Pythagoras Problems 1

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Question 1 of 5

Solve for

$c$ .

Round off answer to

$1$ decimal place

(10.8)cm

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

Use $\large\textbf{+}$

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

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

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

First, find the missing values of the rectangle.

Remember that opposite sides of a rectangle are equal.

Choose one triangle from the rectangle and label the values. Then substitute them into Pythagoras’ Theorem.

$c$

$=?$

$a=6$ cm

$b=9$ cm

${\color{#00880a}{c}}^2$	$=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{c}}^2$	$=$	${\color{#9a00c7}{6}}^2 + {\color{#007DDC}{9}}^2$
$c^2$	$=$	$36+81$
$c^2$	$=$	$117$
$sqrt(c^2)$	$=$	$sqrt117$	Get the square root of both sides
$c$	$=$	$10.817$ cm
$c$	$=$	$10.8$ cm	Round off to $1$ decimal place

$10.8$ cm

Question 2 of 5

Find the height (

$h$ ) of the ramp.

Round off answer to

$1$ decimal place

(2.3)m

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Finding a Side

Use $\large\textbf{-}$

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

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

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

The height of the ramp (

$h$ ) is a side of a right triangle.

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

$c=13$ m

$a=h$

$b=12.8$ m

${\color{#9a00c7}{a}}^2$	$=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{h}}^2$	$=$	${\color{#00880a}{13}}^2-{\color{#007DDC}{12.8}}^2$
$h^2$	$=$	$169-163.84$
$h^2$	$=$	$5.16$
$sqrt(h^2)$	$=$	$sqrt5.16$	Get the square root of both sides
$h$	$=$	$2.27156…$ m
$h$	$=$	$2.3$ m	Round off to $1$ decimal place

$2.3$ m

Question 3 of 5

Jack is holding his kite

$1.3$ meters above the ground. Using the illustration below, find out how high the kite is above the ground.

Round off answer to

$1$ decimal place

(35.9)m

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Finding a Side

Use $\large\textbf{-}$

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

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

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

First, find the value of side

$h$ .

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

$c=41$ m

$a=h$

$b=22$ m

${\color{#9a00c7}{a}}^2$	$=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{h}}^2$	$=$	${\color{#00880a}{41}}^2-{\color{#007DDC}{22}}^2$
$h^2$	$=$	$1681-484$
$h^2$	$=$	$1197$
$sqrt(h^2)$	$=$	$sqrt1197$	Get the square root of both sides
$h$	$=$	$34.59768$ m
$h$	$=$	$34.6$ m	Round off to $1$ decimal place

Finally, add

$1.3$ meters to the length of side

$h$ .

$34.6+1.3$

$=$

$35.9$ m

Therefore, the kite is

$35.9$ meters above the ground.

$35.9$ m

Question 4 of 5

$9$ -metre tree and a

$40$ -metre tree are

$23$ metres apart. Find the distance between the tips of the two trees (

$y$ ).

Round off answer to

$1$ decimal place

(38.6)m

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

Use $\large\textbf{+}$

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

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

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

First, draw and label the right triangle made by the two trees.

Next, find the value of side

$b$ by subtracting the height of the smaller tree from the height of the taller tree.

$b$	$=$	$40-9$
	$=$	$31$ m

Finally, find the value of

$y$ by substituting the known values into Pythagoras’ Theorem.

$c=y$

$a=23$ m

$b=31$ m

${\color{#00880a}{c}}^2$	$=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{y}}^2$	$=$	${\color{#9a00c7}{23}}^2 + {\color{#007DDC}{31}}^2$
$y^2$	$=$	$529+961$
$y^2$	$=$	$1490$
$sqrt(y^2)$	$=$	$sqrt1490$	Get the square root of both sides
$y$	$=$	$38.60051…$ m
$y$	$=$	$38.6$ m	Round off to $1$ decimal place

$38.6$ m

Question 5 of 5

Solve for

$h$ .

Round off answer to

$2$ decimal places

(25.48)cm

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Finding a Side

Use $\large\textbf{-}$

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

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

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

First, find the length of one of the sides of the right triangle by subtracting

$14$ from

$38$ . Label this side as

$b$ .

$b$	$=$	$38-14$
	$=$	$24$ cm

Finally, substitute the known values to Pythagoras’ Theorem (side).

$c=35$ cm

$a=h$

$b=28$ cm

${\color{#9a00c7}{a}}^2$	$=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{h}}^2$	$=$	${\color{#00880a}{35}}^2-{\color{#007DDC}{28}}^2$
$h^2$	$=$	$1225-576$
$h^2$	$=$	$649$
$sqrt(h^2)$	$=$	$sqrt649$	Get the square root of both sides
$h$	$=$	$25.4754…$ cm
$h$	$=$	$25.48$ cm	Round off to $2$ decimal places

$25.48$ cm

Pythagoras Problems 1

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Quiz summary

Information

1. Question

Hint

Great Work!

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

Use $\large\textbf{+}$

2. Question

Hint

Correct!

Finding a Side

Use $\large\textbf{-}$

3. Question

Hint

Fantastic!

Finding a Side

Use $\large\textbf{-}$

4. Question

Hint

Nice Job!

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

Use $\large\textbf{+}$

5. Question

Hint

Great Work!

Finding a Side

Use $\large\textbf{-}$

Quizzes

Pythagoras Problems 1

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Great Work!

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

Use +\large\textbf{+}

2. Question

Hint

Correct!

Finding a Side

Use -\large\textbf{-}

3. Question

Hint

Fantastic!

Finding a Side

Use -\large\textbf{-}

4. Question

Hint

Nice Job!

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

Use +\large\textbf{+}

5. Question

Hint

Great Work!

Finding a Side

Use -\large\textbf{-}

Quizzes

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

Use $\large\textbf{+}$

Use $\large\textbf{-}$

Use $\large\textbf{-}$

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

Use $\large\textbf{+}$

Use $\large\textbf{-}$