Verified Answer :
Let \(a=\)the length of the shorter leg
Let \(b=\)the length of the longer leg
Let \(c=\)the length of the hypotenuse
\(b=a^{2}-5\)
\(c=2a^{2}-13\)
Pythagorean Theorem:
\(c^{2}=a^{2}+b^{2}\)
⇒\((2a^{2}-13)^{2}=a^{2}+(a^{2}-5)^{2}\)
Substitute a new variable \(n\) into \(a^{2}\)
⇒\((2n-13)^{2}=n+(n-5)^{2}\)
⇒\(4n^{2}-52n+169=n+n^{2}-10n+25\)
⇒\(4n^{2}-52n+169=n^{2}-9n+25\)
⇒\(3n^{2}-43n+144=0\)
⇒\((3n-16)(n-9)=0\)
⇒\(n=\frac{16}{3}\) or \(n=9\)
Since 9 is a perfect square, we go with \(n=a^{2}=9\)
⇒\(a=3\)
Plug \(a\) into \(b=a^{2}-5\)
⇒\(b=3^{2}-5\)
⇒\(b=9-5\)
⇒\(b=4\)
Plug \(a\) into \(c=2a^{2}-13\)
⇒\(c=2(3)^{2}-13\)
⇒\(c=18-13\)
⇒\(c=5\)
So the triangle is a 3-4-5 right triangle.
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:
\(y = x^2 - 5\)
\(z = 2x^2 - 13\)
By Pythagoras theorem, we have:
\(z^2 =x^2 + y^2\)
This gives
\((2x^2 - 13)^2 = x^2 + (x^2 - 5)^2\)
Expand the equation
\(4x^4 - 52x^2 + 169 = x^2 + x^4 - 10x^2 + 25\)
Collect like terms
\(4x^4 -x^4 - 52x^2 -x^2 +10x^2 + 169 - 25=0\)
\(3x^4 -43x^2 + 144=0\)
Factorize
\((3x^2 - 16)(x^2 - 9) = 0\)
Split
\(3x^2 = 16\ or\ x^2 = 9\)
The boundaries are whole numbers. So, we discard one of the solutions.
So, we have:
\(x^2 = 9\)
Solve for x
\(x = 3\)
Recall that:
\(y = x^2 - 5\)
\(z = 2x^2 - 13\)
So, we have:
\(y = 9 -5=4\)
\(z = 2(9) - 13 = 5\)
Hence, the lengths of the boundaries are 3, 4 and 5 miles respectively
Read more about Pythagoras theorems at:
https://brainly.com/question/654982