertex then it makes some change. This is not a question to work through conventional methods it chould be done by approximation.

In general, since each side of a square is of length 5 units, the diagonal is of length 5√2.

The point C is center on the diagonal, when the square is rotated 450 the point C is moved to the point C' which is obviously on the side of the square as shown in the figure .

So CB = 5√2/2 and C'B = 5√2/2 , CA = 5/2 and AC' = (5√2/2) - 5/2.

By Pythagoras theorem CC' = (5/2)( √(4-2√2)).

But by plugging method we can understand that the value of CC' lies between 5/2 and 5/√2.

I think that question said: If the square of side 5 units is rotated by 45 degrees about one of its corners, how far would be its new center from the old one.

because, as jennifer pointed out, rotation about the center would hardly make any difference.

Approach:

we know that the diagonal of a square of 'a' units is a√2 and the center is at a distance of a/√2 from the square's corner.

So, if we rotate the square about one of its corners, the new center would still be at a distance of a/√2 units from the corner about which it is rotated.

(imagine making an arc of radius a/√2 units and of 45 degrees measure)

now in that arc, the chord making 45 degrees will be the distance between the two centers.

Drop a perpendicular from the rotated corner on the chord and you get

1/2 (chord distance) = Sin(45/2) * a/√2

=> chord distance/ distance between the centers = Sin 22.5 * a/√2

The original centre was (5/2, 5/2) that is the intersection of the diagonal of the square. The slope is 1 or 45°.
The centre moves through 45° so the new centre will be on y axis coordiantes are (0, 5√2)
distance between c and c1 = √(5/2-5/√2)²+(5/√2)²

I agree on Jeniffer's opinion.
See the above picture that helps you understand.
I think this question wants the distance between c and c' if the rotation is made at any vertex.

Distance between c and c' = √((5/2-0)^2 + (5/2-(5/2)√2)^2)

So, the answer is (5/2)( √(4-2√2)).