The length of one leg of a right triangle is 9 ft. The length of the hypotenuse is 3 feet longer than the other leg. Find the length of the hypotenuse and the other leg.

The length of the hypotenuse is ___ ft.

Respuesta :

Answer:

  • The length of the hypotenuse is 14 ft

[tex] \: [/tex]

Step-by-step explanation:

We'll solve this using the Pythagorean theorem, we know that,

[tex]\\ {\longrightarrow \pmb{\sf {\qquad (Hypotenuse {)}^{2}= (Base) {}^{2} + (Perpendicular {)}^{2} }}}[/tex]

So, As it is given in the question that, the length of one leg of a right triangle is 9 ft and the length of the hypotenuse is 3 feet longer than the other leg and we are to find the length of the hypotenuse and the other leg.

[tex] \: [/tex]

Let us assume the other leg as x ft and therefore the hypotenuse will be (x + 3) ft.

[tex] \: [/tex]

Now, substituting the values in the formula :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad (x + 3 {)}^{2} = (9 {)}^{2} + (x) {}^{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad (x {)}^{2} + 2.x.3 + {(3)}^{2} =81 + (x) {}^{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad \cancel{ (x {)}^{2}} + 2.x.3 +9 =81 + \cancel{(x) {}^{2} }}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 6x + 9 =81 }}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 6x =81 - 9 }}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 6x =72 }}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{72}{6} }}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\frak {\qquad x =12 }}} \\ \\ [/tex]

Therefore,

  • The length of other leg is 12 ft . And the length of hypotenuse is (12 + 2) ft = 14 ft