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50 PTS!!A right triangle has a leg measuring 16in. And 28in.

What is the length of the hypotenuse?

Round your answer to the nearest tenth.

1)23.0 in.
2)32.2 in.
3)44.0 in.
4)1040.0 in

Respuesta :

semsee45

Answer:

2)  32.2 in (nearest tenth)

Step-by-step explanation:

Pythagoras’ Theorem

[tex]\sf a^2+b^2=c^2[/tex]

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

Given:

  • a = 16 in
  • b = 28 in
  • c = hypotenuse

Substituting these values into the formula and solving for c:

[tex]\implies \sf 16^2+28^2=c^2[/tex]

[tex]\implies \sf 256+784=c^2[/tex]

[tex]\implies \sf c^2=1040[/tex]

[tex]\implies \sf c=\pm\sqrt{1040}[/tex]

[tex]\implies \sf c=\pm 32.2\:in\:(nearest\:tenth)[/tex]

As distance is positive, c = 32.2 in (nearest tenth)
  • 32.2 inches. (Option 2)

Step-by-step explanation :

Here, A right angled triangle is given with the measure of two sides and we are to find the measure of the third side.

We'll find the measure of third side with the help of the Pythagorean theorem,

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

Here,

  • The base (BC) is 16 in

  • The perpendicular (AB) is 28 in

  • The hypotenuse is AC.

[tex] \: [/tex]

So, substituting the values in the formula we get :

[tex]\\ {\longrightarrow \pmb{\sf {\qquad (AC {)}^{2}= (16) {}^{2} + (28 {)}^{2} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad (AC {)}^{2}=256 + 784 }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad (AC {)}^{2}=1040 }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad AC = \sqrt{1040} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\frak {\qquad AC = 32.24 }}} \\ \\[/tex]

Therefore,

  • The length of the hypotenuse (AC) is 32.2 in (Rounded to nearest tenth)
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