0Question:
Solution:
OA :
Solution : BC = 4√3
Ptolemy's Theorem :
AB × DC + BC × AD = BD × AC
(4√2 × 4 )+ (4√2 × 4√3) = 8 × AC
AC = (16√2 + 16√6)/8
= 2√2 + 2√6
2√2 ( 1 + √3 )
So 4.
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Solution:
OA :
Solution : BC = 4√3
Ptolemy's Theorem :
AB × DC + BC × AD = BD × AC
(4√2 × 4 )+ (4√2 × 4√3) = 8 × AC
AC = (16√2 + 16√6)/8
= 2√2 + 2√6
2√2 ( 1 + √3 )
So 4.
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