Matemática, perguntado por franciscovictor2, 1 ano atrás

calcule a altura de um triangulo equilatero de area raiz de 15 centimetro quadrado de area

Soluções para a tarefa

Respondido por FibonacciTH
1
Um triangulo equilateral possui todos os lados de mesma medida. Cuja formula da área é:

\mathsf{A=\dfrac{L^2\sqrt{3}}{4}}

É altura dada pela formula:

\mathsf{H=\dfrac{L\sqrt{3}}{2}}

= = = = = 

Substituindo o valor da área na formula:

\mathsf{\sqrt{15}=\dfrac{L^2\sqrt{3}}{4}}\\\\\\\mathsf{4\sqrt{15}=L^2\sqrt{3}}\\\\\\\mathsf{L^2=\dfrac{4\sqrt{15}}{\sqrt{3}}}\\\\\\\mathsf{L^2=4\sqrt{\dfrac{15}{3}}}\\\\\\\mathsf{L^2=4\sqrt{5}}\\\\\\\mathsf{L=\sqrt{4\sqrt{5}}}\\\\\\\mathsf{L=2\sqrt{\sqrt{5}}}

Logo, a altura do triângulo sera:

\mathsf{H=\dfrac{2\sqrt{\sqrt{5}}\cdot \sqrt{3}}{2}}\\\\\\\mathsf{H=\sqrt{\sqrt{5}}\cdot \sqrt{3}}\\\\\\\mathsf{H=\sqrt{5^{\frac{1}{2}}}\cdot \sqrt{3}}\\\\\\\mathsf{H=\left(5^{\frac{1}{2}}\right)^{\frac{1}{2}}\cdot \sqrt{3}}\\\\\\\mathsf{H=5^{\frac{1}{4}}\cdot \sqrt{3}}\\\\\\\mathsf{H=1,49\cdot 1,73}\\\\\\\boxed{\mathsf{H=2,57\:cm}}\: \: \checkmark
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