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Pythagoras' Theorem

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Pythagoras' Theorem Problems

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Pythagoras’ Theorem Problems 2

Pythagoras’ Theorem Problems 2

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Question 1 of 5

Solve for

x

$x$ .

Round off answer to

2

$2$ decimal places

(26.25)m

Incorrect

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

Use $\large\textbf{+}$

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

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

Finding a Side

Use $\large\textbf{-}$

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

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

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

c

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

First, use Pythagoras’ Theorem to find the value of the line in the middle of the shape. Label it as

y

$y$

$c = y$ $c=y$

$a = 14$ $a=14$ m

$b = 23$ $b=23$ m

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{y}}^2$	$=$ $=$	${\color{#9a00c7}{14}}^2 + {\color{#007DDC}{23}}^2$
$y^{2}$ $y^2$	$=$ $=$	$529 + 196$ $529+196$
$y^{2}$ $y^2$	$=$ $=$	$725$ $725$
$\sqrt{y^{2}}$ $sqrt(y^2)$	$=$ $=$	$\sqrt{725}$ $sqrt725$	Get the square root of both sides
$y$ $y$	$=$ $=$	$26.9258$ $26.9258$ m
$y$ $y$	$=$ $=$	$26.93$ $26.93$ m	Round off to $2$ $2$ decimal places

Finally, use Pythagoras’ Theorem (side) to find the value of

x

$x$ .

$c = 26.93$ $c=26.93$ m

$a = x$ $a=x$

$b = 6$ $b=6$ m

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{x}}^2$	$=$ $=$	${\color{#00880a}{26.93}}^2-{\color{#007DDC}{6}}^2$
$x^{2}$ $x^2$	$=$ $=$	$725.22 - 36$ $725.22-36$
$x^{2}$ $x^2$	$=$ $=$	$689.22$ $689.22$
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{689.22}$ $sqrt689.22$	Get the square root of both sides
$x$ $x$	$=$ $=$	$26.2529 \dots$ $26.2529…$ m
$x$ $x$	$=$ $=$	$26.25$ $26.25$ m	Round off to $2$ $2$ decimal places

26.25

$26.25$ m

Question 2 of 5

Solve for

y

$y$ .

Round off answer to

2

$2$ decimal places

(7.41)m

Incorrect

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

Use $\large\textbf{+}$

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

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

Finding a Side

Use $\large\textbf{-}$

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

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

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

c

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

First, use Pythagoras’ Theorem to find the value of the line in the middle of the shape. Label it as

h

$h$ .

$c = h$ $c=h$

$a = 8$ $a=8$ m

$b = 4$ $b=4$ m

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{h}}^2$	$=$ $=$	${\color{#9a00c7}{8}}^2 + {\color{#007DDC}{4}}^2$
$h^{2}$ $h^2$	$=$ $=$	$64 + 16$ $64+16$
$h^{2}$ $h^2$	$=$ $=$	$80$ $80$
$\sqrt{h^{2}}$ $sqrt(h^2)$	$=$ $=$	$\sqrt{80}$ $sqrt80$	Get the square root of both sides
$h$ $h$	$=$ $=$	$8.9442$ $8.9442$ m
$h$ $h$	$=$ $=$	$8.94$ $8.94$ m	Round off to $2$ $2$ decimal place

Finally, use Pythagoras’ Theorem (side) to find the value of

y

$y$ .

$c = 8.94$ $c=8.94$ m

$a = y$ $a=y$

$b = 5$ $b=5$ m

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{y}}^2$	$=$ $=$	${\color{#00880a}{8.94}}^2-{\color{#007DDC}{5}}^2$
$y^{2}$ $y^2$	$=$ $=$	$79.92 - 25$ $79.92-25$
$y^{2}$ $y^2$	$=$ $=$	$54.92$ $54.92$
$\sqrt{y^{2}}$ $sqrt(y^2)$	$=$	$sqrt54.92$	Get the square root of both sides
$y$	$=$	$7.4108$ m
$y$	$=$	$7.41$ m	Round off to $2$ decimal places

$7.41$ m

Question 3 of 5

Solve for

$x$ .

Round off answer to

$2$ decimal places

(8.49)cm

Incorrect

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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.

Since the shape is a square, all of its sides have the same length.

Choose one triangle from the square then substitute the values into Pythagoras’ Theorem.

$c=12$ cm

$a=x$

$b=x$

${\color{#00880a}{c}}^2$	$=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{12}}^2$	$=$	${\color{#9a00c7}{x}}^2 + {\color{#007DDC}{x}}^2$
$144$	$=$	$2x^2$
$144$ $div2$	$=$	$2x^2$ $div2$	Divide both sides by $2$
$72$	$=$	$x^2$
$sqrt72$	$=$	$sqrt(x^2)$	Get the square root of both sides
$8.48528…$ cm	$=$	$x$
$x$	$=$	$8.49$ cm	Round off to $2$ decimal places

$8.49$ cm

Question 4 of 5

Solve for

$y$ .

Round off answer to

$2$ decimal places

(14.42)m

Incorrect

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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 missing values of the triangle.

Remember that opposite sides of a rectangle are equal.

Now, label the values of the right triangle, and then substitute them into Pythagoras’ Theorem (side).

$c=17$ m

$a=y$

$b=9$ m

${\color{#9a00c7}{a}}^2$	$=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{y}}^2$	$=$	${\color{#00880a}{17}}^2-{\color{#007DDC}{9}}^2$
$y^2$	$=$	$289-81$
$y^2$	$=$	$208$
$sqrt(y^2)$	$=$	$sqrt208$	Get the square root of both sides
$y$	$=$	$14.4222$ m
$y$	$=$	$14.42$ m	Round off to $2$ decimal places

$14.42$ m

Question 5 of 5

Solve for

$x$ .

(8)m

Incorrect

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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 sides with the same markers (single line or double lines) have the same lengths.

Now label the values of one of the right triangles, and then substitute them into Pythagoras’ Theorem (side).

$c=10$ m

$a=x$

$b=6$ m

${\color{#9a00c7}{a}}^2$	$=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{x}}^2$	$=$	${\color{#00880a}{10}}^2-{\color{#007DDC}{6}}^2$
$x^2$	$=$	$100-36$
$x^2$	$=$	$64$
$sqrt(x^2)$	$=$	$sqrt64$	Get the square root of both sides
$x$	$=$	$8$ m

$8$ m