For ΔABC, ∠A = 3x, ∠B = 2x - 3, and ∠C = x + 3. If ΔABC undergoes a dilation by a scale factor of 2 to create ΔA'B'C' with ∠A' = 2x + 30, ∠B' = x + 27, and ∠C' = 1 2 x + 18, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

Question

Mathematics

Karissa Dougherty

27 February, 19:14

For ΔABC, ∠A = 3x, ∠B = 2x - 3, and ∠C = x + 3. If ΔABC undergoes a dilation by a scale factor of 2 to create ΔA'B'C' with ∠A' = 2x + 30, ∠B' = x + 27, and ∠C' = 1 2 x + 18, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

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Answers (1)

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Jamir Reyes · Answer 1

To confirm that triangle ABC and triangle A'B'C' are congruent by the AA criterion, we can equate the original angle to the angle after that dilation and solve for x. Then do the same for a different angle and see if the value of x remains the same.

A = A'

3x = 2x + 30

x = 30

B = B'

2x - 3 = x + 27

x = 30

Therefore, ΔABC∼ΔA'B'C by the AA criterion since A = A' and B = B'