○=> Solution (1) :
Given :
▪︎Length of a the hypotenuse of the triangle formed = 17 ft
▪︎Length of one of its leg = 15 ft
Let the length of another side or the diameter of the object be x.
We know that :
[tex]\color{hotpink}\tt{hypotenuse}^{2} \color{plum}= {base}^{2} + {height}^{2} [/tex]
Which means :
[tex] =\tt {x}^{2} + {15}^{2} = {17}^{2} [/tex]
[tex] = \tt {x}^{2} = {17}^{2} - {15}^{2} [/tex]
[tex] =\tt {x}^{2} = 289 - 225[/tex]
[tex] = \tt {x}^{2} = 64[/tex]
[tex] =\tt x = \sqrt{64} [/tex]
[tex]\hookrightarrow\color{plum}\tt x = 8 \: ft[/tex]
▪︎Therefore, the diameter of this object = 8 ft
○=> Solution (2) :
Given :
▪︎Length of one of the legs of this triangle = 34 cm
▪︎Length of another leg of this triangle = 16 cm
Let the length of the hypotenuse of the triangle formed be x.
We know that :
[tex]\color{hotpink}\tt{hypotenuse}^{2} \color{plum}= {base}^{2} + {height}^{2} [/tex]
Which means :
[tex] = \tt {34}^{2} + {16}^{2} = {x}^{2} [/tex]
[tex] = \tt\tt1156 + 256 = {x}^{2} [/tex]
[tex] = \tt1412 = {x}^{2} [/tex]
[tex] =\tt \sqrt{1412} = x[/tex]
[tex]\hookrightarrow\color{plum}\tt 37 \: ft = x[/tex]
▪︎Therefore, the diameter of this object = 37 ft
