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Pythagoras’ Theorem Problems 3Pythagoras’ Theorem Problems 3
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Question 1 of 4
1. Question
Solve for line `MO`.
Round off answer to `1` decimal place- (25.6)cm
Hint
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Keep Going!
Incorrect
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, use Pythagoras’ Theorem (side) to find the value of line `PN`.
`c=17` cm`a=PN``b=9` cm$${\color{#9a00c7}{a}}^2$$ `=` $${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$$ Pythagoras’ Theorem $${\color{#9a00c7}{PN}}^2$$ `=` $${\color{#00880a}{17}}^2-{\color{#007DDC}{9}}^2$$ `PN^2` `=` `289-81` `PN^2` `=` `208` `sqrt(PN^2)` `=` `sqrt208` Get the square root of both sides `PN` `=` `14.422` m `PN` `=` `14.4` m Round off to `1` decimal place Next, use Pythagoras’ Theorem (side) to find the value of line `MP`.
`c=22` cm`a=MP``b=14.42` cm$${\color{#9a00c7}{a}}^2$$ `=` $${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$$ Pythagoras’ Theorem $${\color{#9a00c7}{MP}}^2$$ `=` $${\color{#00880a}{22}}^2-{\color{#007DDC}{14.42}}^2$$ `MP^2` `=` `484-207.936` `MP^2` `=` `276.064` `sqrt(MP^2)` `=` `sqrt276.064` Get the square root of both sides `MP` `=` `16.6151` cm `MP` `=` `16.6` cm Round off to `1` decimal place Finally, add the values of line `MP` and `PO` to get the value of line `MO`.`MO` `=` `MP+PO` `=` `16.6+9` `=` `25.6` cm `25.6` m -
Question 2 of 4
2. Question
Find the perimeter of this shape.
Round off answer to `1` decimal place- (142.4)cm
Hint
Help VideoCorrect
Fantastic!
Incorrect
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.The Perimeter of a shape is the sum of all the side lengths.First, solve for the missing side, `h`, which is the hypotenuse. This means we can use Pythagoras’ Theorem.Start by finding the `2` missing sides of the triangle.Remember that opposite sides of a rectangle are equal.
Then, subtract the two values highlighted below.
`50``-``20` `=` `30` cm 
Now that we know all lengths of the triangle, label the values, then substitute them into Pythagoras’ Theorem.
`c=h``a=30` cm`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` `=` `900+900` `h^2` `=` `1800` `sqrt(h^2)` `=` `sqrt1800` Get the square root of both sides `h` `=` `42.4264…` cm `h` `=` `42.4` cm Round off to `1` decimal place Finally, add all the side lengths of the shape to find the perimeter.
Perimeter `=` `20+30+50+42.4` `=` `142.4` cm `142.4` cm -
Question 3 of 4
3. Question
Find the perimeter of this shape.
Round off answer to `1` decimal place- (33.6)m
Hint
Help VideoCorrect
Correct!
Incorrect
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.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` `=` `6` Since the triangles are equal, we can divide `6` by `2` which is equal to `3`.This means the length of the lower sides of the triangles is `3`m.
Now that we know all lengths of the triangle, label the values, then substitute them into Pythagoras’ Theorem.
`c=``?``a=3` m`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` `=` `9+25` `c^2` `=` `34` `sqrt(c^2)` `=` `sqrt34` Get the square root of both sides `c` `=` `5.83095…` m `c` `=` `5.8` m Round off to `1` decimal place Finally, add all the side lengths of the shape to find the perimeter.
Perimeter `=` `5.8+5.8+8+14` `=` `33.6` m `33.6` m -
Question 4 of 4
4. Question
Find the perimeter of this shape.
Round off answer to `1` decimal place- (38.2)cm
Hint
Help VideoCorrect
Great Work!
Incorrect
Finding the Hypo$$\large\textbf{+}$$enuse
Use $$\large\textbf{+}$$
$${\color{#00880a}{c}}^2={\color{#9a00c7}{a}}^2 \hspace{1mm} \large\textbf{+} \hspace{1mm} \normalsize{{\color{#007DDC}{b}}^2}$$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) 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`.
`c=h``a=10` cm`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` `=` `100+100` `h^2` `=` `200` `sqrt(h^2)` `=` `sqrt200` Get the square root of both sides `h` `=` `14.1421…` cm `h` `=` `14.1` cm Round off to `1` decimal place Next, use Pythagoras’ Theorem (side) to find the value of the rightmost side. Label this side as `y`.
`c=14.1` cm`a=y``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` `=` `200-25` `y^2` `=` `175` `sqrt(y^2)` `=` `sqrt175` Get the square root of both sides `y` `=` `13.22876…` cm `y` `=` `13.2` cm Round off to `1` decimal place Finally, add all the side lengths of the shape to find the perimeter.
Perimeter `=` `10+10+5+13.2` `=` `38.2` cm `38.2` cm
Quizzes
- Find the Hypotenuse 1
- Find the Hypotenuse 2
- Find the Hypotenuse 3
- Find a Side 1
- Find a Side 2
- Find a Side 3
- Pythagoras Mixed Review 1
- Pythagoras Mixed Review 2
- Pythagoras Mixed Review 3
- Pythagoras Mixed Review 4
- Pythagoras’ Theorem Problems 1
- Pythagoras’ Theorem Problems 2
- Pythagoras’ Theorem Problems 3