Pythagoras Theorem Mixed Review 3

Question 1 of 4

ABCD is a square with

B D = 18

$BD=18$ cm. Find the length of

A B

$AB$ .

Round your answer to 2 decimal places

$A B =$ $AB=$ (12.73) $cm$ $\text(cm)$

Incorrect

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Method One

Finding a Side

Use $\large\textbf{-}$

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

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

Labelling each length of the triangle

All sides of a square are equal, so

a = b = A B

$a=b=AB$ .

Use the formula for Finding a Side to solve for

A B

$AB$

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Finding a Side
${\color{#9a00c7}{AB}}^2$	$=$ $=$	${\color{#00880a}{18}}^2-{\color{#007DDC}{AB}}^2$	Plug in the known lengths
$A B^{2}$ $AB^2$ $+ A B^{2}$ $+AB^2$	$=$ $=$	$324 - A B^{2}$ $324-AB^2$ $+ A B^{2}$ $+AB^2$	Add $A B^{2}$ $AB^2$ to both sides
$2 A B^{2}$ $2AB^2$	$=$ $=$	$324$ $324$	$A B^{2} - A B^{2}$ $AB^2-AB^2$ cancels out
$2 A B^{2}$ $2AB^2$ $\div 2$ $divide2$	$=$ $=$	$324$ $324$ $\div 2$ $divide2$	Divide both sides by $2$ $2$
$2$ $2$ $A B^{2}$ $AB^2$ $\div 2$ $divide2$	$=$ $=$	$162$ $162$	$\times 2 \div 2$ $times2divide2$ cancels out
$\sqrt{A B^{2}}$ $sqrt(AB^2)$	$=$ $=$	$\sqrt{162}$ $sqrt162$	Take the square root of both sides
$A B$ $AB$	$=$ $=$	$12.73 cm$ $12.73 \text(cm)$	Rounded to two decimal places

A B = 12.73 cm

$AB=12.73 \text(cm)$

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

All sides of a square are equal, so

a = b = A B

$a=b=AB$ .

We can use the Pythagorean Theorem Formula to solve for $A B$ $AB$

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem Formula
$A B^{2}$ $AB^2$ $+$ $+$ $A B^{2}$ $AB^2$	$=$ $=$	$18^{2}$ $18^2$	Plug in the known lengths
$2 A B^{2}$ $2AB^2$	$=$ $=$	$324$ $324$	Evaluate
$2 A B^{2}$ $2AB^2$ $\div 2$ $divide2$	$=$ $=$	$324$ $324$ $\div 2$ $divide2$	Divide both sides by $2$ $2$
$2$ $2$ $A B^{2}$ $AB^2$ $\div 2$ $divide2$	$=$ $=$	$162$ $162$	$\times 2 \div 2$ $times2divide2$ cancels out
$\sqrt{A B^{2}}$ $sqrt(AB^2)$	$=$ $=$	$\sqrt{162}$ $sqrt162$	Take the square root of both sides
$A B$ $AB$	$=$ $=$	$12.73 cm$ $12.73 \text(cm)$	Rounded to two decimal places

A B = 12.73 cm

$AB=12.73 \text(cm)$

Question 2 of 4

Find the value of the missing length

The given measurements are in units

Round your answer to one decimal place

$missing length =$ $\text(missing length )=$ (10.2) $units$ $\text(units)$

Incorrect

Method One

Finding a Side

Use $\large\textbf{-}$

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

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

Labelling each length of the triangle

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

${\color{#007DDC}{b}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#9a00c7}{a}}^2$	Finding a Side
${\color{#007DDC}{b}}^2$	$=$ $=$	${\color{#00880a}{10.6}}^2-{\color{#9a00c7}{2.9}}^2$	Plug in the known lengths
$b^{2}$ $b^2$	$=$ $=$	$112.36 - 8.41$ $112.36-8.41$	Evaluate
$b^{2}$ $b^2$	$=$ $=$	$103.95$ $103.95$
$\sqrt{b^{2}}$ $sqrt(b^2)$	$=$ $=$	$\sqrt{103.95}$ $sqrt103.95$	Take the square root of both sides
$b$ $b$	$=$ $=$	$10.2 units$ $10.2 \text(units)$	Rounded to one decimal place

b = 10.2 units

$b=10.2 \text(units)$

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 $b$ $b$

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem Formula
${2.9}^{2}$ $2.9^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	${10.6}^{2}$ $10.6^2$	Plug in the known lengths
$8.41 + b^{2}$ $8.41+b^2$	$=$ $=$	$112.36$ $112.36$	Evaluate
$8.41 + b^{2}$ $8.41+b^2$ $- 8.41$ $-8.41$	$=$ $=$	$112.36$ $112.36$ $- 8.41$ $-8.41$	Subtract $8.41$ $8.41$ from both sides
$8.41$ $8.41$ $+ b^{2}$ $+b^2$ $- 8.41$ $-8.41$	$=$ $=$	$103.95$ $103.95$	$8.41 - 8.41$ $8.41-8.41$ cancels out
$\sqrt{b^{2}}$ $sqrt(b^2)$	$=$ $=$	$\sqrt{103.95}$ $sqrt103.95$	Take the square root of both sides
$b$ $b$	$=$ $=$	$10.2 units$ $10.2 \text(units)$	Rounded to one decimal place

b = 10.2 units

$b=10.2 \text(units)$

Question 3 of 4

The foot of a ladder is

3

$3$ metres from the base of a brick wall. If the ladder reaches

7

$7$ metres up the wall, find its length.

Round your answer to a whole number

$x =$ $x=$ (8) $m$ $\text(m)$

Question 4 of 4

Find the value of the missing length

b

$b$

a = 23.5

$a = 23.5$

b = ?

$b=?$

c = 32.4

$c=32.4$

The given measurements are in units

Round your answer to one decimal place

$b =$ $b=$ (22.3) $units$ $\text(units)$

Incorrect

Method One

Finding a Side

Use $\large\textbf{-}$

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

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

Labelling each length of the triangle

$a = 23.5$ $a=23.5$

$c = 32.4$ $c=32.4$

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

${\color{#007DDC}{b}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#9a00c7}{a}}^2$	Finding a Side
${\color{#007DDC}{b}}^2$	$=$ $=$	${\color{#00880a}{32.4}}^2-{\color{#9a00c7}{23.5}}^2$	Plug in the known lengths
$b^{2}$ $b^2$	$=$ $=$	$1049.76 - 552.25$ $1049.76-552.25$	Evaluate
$b^{2}$ $b^2$	$=$ $=$	$497.51$ $497.51$
$\sqrt{b^{2}}$ $sqrt(b^2)$	$=$ $=$	$\sqrt{497.51}$ $sqrt497.51$	Take the square root of both sides
$b$ $b$	$=$ $=$	$22.3 units$ $22.3 \text(units)$	Rounded to one decimal place

b = 22.3 units

$b=22.3 \text(units)$

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

$a = 23.5$ $a=23.5$

$c = 32.4$ $c=32.4$

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

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem Formula
${23.5}^{2}$ $23.5^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	${32.4}^{2}$ $32.4^2$	Plug in the known lengths
$552.25+b^2$	$=$	$1049.76$	Evaluate
$552.25+b^2$ $-552.25$	$=$	$1049.76$ $-552.25$	Subtract $552.25$ from both sides
$552.25$ $+b^2$ $-552.25$	$=$	$497.51$	$552.25-552.25$ cancels out
$sqrt(b^2)$	$=$	$sqrt497.51$	Take the square root of both sides
$b$	$=$	$22.3 \text(units)$	Rounded to one decimal place

$b=22.3 \text(units)$

Pythagoras Theorem Mixed Review 3

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Quiz summary

Information

1. Question

Hint

Exceptional!

Finding a Side

Use $\large\textbf{-}$

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

2. Question

Fantastic!

Finding a Side

Use $\large\textbf{-}$

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

3. Question

Hint

Excellent!

Pythagoras’ Theorem Formula

Labelling each length of the triangle

4. Question

Excellent!

Finding a Side

Use $\large\textbf{-}$

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

Quizzes

Pythagoras Theorem Mixed Review 3

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Exceptional!

Finding a Side

Use --\large\textbf{-}

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

2. Question

Fantastic!

Finding a Side

Use --\large\textbf{-}

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

3. Question

Hint

Excellent!

Pythagoras’ Theorem Formula

Labelling each length of the triangle

4. Question

Excellent!

Finding a Side

Use --\large\textbf{-}

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

Quizzes

Use $\large\textbf{-}$

Use $\large\textbf{-}$

Use $\large\textbf{-}$