In the diagram below, Point A is sqrt(3) units above O. Whenever Tasha moves a point, she translates the point sqrt(2) units to the right. Whenever Richard moves a point, he rotates the point clockwise about point O by 90˚
In one round of transformations, Tasha moves A to A' and then Richard moves A' to A'' In another round of transformations, Richard moves A to A_1 and then Tasha moves A_1 to A_2. Find A''A_2
A
.


 sqrt(3)


.
O
Understanding the Transformations:
Tasha's Transformation: Moves the point sqrt(2) units to the right.
Richard's Transformation: Rotates the point clockwise about point O by 90 degrees.
Round 1:
A' is sqrt(2) units to the right of A (Tasha's translation).
A" is obtained by rotating A' by 90 degrees clockwise around O (Richard's rotation).
Round 2:
A1 is obtained by rotating A by 90 degrees clockwise around O (Richard's rotation).
A2 is sqrt(2) units to the right of A1 (Tasha's translation).
Finding A"A2:
To find the distance between A" and A2, we can consider the following:
A" and A1 are the same distance away from O (since both are obtained by a 90degree rotation from A and they share the same original distance from O).
A'A2 forms a right triangle with A"A1 as the hypotenuse (due to the 90degree rotations).
Pythagorean Theorem:
We can use the Pythagorean theorem to find A"A2:
a^2 + b^2 = c^2
where:
a = A'A2 (we know this is sqrt(2) from Tasha's translation)
b = A"A1 (distance between A" and A1, which is the same as the distance between A and O, which is sqrt(3))
c = A"A2 (what we want to find)
Substituting Values:
(sqrt(2))^2 + (sqrt(3))^2 = (A"A2)^2
2 + 3 = (A"A2)^2
5 = (A"A2)^2
Taking the Square Root (consider both positive and negative):
A"A2 = ±√5
Since distance cannot be negative, we take the positive square root:
A"A2 = √5
Therefore, the distance between A" and A2 is √5 units.