Given right triangle ABC, with right angle at C, if a = 5 and b = 11 use the pythagorean theorem to solve for b?

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Given right triangle ABC, with right angle at C, if a = 5 and b = 11 use the pythagorean theorem to solve for b?

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Answer 1 · 2016-06-12T12:32:25

Error in question:
If $b = 11$ $b=11$ and we are to solve for $c$ $color(red)(c)$ then $c = \sqrt{146} \approx 12.08305$ $color(green)(c=sqrt(146)~~12.08305)$
If $c = 11$ $color(red)(c)=11$ and we are to solve for $b$ $b$ then $b = 4 \sqrt{6} \approx 9.797959$ $color(green)(b=4sqrt(6)~~9.797959)$

Explanation:

By Pythagorean Theorem (since $c$ $c$ is the hypotenuse)
$XXX a^{2} + b^{2} = c^{2}$ $color(white)("XXX")a^2+b^2=c^2$

If we are trying to find the value of $b$ $b$ , then
$XXX b = \sqrt{c^{2} - a^{2}}$ $color(white)("XXX")b=sqrt(c^2-a^2)$

If we are trying to find the value of $c$ $c$ , then
$XXX c = \sqrt{a^{2} + b^{2}}$ $color(white)("XXX")c=sqrt(a^2+b^2)$

Simply insert whichever values were intended and perform (or have your calculator perform) the arithmetic.

Answer 2 · 2016-06-12T12:37:24

The question needs to be clarified... b appears twice.
Either: $c = 12.08$ $c = 12.08$ or $b = 9.80 or 4 \sqrt{6}$ $b= 9.80 " or4sqrt6$

Explanation:

The small letters represent the sides opposite the vertices with the same capital letter.
c would therefore be the hypotenuse.

This would involved squaring and adding the given sides.

$c^{2} = 5^{2} + 11^{2} = 25 + 121$ $c^2 = 5^2 + 11^2 = 25 + 121$
If $c^{2} = 146, \Rightarrow c = \sqrt{146}$ $c^2 = 146, " " rArr c = sqrt146$
$c = 12.08$ $c = 12.08$

However, if b=11 is meant to be c = 11, it means we are trying to find one of the shorter sides (b), which would involve subtracting:

$b^{2} = 11^{2} - 5^{2} = 121 - 25$ $b^2 = 11^2 -5^2 = 121 - 25$
if $b^{2} = 96, \Rightarrow b = \sqrt{96}$ $b^2 = 96, " " rArr b = sqrt96$
$b = 9.80 or 4 \sqrt{6}$ $b= 9.80 " " or4sqrt6$

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Given right triangle ABC, with right angle at C, if a = 5 and b = 11 use the pythagorean theorem to solve for b?

2 Answers

Explanation:

Explanation:

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