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Pythagoras’ Theorem Problems 1Pythagoras’ Theorem Problems 1
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Question 1 of 5
1. Question
Solve for `c`.
Round off answer to `1` decimal place- (10.8)cm
Hint
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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
2. Question
Find the height (`h`) of the ramp.
Round off answer to `1` decimal place- (2.3)m
Hint
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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
3. Question
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
Hint
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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, 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
4. Question
A `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
Hint
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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.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
5. Question
Solve for `h`.
Round off answer to `2` decimal places- (25.48)cm
Hint
Help VideoCorrect
Great Work!
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, 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
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