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# Question

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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

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|  sqrt(3)

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O

Apr 13, 2024

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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 90-degree 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 90-degree 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.

Apr 14, 2024