Question

For ΔABC, ∠A = 4x + 7, ∠B = 2x + 3, and ∠C = 6x - 10. If ΔABC undergoes a dilation by a scale factor of 2 to create ΔA'B'C' with ∠A' = 5x - 8, ∠B' = 3x - 12, and ∠C' = 7x - 25, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

Answer 1

Bizzare problem. What are the choices?

The triangle angles add to 180.

4x + 7 + 2x +3 + 6x - 10 = 180

12x = 180

x = 15

Dilation doesn't change the angles so we must have

4x + 7 = 5x - 8 and 2x + 3 = 3x - 12 and 6x - 10 = 7x - 25

Fortunately x=15 is the solution to all of these so it's consistent.

Answer 2

The value of x is 15.

Explanation:

It is given that in ΔABC, ∠A = 4x + 7, ∠B = 2x + 3, and ∠C = 6x - 10.

According to angle sum property of triangle, the sum of all interior angles of a triangle is 180°.

In ΔABC,

`\angle A+\angle B+\angle C=180`

`(4x+7)+(2x+3)+(6x-10)=180`

Combine like terms.

`(4x+2x+6x)+(7+3-10)=180`

`12x=180`

Divide both sides by 12.

`x=15`

The value of x is 15.

`\angle A=4x+7\Rightarrow 4(15)+7=67^(\circ)`

`\angle B=2x+3\Rightarrow 2(15)+3=33^(\circ)`

`\angle C=6x-10\Rightarrow 6(15)-10=80^(\circ)`

`\angle A'=5x-8\Rightarrow 5(15)-8=67^(\circ)`

`\angle B'=3x-12\Rightarrow 3(15)-12=33^(\circ)`

`\angle C'=7x-25\Rightarrow 7(15)-25=80^(\circ)`

We get

`\angle A\cong \angle A'`

`\angle B\cong \angle B'`

Two corresponding angles of triangles are congruent. By AA criterion

`\triangle ABC\cong \triangle A'B'C'`

Hence proved.