First
[tex]\\ \sf\longmapsto BD^2=AD\times DC[/tex]
[tex]\\ \sf\longmapsto BD^2=18^2+7^2[/tex]
[tex]\\ \sf\longmapsto BD^2=324+49[/tex]
[tex]\\ \sf\longmapsto BD^2=363[/tex]
[tex]\\ \sf\longmapsto BD=\sqrt{363}[/tex]
[tex]\\ \sf\longmapsto BD=19.2[/tex]
Now
Using Pythagorean theorem
[tex]\\ \sf\longmapsto BD^2+CD^2=m^2[/tex]
[tex]\\ \sf\longmapsto m^2=7^2+19.2^2[/tex]
[tex]\\ \sf\longmapsto m^2=49+363[/tex]
[tex]\\ \sf\longmapsto m^2=412[/tex]
[tex]\\ \sf\longmapsto m=\sqrt{412}[/tex]
[tex]\\ \sf\longmapsto m=20.3[/tex]
Nearest value in options is 18
Hence option a is correct
