Given that the triangle MAH is similar to triangle WCF.
When two triangles are similar, then all three pairs of corresponding sides in the same ratio.
Similarly, [tex] \triangle MAH \sim \triangle WCF [/tex]
Therefore, [tex] \frac{MA}{WC}=\frac{AH}{CF}=\frac{MH}{WF} [/tex]
Now, substituting the measurements of the sides.
[tex] \frac{62}{15.5}=\frac{AH}{CF}=\frac{92}{x} [/tex]
Equating the first and last ratio of the corresponding sides.
[tex] \frac{62}{15.5}=\frac{92}{x} [/tex]
Cross multiplying in the above equation,
[tex] 62 \times x = 92 \times 15.5 [/tex]
[tex] 62 \times x = 1426 [/tex]
[tex] x =\frac{1426}{62} [/tex]
x = 23.
Therefore, the value of x is 23.
