Answer:
Step-by-step explanation:
(1)
[tex]\triangle ACD[/tex] has the constant ratio of the sides. AB/AC = AE/ED
So solving for x is simple:
AB/AC = 6/2 = 3 = AE/x = 9/x
9/x = 3
9 = 3x
x = 3
(2)
[tex]\triangle GIF[/tex] is similar to [tex]\triangle IJH[/tex]
To be similar, the ratio between the respective sides must be equal =>
IJ/IF = IH/IG = HJ/GF.
So we plug in the value of the sides.
10/5 = x/12
x/12 = 10/5
x/12 = 2
x = 24
(3)
I don't understand the picture (Sorry)
(4)
[tex]\triangle RPS[\tex] is similar to [tex]\triangle QPT[/tex]
Which implies:
PQ/QR = QT/RS
5/7 = 4/x
[4/x = 5/7] * x
4 = (5/7)x
x = 4 * 7/5 = 28/5 = 5 3/5
