Respuesta :
Answer:
Hello,
a)18
Step-by-step explanation:
[tex]p=27, int[p/2]=13\\\\\begin{array}{c|c|c|c|c}a&p-a&b&c\\---&---&---&---\\1&26&13&13\\2&25&12&13\\3&24&12&12\\4&23&10&13\\&&11&12\\5&22&9&13\\&&10&12\\&&11&11\\6&21&8&13\\&&9&12\\&&10&11\\7&20&7&13\\&&8&12\\&&9&11\\&&10&10\\8&19&8&11\\&&9&10\\9&18&9&9\\---&---&---&---\\\end{array}[/tex]
number of triangle=18
b)
Side of the equilateral triangle =2
Height=√3 (using Pythagorean's theorem) (tr ACP)
Triangle ABJ has the same area as the triangle ABC
Uisng Thalès's theorem,
[tex]\dfrac{2-v}{2}=\dfrac{v}{\sqrt{3} } \\\\\\v=\dfrac{2\sqrt{3} }{2+\sqrt{3} } \\\\\\v=\dfrac{2\sqrt{3}*(2-\sqrt{3}) }{1} } \\\\v=4\sqrt{3}-6\\\\v^2=84-48\sqrt{3} \\Area =a\sqrt{b}-c= \sqrt{3} -(84-48\sqrt{3} )=49\sqrt{3} -84\approx{0,87049...}\\\\So:\\b=3\\a=49\\c=84\\\\a+b+c=3+49+84=136\\[/tex]
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