Question

For ΔABC, ∠A = x - 2, ∠B = x + 2, and ∠C = 2x - 20. If ΔABC undergoes a dilation by a scale factor of 2 to create ΔA'B'C' with ∠A' = 98 - x, ∠B' =
x
2
+ 27, and ∠C' = x + 30, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

Answer 1

Answer:

The triangle ABC and A'B'C' are similar to each other by AA criterion.

Step-by-step explanation:

The dilation is the enlargement and compression of the a figure. The angles remains same but the sides changes according to the scale factor.

Therefore, the preimage and image are always similar.

According to the angle sum property the sum of all interiors angles of a triangle is always 180 degree.

$\angle A+\angle B+\angle C=180^{\circ}$

$(x-2)+(x+2)+(2x-20)=180$

$4x-20=180$

$4x=200$

$x=50$

Since the value of x is 50, therefore the measure of angles A,B and C are 48, 52 and 80.

Apply angle sum property is A'B'C'.

$\angle A'+\angle B'+\angle C'=180^{\circ}$

$(98-x)+(\frac{x}{2}+27)+(x+30)=180$

$\frac{x}{2}+155=180$

$\frac{x}{2}=25$

$x=50$

Since the value of x is 50, therefore the measure of angles A',B' and C' are 48, 52 and 80.

$\angel A=\angle A'$

$\angel B=\angle B'$

By AA criterion we can say that ΔABC∼ΔA'B'C.

Answer 2

I think you'll understand when I say that when a triangle gets dilated the angle measures don't change