Given
AB = 18 , CD = 10
diameter = 30
Find
How much closer is AB than CD to point O.
Explanation
Radius = 15
AO = 15 , CO = 15
In right angle triangle , OFB ,
[tex]\begin{gathered} AO^2=OP^2+FB^2 \\ 15^2=OP^2+9^2 \\ 225-81=OF^2 \\ 144=OF^2 \\ OF=12 \end{gathered}[/tex]
and in right triangle OED
[tex]\begin{gathered} OD^2=OE^2+ED^2 \\ 15^2=5^2+OE^2 \\ 225-25=OE^2 \\ 200=OE^2 \\ OE=14.1421356237 \\ \end{gathered}[/tex]
subtract the length of OF - OE = 14.1421356237 - 12 = 2.14213562373
Final Answer
Hence , the length AB is 2.14 cm closer than CD to point O
