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Sagot :
[see the figure attached]
Let's take a look at a vertical section of the cone.
By Thales' theorem, the height is [tex]\dfrac{BD}{CE}=\dfrac{AB}{AC}[/tex] hence [tex]h=AC=\dfrac{AB\cdot CE}{BD}=\dfrac{(h-8)\cdot6}{4}[/tex] thus [tex]\boxed{h=24}[/tex]
Answer E
Let's take a look at a vertical section of the cone.
By Thales' theorem, the height is [tex]\dfrac{BD}{CE}=\dfrac{AB}{AC}[/tex] hence [tex]h=AC=\dfrac{AB\cdot CE}{BD}=\dfrac{(h-8)\cdot6}{4}[/tex] thus [tex]\boxed{h=24}[/tex]
Answer E

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