Pythagoras Problems 2
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Question 1 of 5
1. Question
Solve for `x`.
Round off answer to `2` decimal places- (26.25)m
Hint
Help VideoCorrect
Keep Going!
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.First, use Pythagoras’ Theorem to find the value of the line in the middle of the shape. Label it as `y`
`c=y``a=14` m`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` `=` `529+196` `y^2` `=` `725` `sqrt(y^2)` `=` `sqrt725` Get the square root of both sides `y` `=` `26.9258` m `y` `=` `26.93` m Round off to `2` decimal places Finally, use Pythagoras’ Theorem (side) to find the value of `x`.
`c=26.93` m`a=x``b=6` m$${\color{#9a00c7}{a}}^2$$ `=` $${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$$ Pythagoras’ Theorem $${\color{#9a00c7}{x}}^2$$ `=` $${\color{#00880a}{26.93}}^2-{\color{#007DDC}{6}}^2$$ `x^2` `=` `725.22-36` `x^2` `=` `689.22` `sqrt(x^2)` `=` `sqrt689.22` Get the square root of both sides `x` `=` `26.2529…` m `x` `=` `26.25` m Round off to `2` decimal places `26.25` m -
Question 2 of 5
2. Question
Solve for `y`.
Round off answer to `2` decimal places- (7.41)m
Hint
Help VideoCorrect
Exceptional!
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.First, use Pythagoras’ Theorem to find the value of the line in the middle of the shape. Label it as `h`.
`c=h``a=8` m`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` `=` `64+16` `h^2` `=` `80` `sqrt(h^2)` `=` `sqrt80` Get the square root of both sides `h` `=` `8.9442` m `h` `=` `8.94` m Round off to `2` decimal place Finally, use Pythagoras’ Theorem (side) to find the value of `y`.
`c=8.94` m`a=y``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` `=` `79.92-25` `y^2` `=` `54.92` `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
3. Question
Solve for `x`.
Round off answer to `2` decimal places- (8.49)cm
Hint
Help VideoCorrect
Excellent!
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.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
4. Question
Solve for `y`.
Round off answer to `2` decimal places- (14.42)m
Hint
Help VideoCorrect
Well Done!
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 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
5. Question
Solve for `x`.
- (8)m
Hint
Help VideoCorrect
Fantastic!
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.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
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 Problems 1
- Pythagoras Problems 2
- Pythagoras Problems 3
- Pythagoras Theorem Mixed Review 1
- Pythagoras Theorem Mixed Review 2
- Pythagoras Theorem Mixed Review 3
- Pythagoras Theorem Mixed Review 4