Question

For ΔABC, ∠A = 8x - 10, ∠B = 10x - 40, and ∠C = 3x + 20. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C' with ∠A' = 6x + 10, ∠B' = 70 - x, and ∠C' = 10x 2 , which confirms that ΔABC∼ΔA'B'C by the AA criterion?

Answer 1

Answer: It does not confirm that ΔABC∼ΔA'B'C by the AA criterion as the measure of angle B is negative which is not possible.

Step-by-step explanation:

Since we have given that

In ΔABC,

∠A = 8x - 10, ∠B = 10x - 40, and ∠C = 3x + 20

In ΔA'B'C',

∠A' = 6x + 10, ∠B' = 70 - x, and ∠C' = 10x 2

Since ΔABC gets a dilation by a scale factor of $\dfrac{1}{2}$

So, it becomes,

$\dfrac{\angle A}{\angle A'}=\dfrac{1}{2}\\\\\dfrac{8x-10}{6x+10}=\dfrac{1}{2}\\\\2(8x-10)=6x+10\\\\16x-20=6x+10\\\\16x-6x=10+20\\\\10x=30\\\\x=\dfrac{30}{10}=3$

Now, put the value of

$\angle B=10(3)-40=30-40=-10\\\\and \angle B'=70-30=40$

It does not confirm that ΔABC∼ΔA'B'C by the AA criterion as the measure of angle B is negative which is not possible.