Question

I need help with this because I don’t get it so if you can help me please let me know

Answer 1

Explanation:

(1)

`\triangle ACD` 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)

`\triangle GIF` is similar to
`\triangle IJH`

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)

`\triangle RPS[\tex] is similar to [tex]\triangle QPT`

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