If a, b and c are a Pythagorean triple, then:
[tex]a^2+b^2=c^2[/tex]
for a ≤ b < c.
We have:
A) 11, 60, 61
check:
[tex]11^2+60^2=121+3600=3721\\\\61^2=3721[/tex]
CORRECT
B) 6, 8, 15
These are not the sides of the triangle because 6 + 8 = 14 < 15
C) 5, 11, 12
check
[tex]5^2+11^2=25+121=146\\\\12^2=144\\\\146\neq144[/tex]
INCORRECT
D) 9, 24, 25
[tex]9^2+24^2=81+576=657\\\\25^2=625\\\\657\neq625[/tex]
INCORRECT
Answer: A) 11, 60, 61
2)
If a, b, c are the sides of the triangle, then:
a + b > c and a + c > b and b + c > a.
We have 14, 48, x. Therefore:
x < 14 + 48
x < 62
and
x + 14 > 48
x > 34
therefore 34 < x < 62
If 34 < x < 48, then:
[tex]14^2+x^2=48^2[/tex]
[tex]196+x^2=2304[/tex] subtract 196 from both sides
[tex]x^2=2108\to x=\sqrt{2108}\\\\x=\sqrt{4\cdot527}\\\\x=\sqrt4\cdot\sqrt{527}\\\\\boxed{x=2\sqrt{527}}[/tex]
If 48 < x < 62, then:
[tex]x^2=14^2+48^2[/tex]
[tex]x^2=196+2304[/tex]
[tex]x^2=2500\to x=\sqrt{2500}\\\\\boxed{x=50}[/tex]
Answer: D) 50
