Pythagoras Mixed Review 4

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Question 1 of 4

Find the value of the missing length

c

$c$

a = 30.9

$a = 30.9$

b = 27.9

$b=27.9$

c = ?

$c=?$

The given measurements are in units

Round your answer to one decimal place

$c =$ $c=$ (41.6) $units$ $\text(units)$

Question 2 of 4

Find the value of the missing length

c

$c$

a = 3.5

$a = 3.5$

b = 3.7

$b=3.7$

c = ?

$c=?$

The given measurements are in units

Round your answer to one decimal place

$c =$ $c=$ (5.1) $units$ $\text(units)$

Question 3 of 4

One wall is

16 m

$16 m$ tall while the other is

10 m

$10 m$ tall. They stand

8 m

$8m$ apart on a horizontal ground. A roof rests on top of both walls. Find the length of the roof.

$c =$ $c=$ (10) $m$ $\text(m)$

Question 4 of 4

A thin piece of wire

41

$41$ metres long is attached to the top of a flag pole. The other end is fixed to the ground at a distance of

15

$15$ metres from the base of the flag pole. Find the height of the flag pole.

The given measurements are in metres

Round your answer to 2 decimal places

$h =$ $h=$ (38.16) $m$ $\text(m)$

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Method One

Finding a Side

Use $\large\textbf{-}$

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

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

Labelling each length of the triangle

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

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Finding a Side
${\color{#9a00c7}{h}}^2$	$=$ $=$	${\color{#00880a}{41}}^2-{\color{#007DDC}{15}}^2$	Plug in the known lengths
$h^{2}$ $h^2$	$=$ $=$	$1681 - 225$ $1681-225$	Evaluate
$h^{2}$ $h^2$	$=$ $=$	$1456$ $1456$
$\sqrt{h^{2}}$ $sqrt(h^2)$	$=$ $=$	$\sqrt{1456}$ $sqrt1456$	Take the square root of both sides
$h$ $h$	$=$ $=$	$38.16 m$ $38.16 \text(m)$	Rounded to two decimal places

h = 38.16 m

$h=38.16 \text(m)$

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 $h$ $h$

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem Formula
$h^{2}$ $h^2$ $+$ $+$ $15^{2}$ $15^2$	$=$ $=$	$41^{2}$ $41^2$	Plug in the known lengths
$h^{2} + 225$ $h^2+225$	$=$ $=$	$1681$ $1681$	Evaluate
$h^{2} + 225$ $h^2+225$ $- 225$ $-225$	$=$ $=$	$1681$ $1681$ $- 225$ $-225$	Subtract $225$ $225$ from both sides
$h^{2}$ $h^2$ $+ 225$ $+225$ $- 225$ $-225$	$=$ $=$	$1456$ $1456$	$225 - 225$ $225-225$ cancels out
$\sqrt{h^{2}}$ $sqrt(h^2)$	$=$ $=$	$\sqrt{1456}$ $sqrt1456$	Take the square root of both sides
$h$ $h$	$=$ $=$	$38.16 m$ $38.16 \text(m)$	Rounded to two decimal places

h = 38.16 m

$h=38.16 \text(m)$

Pythagoras Mixed Review 4

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Quiz summary

Information

1. Question

Correct!

Pythagoras’ Theorem Formula

2. Question

Great Work!

Pythagoras’ Theorem Formula

3. Question

Hint

Well Done!

Pythagoras’ Theorem Formula

Labelling each length of the triangle

4. Question

Hint

Nice Job!

Finding a Side

Use $\large\textbf{-}$

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

Quizzes

Pythagoras Mixed Review 4

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Correct!

Pythagoras’ Theorem Formula

2. Question

Great Work!

Pythagoras’ Theorem Formula

3. Question

Hint

Well Done!

Pythagoras’ Theorem Formula

Labelling each length of the triangle

4. Question

Hint

Nice Job!

Finding a Side

Use --\large\textbf{-}

Labelling each length of the triangle

Pythagoras’ Theorem Formula

Labelling each length of the triangle

Quizzes

Use $\large\textbf{-}$