For δabc, ∠a = 4x - 4, ∠b = 6x - 1, and ∠c = 8x - 13. if δabc undergoes a dilation by a scale factor of 2 to create δa'b'c' with ∠a' = 51 - x, ∠b' = 4x + 21, and ∠c' = 6x + 9, which confirms that δabc∼δa'b'c by the aa criterion?

Solution:

As, Triangle ABC and Triangle A'B'C' will be similar to each other.So,their corresponding interior angles will be equal.

So, ∠A=∠A'

∠B=∠B'

∠C=∠C'

1. 4 x -4=51-x

4x +x=51+4

5 x=55

Dividing both sides by 11

x=11

2. 6 x-1= 4 x +21

6x-4x=21+1

2x=22

Dividing both sides by 2, gives

x=11

So, when x=11,the two triangles ΔABC∼ΔA'B'C by the AA criterion.

X = 60° (corresponding angles are equal )