Question

POR FAVOR, ME AJUDEM a fazer essas contas com as respostas ja prontas
a) o perimetro do triangulo:
p= $2 \sqrt[4]{5} + 2 \sqrt[4]{80} + \sqrt[4]{405}$

resp: $9 \sqrt[5]{4}$

b) a area do triangulo:
a= $2\sqrt[4]{80} . \frac{3}{8} \sqrt[4]{1125}$
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2
resp:$\frac{15}{4} \sqrt{3}$

Answer 1

Ola luiza

2⁴√5 + 2⁴√80 + ⁴√405 

2⁴√5 + 2⁴√(16*5)+ ⁴√(81*5) 

2⁴√5 + 4⁴√5 + 3⁴√5 = 9⁴√5 

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A = 2⁴√80 * (3/8) ⁴√1125/2 

A = (3/8) * ⁴√(80*1125) = (3/8)*10√3 = 15√3/4

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