We can see the figure of triangles in attachment as shown
In triangles UVW and VZW
∠UVW = ∠VZW (both are equal to 90°)
∠VWU = ∠VWZ ( common angle W in both triangles)
So by AA property Δ UVW ~ ΔVZW
So its coressponding sides will have equal ratio
So [tex] \frac{UW}{VW} = \frac{VW}{ZW} [/tex]
Now plugging 3 in VW place we get
[tex] \frac{UV}{3} = \frac{3}{ZW} [/tex]
Now on cross multiplying we get
UV × ZW = 3 × 3
UV × ZW = 9
so clearly all choices (1),(2) and (4) are incorrect as they state
UV × ZW = 4
As Δ UVW ~ ΔVZW and so corresponding sides will have equal ratios so we can also have
[tex] \frac{UW}{VW} = \frac{VW}{ZW} [/tex]
Now if you plug 3 in VW place here we get
[tex] \frac{UW}{3} = \frac{3}{ZW} [/tex]
Now cross multiply
UW × ZW = 3 × 3
UW × ZW = 9 ----------------------(1)
Please recheck choice (3) second statement must have been
Segment UW • segment ZW = 9, instead of VW its must be ZW
So now lets check choice (3) first statement
For that we can use triangles UVW and UZV
Here ∠UVW = ∠UZV (both are equal to 90°)
∠VUW = ∠VUZ ( common angle U in both triangles)
So by AA property Δ UVW ~ ΔUZV
So its coressponding sides will have equal ratio
So [tex] \frac{UW}{UV} = \frac{UV}{UZ} [/tex]
now plug 2 in UV place here
[tex] \frac{UW}{2} = \frac{2}{UZ} [/tex]
now cross multiply
UW × UZ = 2 × 2
UW × UZ = 4 -------------------(2)
which is there in choice (3) first statement
Segment UW • segment UZ = 4
From equation (1)
UW × ZW = 9
plug x in UW place and solve for ZW
x × ZW = 9-----------------------------------(3)
similarly plug x in UW place in equation (2) also so we get
x × UZ = 4 -----------------------------(4)
Now add equations (3) and (4)
x × ZW + x × UZ = 9 + 4
x × ZW + x × UZ = 13
x is common in both terms so can take x common out
x (ZW +UZ) = 13
now ZW+UZ = UW, can see in figure in attachment
so x(UW) = 13
again plug x in UW place
so x(x) =13
hence [tex] x^{2} =13 [/tex] is proved
so choice (3) is the right answer
Segment UW • segment UZ = 4
Segment UW • segment ZW = 9
