[tex]\displaystyle \sf S_{BCD} = \frac{15\cdot BD }{2} \ \ ;\ \ S_{ABD} = \frac{45\cdot h }{2} \\\\\\ \frac{S_{ABD}}{S_{BCD}} = \frac{\displaystyle \frac{45\cdot BD }{2} }{\displaystyle \frac{15\cdot BD }{2} } \\\\\\ \frac{S_{ABD}}{S_{BCD}} = \frac{45}{15} = 3 \\\\\\ \huge\boxed{\sf S_{ABD} = 3\cdot S_{BCD} }\checkmark[/tex]
logo a área do triangulo ABD NÃO É o dobro da área do triangulo BCD
das outras alternativas :
[tex]\displaystyle \sf \text{b) ABC {\'e} is{\'o}sceles } \ (VERDADEIRO) \\\\ \text{vamos partir da seguinte ideia} : \\\\ \text{Pelo teorema do angulo externo, sabemos que} \\\\ \theta = \frac{\theta}{2} + \alpha \\\\\\ \boxed{\sf \alpha = \frac{\theta }{2} }\checkmark[/tex]
Portanto AC = BD. Daí : [tex]\displaystyle \sf (\Delta_{BDC}) \\\\ Pitagoras : \\\\ BC^2= CD^2+BD^2 \\\\ 30^2 = 15^2 +BD^2 \\\\ BD^2 = 30^2-15^2 \\\\ BD^2 = (30+15)\cdot (30-15) \\\\ BD^2 = 45\cdot 15 = 9\cdot 5\cdot 5\cdot 3 = 3^2\cdot 5^2\cdot 3 \\\\ BD =\sqrt{3^2\cdot 5^2\cdot 3} \\\\ BD = 3\cdot 5\cdot \sqrt{3} \\\\ BD = 15\sqrt{3}[/tex]
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letra (d) é a alternativa falsa
[tex]\displaystyle \sf S_{BCD} = \frac{15\cdot BD }{2} \ \ ;\ \ S_{ABD} = \frac{45\cdot h }{2} \\\\\\ \frac{S_{ABD}}{S_{BCD}} = \frac{\displaystyle \frac{45\cdot BD }{2} }{\displaystyle \frac{15\cdot BD }{2} } \\\\\\ \frac{S_{ABD}}{S_{BCD}} = \frac{45}{15} = 3 \\\\\\ \huge\boxed{\sf S_{ABD} = 3\cdot S_{BCD} }\checkmark[/tex]
logo a área do triangulo ABD NÃO É o dobro da área do triangulo BCD
das outras alternativas :
[tex]\displaystyle \sf \text{b) ABC {\'e} is{\'o}sceles } \ (VERDADEIRO) \\\\ \text{vamos partir da seguinte ideia} : \\\\ \text{Pelo teorema do angulo externo, sabemos que} \\\\ \theta = \frac{\theta}{2} + \alpha \\\\\\ \boxed{\sf \alpha = \frac{\theta }{2} }\checkmark[/tex]
Portanto AC = BD. Daí :
[tex]\displaystyle \sf (\Delta_{BDC}) \\\\ Pitagoras : \\\\ BC^2= CD^2+BD^2 \\\\ 30^2 = 15^2 +BD^2 \\\\ BD^2 = 30^2-15^2 \\\\ BD^2 = (30+15)\cdot (30-15) \\\\ BD^2 = 45\cdot 15 = 9\cdot 5\cdot 5\cdot 3 = 3^2\cdot 5^2\cdot 3 \\\\ BD =\sqrt{3^2\cdot 5^2\cdot 3} \\\\ BD = 3\cdot 5\cdot \sqrt{3} \\\\ BD = 15\sqrt{3}[/tex]