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2 years ago

An Isosceles Triunde

Mathematics

2 answers:

Amanda [17]2 years ago

4 0

Answer:

base = $x^4 - 3x^3+11x-4$

Step-by-step explanation:

An isosceles triangle has 2 sides of equal length (a) and base (b)

⇒ perimeter of an isosceles = 2a + b

Given:

$\textsf{leg (a)}=3x^4 - 6x-1$

$\textsf{perimeter}=7x^4 - 3x^3-x-6$

$\implies \textsf{perimeter}=2a+b$

$\implies 7x^4 - 3x^3-x-6=2(3x^4 - 6x-1)+b$

$\implies 7x^4 - 3x^3-x-6=6x^4 - 12x-2+b$

$\implies x^4 - 3x^3+11x-4=b$

Therefore, base = $x^4 - 3x^3+11x-4$

Mashcka [7]2 years ago

3 0

Answer:

Perimeter=sum of sides

Let base be y

$\\ \tt\longmapsto 3x^4-6x-1+3x^4-6x-1+y=7x^4-3x^3-x-6$

$\\ \tt\longmapsto 6x^4-12x-2+y=7x^4-3x^3-x-6$

$\\ \tt\longmapsto y=x^4-3x^3+11x-4$

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