Pythagoras Mixed Review 2

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Question 1 of 4

Find the value of the missing length

x

$x$

Round your answer to 2 decimal places

$x =$ $x=$ (42.20) $m$ $\text(m)$

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Method One

Finding a Side

Use $\large\textbf{-}$

a^{2} = c^{2} - b^{2}

${\color{#9a00c7}{a}}^2={\color{#00880a}{c}}^2 \hspace{1mm} \large\textbf{-} \hspace{1mm} \normalsize{\color{#007DDC}{b}}^2$

Use the formula for Finding a Side to solve for $x$ $x$

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Finding a Side
${\color{#9a00c7}{x}}^2$	$=$ $=$	${\color{#00880a}{75}}^2-{\color{#007DDC}{62}}^2$	Plug in the known lengths
$x^{2}$ $x^2$	$=$ $=$	$5625 - 3844$ $5625-3844$	Evaluate
$x^{2}$ $x^2$	$=$ $=$	$1781$ $1781$
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{1781}$ $sqrt1781$	Take the square root of both sides
$x$ $x$	$=$ $=$	$42.20 m$ $42.20 \text(m)$	Rounded to two decimal places

x = 42.20 m

$x=42.20 \text(m)$

Method Two

Pythagoras’ Theorem Formula

$a^{2}$ $a^2$

+

$+$ $b^{2}$ $b^2$

=

$=$ $c^{2}$ $c^2$

a

$a$ and

b

$b$ can be switched as they are both sides

Labelling each length of the triangle

Use the Pythagorean Theorem Formula to solve for $x$ $x$

$a^{2}$ $a^2$ $+$ $+$ $b^2$	$=$	$c^2$	Pythagoras’ Theorem Formula
$x^2$ $+$ $62^2$	$=$	$75^2$	Plug in the known lengths
$x^2+3844$	$=$	$5625$	Evaluate
$x^2+144$ $-3844$	$=$	$5625$ $-3844$	Subtract $3844$ from both sides
$x^2$ $+3844$ $-3844$	$=$	$1781$	$3844-3844$ cancels out
$sqrt(x^2)$	$=$	$sqrt1781$	Take the square root of both sides
$x$	$=$	$42.20 \text(m)$	Rounded to two decimal places

$x=42.20 \text(m)$

Question 2 of 4

Find the value of the missing length

$c$

The given measurements are in units

Round your answer to one decimal place

$c=$ (9.4) $\text(units)$

Question 3 of 4

Find the value of the missing length.

The given measurements are in units

Round your answer to one decimal place

$\text(missing length )=$ (10.0) $\text(units)$

Incorrect

Method One