Resposta:
Sabemos que [tex]AB = 8\,\,m[/tex] e [tex]BC = 10\,\,m.[/tex] Como o [tex]\triangle ABC[/tex] é retângulo em [tex]\^A[/tex], calculemos o comprimento AC por Pitágoras:
[tex]BC^2 = AB^2 + AC^2\\\\\Longleftrightarrow 10^2 = 8^2 + AC^2\\\\\Longleftrightarrow AC^2 = 36\\\\\Longleftrightarrow AC = 6\,\,m.[/tex]
Calculemos a área total do [tex]\triangle ABC[/tex], através da Fórmula de Heron.
Calculemos inicialmente seu semiperímetro:
[tex]p = \dfrac{a+b+c}{2} = \dfrac{10 + 6 + 8}{2} = 12\,\,m[/tex]
[tex]A_{\triangle ABC} = \sqrt{p \cdot (p-a) \cdot (p-b) \cdot (p-c)}\\\\\Longleftrightarrow A_{\triangle ABC} = \sqrt{12 \cdot (12 - 10) \cdot (12 - 6) \cdot (12 - 8)}\\\\\Longleftrightarrow A_{\triangle ABC} = \sqrt{12 \cdot 2 \cdot 6 \cdot 4}\\\\\Longleftrightarrow A_{\triangle ABC} = \sqrt{576}\\\\\Longleftrightarrow A_{\triangle ABC} = 24\,\,m^2[/tex]
Calculemos agora a área [tex]A_I[/tex].
Inicialmente, calculemos a tangente do ângulo [tex]\^B:[/tex]
[tex]tg\, \^B = \dfrac{AC}{AB} = \dfrac{6}{8} = \dfrac{3}{4}.[/tex]
Calculemos o comprimento PM, que é a altura do [tex]\triangle BPM[/tex]:
[tex]PM = tg\, \^B \cdot BM\\\\\Longleftrightarrow PM = \dfrac{3}{4} \cdot 5\\\\\Longleftrightarrow PM = \dfrac{15}{4}\,\,m[/tex]
[tex]A_I = \dfrac{1}{2} \cdot BM \cdot PM\\\\\Longleftrightarrow A_I = \dfrac{1}{2} \cdot 5 \cdot \dfrac{15}{4}\\\\\Longleftrightarrow A_I = \dfrac{75}{8}\,\,m^2[/tex]
Assim:
[tex]A_{II} = A_{\triangle ABC} - A_I\\\\\Longleftrightarrow A_{II} = 24 - \dfrac{75}{8}\\\\\Longleftrightarrow A_{II} = \dfrac{17}{8}\,\,m^2[/tex]
Determinemos a razão entre as áreas [tex]A_I[/tex] e [tex]A_{II}[/tex]:
[tex]\dfrac{A_I}{A_{II}} = \dfrac{75 / 8}{117 / 8} =\dfrac{75}{117} = \boxed{\dfrac{25}{39}}[/tex]
Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\sf (\overline{\rm BC})^2 = (\overline{\rm AB})^2 + (\overline{\rm AC})^2[/tex]
[tex]\sf (10)^2 = (8)^2 + (\overline{\rm AC})^2[/tex]
[tex]\sf (\overline{\rm AC})^2 = 100 - 64[/tex]
[tex]\sf (\overline{\rm AC})^2 = 36[/tex]
[tex]\sf \overline{\rm AC} = 6\:m[/tex]
[tex]\boxed{\sf \overline{\rm PC} = \overline{\rm PB}}[/tex]
[tex]\sf (\overline{\rm PC})^2 = (\overline{\rm PA})^2 + (\overline{\rm AC})^2[/tex]
[tex]\sf (8 - \overline{\rm PA})^2 = (\overline{\rm PA})^2 + (6)^2[/tex]
[tex]\sf 64 - 16\overline{\rm PA} + (\overline{\rm PA})^2 = (\overline{\rm PA})^2 + 36[/tex]
[tex]\sf 16\overline{\rm PA} = 28[/tex]
[tex]\sf \overline{\rm PA} = \dfrac{7}{4}\:m[/tex]
[tex]\sf (\overline{\rm PM})^2 = (\overline{\rm PA})^2 - (\overline{\rm MB})^2[/tex]
[tex]\sf (\overline{\rm PM})^2 = \left(8 - \dfrac{7}{4}\right)^2 - (5)^2[/tex]
[tex]\sf (\overline{\rm PM})^2 = \dfrac{625}{16} - 25[/tex]
[tex]\sf \overline{\rm PM} = \dfrac{15}{4}\:m[/tex]
[tex]\sf r = \dfrac{\Delta_{BPM}}{\Delta_{PMC} + \Delta_{PAC}}[/tex]
[tex]\sf r = \dfrac{\dfrac{\overline{\rm PM} \:.\:\overline{\rm MB}}{2}}{\dfrac{\overline{\rm PM} \:.\:\overline{\rm MC}}{2} + \dfrac{\overline{\rm PA} \:.\:\overline{\rm AC}}{2}}[/tex]
[tex]\sf r = \dfrac{\dfrac{75}{8}}{\dfrac{75}{8} + \dfrac{42}{8}}[/tex]
[tex]\boxed{\boxed{\sf r = \dfrac{25}{39}}}[/tex]
