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Pythagoras Mixed Review 2Pythagoras Mixed Review 2
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Question 1 of 4
1. Question
Find the value of the missing length `x`
Round your answer to 2 decimal places- `x=` (42.20)` \text(m)`
Hint
Help VideoCorrect
Correct!
Incorrect
Method One
Finding a Side
Use $$\large\textbf{-}$$
$${\color{#9a00c7}{a}}^2={\color{#00880a}{c}}^2 \hspace{1mm} \large\textbf{-} \hspace{1mm} \normalsize{\color{#007DDC}{b}}^2$$Use the formula for Finding a Side to solve for `x`$${\color{#9a00c7}{a}}^2$$ `=` $${\color{#00880a}{c}}^2-{\color{#007DDC}{b}}^2$$ Finding a Side $${\color{#9a00c7}{x}}^2$$ `=` $${\color{#00880a}{75}}^2-{\color{#007DDC}{62}}^2$$ Plug in the known lengths `x^2` `=` `5625-3844` Evaluate `x^2` `=` `1781` `sqrt(x^2)` `=` `sqrt1781` Take the square root of both sides `x` `=` `42.20 \text(m)` Rounded to two decimal places `x=42.20 \text(m)`Method Two
Pythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` can be switched as they are both sidesLabelling each length of the triangle
Use the Pythagorean Theorem Formula to solve for `x``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem Formula `x^2``+``62^2` `=` `75^2` Plug in the known lengths `x^2+3844` `=` `5625` Evaluate `x^2+144` `-3844` `=` `5625` `-3844` Subtract `3844` from both sides `x^2``+3844` `-3844` `=` `1781` `3844-3844` cancels out `sqrt(x^2)` `=` `sqrt1781` Take the square root of both sides `x` `=` `42.20 \text(m)` Rounded to two decimal places `x=42.20 \text(m)` -
Question 2 of 4
2. Question
Find the value of the missing length `c`
The given measurements are in unitsRound your answer to one decimal place- `c=` (9.4)` \text(units)`
Correct
Fantastic!
Incorrect
Pythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` are the two sides, and `c` is the hypotenuseLabelling each length of the triangle
Use the Pythagorean Theorem Formula to solve for `c``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem Formula `8^2``+``5^2` `=` `c^2` Plug in the known lengths `64+25` `=` `c^2` Evaluate `sqrt(c^2)` `=` `sqrt89` Take the square root of both sides `c` `=` `9.4 \text(units)` Rounded to one decimal place `c=9.4 \text(units)` -
Question 3 of 4
3. Question
Find the value of the missing length.
The given measurements are in unitsRound your answer to one decimal place- `\text(missing length )=` (10.0)` \text(units)`
Correct
Keep Going!
Incorrect
Method OneFinding a Side
Use $$\large\textbf{-}$$
$${\color{#007DDC}{b}}^2={\color{#00880a}{c}}^2 \hspace{1mm} \large\textbf{-} \hspace{1mm} \normalsize{\color{#9a00c7}{a}}^2$$Labelling each length of the triangle
Use the formula for Finding a Side to solve for `b`$${\color{#007DDC}{b}}^2$$ `=` $${\color{#9a00c7}{c}}^2-{\color{#00880a}{a}}^2$$ Finding a Side $${\color{#007DDC}{b}}^2$$ `=` $${\color{#9a00c7}{12.9}}^2-{\color{#00880a}{8.1}}^2$$ Plug in the known lengths `b^2` `=` `166.41-65.61` Evaluate `b^2` `=` `100.8` `sqrt(b^2)` `=` `sqrt100.8` Take the square root of both sides `b` `=` `10.0 \text(units)` Rounded to one decimal place `b=10.0 \text(units)`Method TwoPythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` can be switched as they are both sidesLabelling each length of the triangle
Use the Pythagorean Theorem Formula to solve for `b``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem Formula `8.1^2``+``b^2` `=` `12.9^2` Plug in the known lengths `65.61+b^2` `=` `166.41` Evaluate `65.61+b^2` `-65.61` `=` `166.41` `-65.61` Subtract `65.61` from both sides `65.61``+b^2` `-65.61` `=` `100.8` `65.61-65.61` cancels out `sqrt(b^2)` `=` `sqrt100.8` Take the square root of both sides `b` `=` `10.0 \text(units)` Rounded to one decimal place `b=10.0 \text(units)` -
Question 4 of 4
4. Question
Find the value of the missing length `c`
The given measurements are in unitsRound your answer to one decimal place- `c=` (7.3)` \text(units)`
Correct
Keep Going!
Incorrect
Pythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` are the two sides, and `c` is the hypotenuseLabelling each length of the triangle
Use the Pythagorean Theorem Formula to solve for `c``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem Formula `5.2^2``+``5.1^2` `=` `c^2` Plug in the known lengths `27.04+26.01` `=` `c^2` Evaluate `sqrt(c^2)` `=` `sqrt53.05` Take the square root of both sides `c` `=` `7.3 \text(units)` Rounded to one decimal place `c=7.3 \text(units)`
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