Find the Hypotenuse 3

Topics

Trigonometry

Pythagoras' Theorem

Find the Hypotenuse

Find the Hypotenuse 3

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Question 1 of 3

Solve for

x

$x$ .

Round off answer to

1

$1$ decimal place

(10.6)m

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Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

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

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

The longest side of a right triangle is called a hypotenuse (

c

$c$ ). It is also the side opposite the right angle.

Label the values, and then substitute them into Pythagoras’ Theorem.

$c = x$ $c=x$

$a = 8$ $a=8$ m

$b = 7$ $b=7$ m

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{x}}^2$	$=$ $=$	${\color{#9a00c7}{8}}^2 + {\color{#007DDC}{7}}^2$
$x^{2}$ $x^2$	$=$ $=$	$64 + 49$ $64+49$
$x^{2}$ $x^2$	$=$ $=$	$113$ $113$
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{113}$ $sqrt113$	Get the square root of both sides
$x$ $x$	$=$ $=$	$10.63014$ $10.63014$ m
$x$ $x$	$=$ $=$	$10.6$ $10.6$ m	Round off to $1$ $1$ decimal place

10.6

$10.6$ m

Question 2 of 3

Solve for

y

$y$ .

Round off answer to

2

$2$ decimal places

(21.93)m

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Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

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

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

The longest side of a right triangle is called a hypotenuse (

c

$c$ ). It is also the side opposite the right angle.

Label the values, and then substitute them into Pythagoras’ Theorem.

$c = y$ $c=y$

$a = 9$ $a=9$ m

$b = 20$ $b=20$ m

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{y}}^2$	$=$ $=$	${\color{#9a00c7}{9}}^2 + {\color{#007DDC}{20}}^2$
$y^{2}$ $y^2$	$=$ $=$	$81 + 400$ $81+400$
$y^{2}$ $y^2$	$=$ $=$	$481$ $481$
$\sqrt{y^{2}}$ $sqrt(y^2)$	$=$ $=$	$\sqrt{481}$ $sqrt481$	Get the square root of both sides
$y$ $y$	$=$ $=$	$21.9317 \dots$ $21.9317…$ m
$y$ $y$	$=$ $=$	$21.93$ $21.93$ m	Round off to $1$ $1$ decimal place

21.93

$21.93$ m

Question 3 of 3

Solve for

m

$m$ .

Round off answer to

2

$2$ decimal places

(30.61)km

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Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

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

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

The longest side of a right triangle is called a hypotenuse (

c

$c$ ). It is also the side opposite the right angle.

Label the values, and then substitute them into Pythagoras’ Theorem.

$c = m$ $c=m$

$a = 19$ $a=19$ km

$b = 24$ $b=24$ km

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{m}}^2$	$=$ $=$	${\color{#9a00c7}{19}}^2 + {\color{#007DDC}{24}}^2$
$m^{2}$ $m^2$	$=$ $=$	$361 + 576$ $361+576$
$m^{2}$ $m^2$	$=$ $=$	$937$ $937$
$\sqrt{m^{2}}$ $sqrt(m^2)$	$=$ $=$	$\sqrt{937}$ $sqrt937$	Get the square root of both sides
$m$ $m$	$=$ $=$	$30.610457 \dots$ $30.610457…$ km
$m$ $m$	$=$ $=$	$30.61$ $30.61$ km	Round off to $1$ $1$ decimal place

30.61

$30.61$ km

Find the Hypotenuse 3

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Quiz summary

Information

1. Question

Hint

Well Done!

Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

2. Question

Hint

Fantastic!

Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

3. Question

Hint

Correct!

Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

Quizzes

Find the Hypotenuse 3

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Quiz summary

Information

1. Question

Hint

Well Done!

Finding the Hypo++\large\textbf{+}enuse

Use ++\large\textbf{+}

2. Question

Hint

Fantastic!

Finding the Hypo++\large\textbf{+}enuse

Use ++\large\textbf{+}

3. Question

Hint

Correct!

Finding the Hypo++\large\textbf{+}enuse

Use ++\large\textbf{+}

Quizzes

Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$

Finding the Hypo $\large\textbf{+}$ enuse

Use $\large\textbf{+}$