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Sagot :
Answer:
[tex] \sqrt{66} [/tex]
Step-by-step explanation:
By geometric mean property:
[tex]BD = \sqrt{6 \times 5} = \sqrt{30} \\ \\ In \: \triangle ABD: \\ by \: pythagoras \: theorem : \\ \\ AD^2 = AB^2 + BD^2 \\ \\ {x}^{2} = {6}^{2} + {( \sqrt{30}) }^{2} \\ \\ {x}^{2} = 36 + 30 \\ \\ {x}^{2} = 66 \\ \\ x = \sqrt{66} [/tex]
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