Pythagoras Problems 3

Question 1 of 4

Solve for line

M O

$MO$ .

Round off answer to

1

$1$ decimal place

(25.6)cm

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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 (side) to find the value of line

P N

$PN$ .

$c = 17$ $c=17$ cm

$a = P N$ $a=PN$

$b = 9$ $b=9$ cm

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{PN}}^2$	$=$ $=$	${\color{#00880a}{17}}^2-{\color{#007DDC}{9}}^2$
$P N^{2}$ $PN^2$	$=$ $=$	$289 - 81$ $289-81$
$P N^{2}$ $PN^2$	$=$ $=$	$208$ $208$
$\sqrt{P N^{2}}$ $sqrt(PN^2)$	$=$ $=$	$\sqrt{208}$ $sqrt208$	Get the square root of both sides
$P N$ $PN$	$=$ $=$	$14.422$ $14.422$ m
$P N$ $PN$	$=$ $=$	$14.4$ $14.4$ m	Round off to $1$ $1$ decimal place

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

M P

$MP$ .

$c = 22$ $c=22$ cm

$a = M P$ $a=MP$

$b = 14.42$ $b=14.42$ cm

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{MP}}^2$	$=$ $=$	${\color{#00880a}{22}}^2-{\color{#007DDC}{14.42}}^2$
$M P^{2}$ $MP^2$	$=$ $=$	$484 - 207.936$ $484-207.936$
$M P^{2}$ $MP^2$	$=$ $=$	$276.064$ $276.064$
$\sqrt{M P^{2}}$ $sqrt(MP^2)$	$=$ $=$	$\sqrt{276.064}$ $sqrt276.064$	Get the square root of both sides
$M P$ $MP$	$=$ $=$	$16.6151$ $16.6151$ cm
$M P$ $MP$	$=$ $=$	$16.6$ $16.6$ cm	Round off to $1$ $1$ decimal place

Finally, add the values of line

M P

$MP$ and

P O

$PO$ to get the value of line

M O

$MO$ .

$M O$ $MO$	$=$ $=$	$M P + P O$ $MP+PO$
	$=$ $=$	$16.6 + 9$ $16.6+9$
	$=$ $=$	$25.6$ $25.6$ cm

25.6

$25.6$ m

Question 2 of 4

Find the perimeter of this shape.

Round off answer to

1

$1$ decimal place

(142.4)cm

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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.

The Perimeter of a shape is the sum of all the side lengths.

First, solve for the missing side,

h

$h$ , which is the hypotenuse. This means we can use Pythagoras’ Theorem.

Start by finding the

2

$2$ missing sides of the triangle.

Remember that opposite sides of a rectangle are equal.

Then, subtract the two values highlighted below.

$50$ $50$

-

$-$ $20$ $20$

=

$=$

30

$30$ cm

Now that we know all lengths of the triangle, label the values, then substitute them into Pythagoras’ Theorem.

$c = h$ $c=h$

$a = 30$ $a=30$ cm

$b = 30$ $b=30$ cm

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{h}}^2$	$=$ $=$	${\color{#9a00c7}{30}}^2 + {\color{#007DDC}{30}}^2$
$h^{2}$ $h^2$	$=$ $=$	$900 + 900$ $900+900$
$h^{2}$ $h^2$	$=$ $=$	$1800$ $1800$
$\sqrt{h^{2}}$ $sqrt(h^2)$	$=$ $=$	$\sqrt{1800}$ $sqrt1800$	Get the square root of both sides
$h$ $h$	$=$ $=$	$42.4264 \dots$ $42.4264…$ cm
$h$ $h$	$=$ $=$	$42.4$ $42.4$ cm	Round off to $1$ $1$ decimal place

Finally, add all the side lengths of the shape to find the perimeter.

Perimeter	$=$ $=$	$20 + 30 + 50 + 42.4$ $20+30+50+42.4$
	$=$ $=$	$142.4$ $142.4$ cm

142.4

$142.4$ cm

Question 3 of 4

Find the perimeter of this shape.

Round off answer to

1

$1$ decimal place

(33.6)m

Incorrect

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

Use $\large\textbf{+}$

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

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

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

c

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

The Perimeter of a shape is the sum of all the side lengths.

The sides with the same markers (single line) have the same length.

Also, since the trapezoid has a horizontal base, this means the two right triangles at both sides of the trapezoid are equal.

Find the value of the lower sides of the triangles by subtracting the length of the shorter base of the trapezoid from the longer base.

14 - 8

$14-8$

=

$=$

6

$6$

Since the triangles are equal, we can divide

6

$6$ by

2

$2$ which is equal to

3

$3$ .

This means the length of the lower sides of the triangles is

3

$3$ m.

Now that we know all lengths of the triangle, label the values, then substitute them into Pythagoras’ Theorem.

$c =$ $c=$

?

$?$

$a = 3$ $a=3$ m

$b = 5$ $b=5$ m

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{3}}^2 + {\color{#007DDC}{5}}^2$
$c^{2}$ $c^2$	$=$ $=$	$9 + 25$ $9+25$
$c^{2}$ $c^2$	$=$ $=$	$34$ $34$
$\sqrt{c^{2}}$ $sqrt(c^2)$	$=$ $=$	$\sqrt{34}$ $sqrt34$	Get the square root of both sides
$c$ $c$	$=$ $=$	$5.83095 \dots$ $5.83095…$ m
$c$ $c$	$=$ $=$	$5.8$ $5.8$ m	Round off to $1$ $1$ decimal place

Finally, add all the side lengths of the shape to find the perimeter.

Perimeter	$=$ $=$	$5.8 + 5.8 + 8 + 14$ $5.8+5.8+8+14$
	$=$ $=$	$33.6$ $33.6$ m

33.6

$33.6$ m

Question 4 of 4

Find the perimeter of this shape.

Round off answer to

1

$1$ decimal place

(38.2)cm

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.

The sides with the same markers (single line) have the same length.

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

h

$h$ .

$c = h$ $c=h$

$a = 10$ $a=10$ cm

$b = 10$ $b=10$ cm

${\color{#00880a}{c}}^2$	$=$ $=$	${\color{#9a00c7}{a}}^2 + {\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#00880a}{h}}^2$	$=$ $=$	${\color{#9a00c7}{10}}^2 + {\color{#007DDC}{10}}^2$
$h^{2}$ $h^2$	$=$ $=$	$100 + 100$ $100+100$
$h^{2}$ $h^2$	$=$ $=$	$200$ $200$
$\sqrt{h^{2}}$ $sqrt(h^2)$	$=$ $=$	$\sqrt{200}$ $sqrt200$	Get the square root of both sides
$h$ $h$	$=$ $=$	$14.1421 \dots$ $14.1421…$ cm
$h$ $h$	$=$ $=$	$14.1$ $14.1$ cm	Round off to $1$ $1$ decimal place

Next, use Pythagoras’ Theorem (side) to find the value of the rightmost side. Label this side as

y

$y$ .

$c = 14.1$ $c=14.1$ cm

$a = y$ $a=y$

$b = 5$ $b=5$ cm

${\color{#9a00c7}{a}}^2$	$=$ $=$	${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$	Pythagoras’ Theorem
${\color{#9a00c7}{y}}^2$	$=$ $=$	${\color{#00880a}{14.1}}^2-{\color{#007DDC}{5}}^2$
$y^{2}$ $y^2$	$=$ $=$	$200 - 25$ $200-25$
$y^{2}$ $y^2$	$=$ $=$	$175$ $175$
$\sqrt{y^{2}}$ $sqrt(y^2)$	$=$ $=$	$\sqrt{175}$ $sqrt175$	Get the square root of both sides
$y$ $y$	$=$ $=$	$13.22876 \dots$ $13.22876…$ cm
$y$ $y$	$=$ $=$	$13.2$ $13.2$ cm	Round off to $1$ $1$ decimal place

Finally, add all the side lengths of the shape to find the perimeter.

Perimeter	$=$ $=$	$10 + 10 + 5 + 13.2$ $10+10+5+13.2$
	$=$ $=$	$38.2$ $38.2$ cm

38.2

$38.2$ cm

Pythagoras Problems 3

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Quiz summary

Information

1. Question

Hint

Keep Going!

Finding a Side

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 $\large\textbf{+}$

4. Question

Hint

Great Work!

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

Use $\large\textbf{+}$

Finding a Side

Use $\large\textbf{-}$

Quizzes

Pythagoras Problems 3

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Quiz summary

Information

1. Question

Hint

Keep Going!

Finding a Side

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{+}

4. Question

Hint

Great Work!

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

Use ++\large\textbf{+}

Finding a Side

Use --\large\textbf{-}

Quizzes

Use $\large\textbf{-}$

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 $\large\textbf{+}$

Use $\large\textbf{-}$