Sagot :
[tex]b=a+3\\
c=a+6\\
a^2+b^2=c^2\\\\
a^2+(a+3)^2=(a+6)^2\\
a^2+a^2+6a+9=a^2+12a+36\\
a^2-6a-27=0\\
a^2-9a+3a-27=0\\
a(a-9)+3(a-9)=0\\
(a+3)(a-9)=0\\
a=-3 \vee a=9\\
[/tex]
The length can't be negative, so the answer is 9 cm.
The length can't be negative, so the answer is 9 cm.
Let the smaller leg be x.
Thus, larger leg = 3 +x
Thus, hypotenuse = 6 + x
Applying pythagoras theorum ;
[tex](6+x)^2 = x^2 + (3+x)^2 \\ \\ or, 6^2 + 12x + x^2 = 3^2 + 6x + x^2 \\ \\ i.e., 36 + 12x = 9 + 6x \\ \\ =>6x = 45 \\ \\ => x = 9[/tex]
Thus, the length of smaller peg is 9 cm.
Thus, larger leg = 3 +x
Thus, hypotenuse = 6 + x
Applying pythagoras theorum ;
[tex](6+x)^2 = x^2 + (3+x)^2 \\ \\ or, 6^2 + 12x + x^2 = 3^2 + 6x + x^2 \\ \\ i.e., 36 + 12x = 9 + 6x \\ \\ =>6x = 45 \\ \\ => x = 9[/tex]
Thus, the length of smaller peg is 9 cm.