Problem 1
The triangles share an overlapping angle at the very top. This is one pair of congruent angles. Another pair could be any of the corresponding angles formed by the parallel lines, and the transversals. We only need 2 pairs of congruent angles to use the AA (angle angle) similarity theorem.
Once we know the triangles are similar, we set up the proportion below and solve for x.
6/(x-1) = (6+x)/(2x+4)
6(2x+4) = (x-1)(6+x)
12x+24 = 6x+x^2-6-x
12x+24 = x^2+5x-6
0 = x^2+5x-6-12x-24
x^2-7x-30 = 0
(x-10)(x+3) = 0
x-10 = 0 or x+3 = 0
x = 10 or x = -3
Ignore negative x values because we cannot have negative lengths.
Therefore, x = 10 is the only plausible solution.
Answer: x = 10
=================================================
Problem 2
Notice how 12/9 = (4*3)/(3*3) = 4/3
While 28/21 = (4*7)/(3*7) = 4/3
We end up with 4/3 for both which allows us to say 12/9 = 28/21 is a true statement. From this, we know the triangles are similar due to the SAS similarity theorem. The last bit of info needed really is the set of congruent vertical angles in between the mentioned sides.
Let's set up a proportion to solve for x.
(4x-4)/18 = 12/9
9(4x-4) = 18*12
36x-36 = 216
36x = 216+36
36x = 252
x = 252/36
x = 7
Answer: x = 7
=================================================
Problem 3
Both triangles are equilateral. This is due to each triangle having the same tickmarks for all three sides. Every equilateral triangle has all three angles of 60 degrees. We then use the AA theorem to prove any two equilateral triangles are similar.
Let's solve for x.
(10x-10)/(2x) = 10/4
4(10x-10) = 2x*10
40x-40 = 20x
40x-20x = 40
20x = 40
x = 40/20
x = 2
Answer: x = 2
