Question

An Isosceles Triunde

has legs measure
3x^4 - 6x-1 unit.
The Perimeter of the
triangle is
7x^4 - 3x^3-x-6
Find the Base.

Answer 1

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`

Answer 2

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

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`