Solved Examples 2 and 3 on page `1//50` give us the following data :
Distance of moon from earht, `ME = 3.84xx10^8m`
Distance fo sun from earht, `ME = 1.496xx10^(11)m`.
Diameter of sun `AB = 1.39xx10^9m`
The situation during total solar eclipes is shwon in fig.
As `Delta^s` ABE of CDE are similar, therefore,
`CD = ABxx(ME)/(SE) = (1.39xx10^9xx3.84xx10^8)/(1.496xx10^(11)) = 3.5679xx10^6m`
` = 3567.9km`
This is the diameter of the moon.