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Vista county is setting aside a large parcel of land to preserve it as open space. the county has hired Meghan's surveying firm to survey the parcel which is in the shape of a right triangle. the longer leg of the triangle measures 5 miles less than the square of the shorter leg, and the hypotenuse of the triangle measures 13 miles less than twice the square of the shorter leg. the length of each boundary is a whole number. find the length of each boundary.

Respuesta :

bartdrinksmalk
Let [tex]a=[/tex]the length of the shorter leg
Let [tex]b=[/tex]the length of the longer leg
Let [tex]c=[/tex]the length of the hypotenuse

[tex]b=a^{2}-5[/tex]
[tex]c=2a^{2}-13[/tex]
Pythagorean Theorem:
[tex]c^{2}=a^{2}+b^{2}[/tex]
⇒[tex](2a^{2}-13)^{2}=a^{2}+(a^{2}-5)^{2}[/tex]
Substitute a new variable [tex]n[/tex] into [tex]a^{2}[/tex]
⇒[tex](2n-13)^{2}=n+(n-5)^{2}[/tex]
⇒[tex]4n^{2}-52n+169=n+n^{2}-10n+25[/tex]
⇒[tex]4n^{2}-52n+169=n^{2}-9n+25[/tex]
⇒[tex]3n^{2}-43n+144=0[/tex]
⇒[tex](3n-16)(n-9)=0[/tex]
⇒[tex]n=\frac{16}{3}[/tex] or [tex]n=9[/tex]
Since 9 is a perfect square, we go with [tex]n=a^{2}=9[/tex]
⇒[tex]a=3[/tex]

Plug [tex]a[/tex] into [tex]b=a^{2}-5[/tex]
⇒[tex]b=3^{2}-5[/tex]
⇒[tex]b=9-5[/tex]
⇒[tex]b=4[/tex]

Plug [tex]a[/tex] into [tex]c=2a^{2}-13[/tex]
⇒[tex]c=2(3)^{2}-13[/tex]
⇒[tex]c=18-13[/tex]
⇒[tex]c=5[/tex]

So the triangle is a 3-4-5 right triangle.
MrRoyal

The lengths of the boundaries are 3, 4 and 5 miles respectively

Represent the boundaries with x, y and z, where:

  • x represents the shorter leg
  • z represents the hypotenuse

So, we have:

[tex]y = x^2 - 5[/tex]

[tex]z = 2x^2 - 13[/tex]

By Pythagoras theorem, we have:

[tex]z^2 =x^2 + y^2[/tex]

This gives

[tex](2x^2 - 13)^2 = x^2 + (x^2 - 5)^2[/tex]

Expand the equation

[tex]4x^4 - 52x^2 + 169 = x^2 + x^4 - 10x^2 + 25[/tex]

Collect like terms

[tex]4x^4 -x^4 - 52x^2 -x^2 +10x^2 + 169 - 25=0[/tex]

[tex]3x^4 -43x^2 + 144=0[/tex]

Factorize

[tex](3x^2 - 16)(x^2 - 9) = 0[/tex]

Split

[tex]3x^2 = 16\ or\ x^2 = 9[/tex]

The boundaries are whole numbers. So, we discard one of the solutions.

So, we have:

[tex]x^2 = 9[/tex]

Solve for x

[tex]x = 3[/tex]

Recall that:

[tex]y = x^2 - 5[/tex]

[tex]z = 2x^2 - 13[/tex]

So, we have:

[tex]y = 9 -5=4[/tex]

[tex]z = 2(9) - 13 = 5[/tex]

Hence, the lengths of the boundaries are 3, 4 and 5 miles respectively

Read more about Pythagoras theorems at:

https://brainly.com/question/654982

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