QUESTION IMAGE
Question
date: 2/10/26
class:
notes/examples
- used to find the missing ____ of a ____ triangle.
- sides $\boldsymbol{a}$ and $\boldsymbol{b}$ are called ______.
- side $\boldsymbol{c}$ is called the ______.
- for any right triangle: ______.
directions: find the value of $x$. round each answer to the nearest tenth.
1.
2.
3.
4.
5.
6.
7.
Step1: Fill in Pythagorean notes
- Used to find the missing side of a right triangle.
- Sides $\boldsymbol{a}$ and $\boldsymbol{b}$ are called legs.
- Side $\boldsymbol{c}$ is called the hypotenuse.
- For any right triangle: $a^2 + b^2 = c^2$
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Step2: Solve Q1 (find hypotenuse)
Use Pythagorean theorem:
$$x = \sqrt{8^2 + 13^2} = \sqrt{64 + 169} = \sqrt{233} \approx 15.3$$
Step3: Solve Q2 (find hypotenuse)
Use Pythagorean theorem:
$$x = \sqrt{27^2 + 22^2} = \sqrt{729 + 484} = \sqrt{1213} \approx 34.8$$
Step4: Solve Q3 (find leg)
Rearrange Pythagorean theorem:
$$x = \sqrt{9^2 - 7^2} = \sqrt{81 - 49} = \sqrt{32} \approx 5.7$$
Step5: Solve Q4 (find leg)
Rearrange Pythagorean theorem:
$$x = \sqrt{30.5^2 - 19.1^2} = \sqrt{930.25 - 364.81} = \sqrt{565.44} \approx 23.8$$
Step6: Solve Q5 (find triangle height)
Split base into two 11-unit segments, use Pythagorean theorem:
$$x = \sqrt{24^2 - 11^2} = \sqrt{576 - 121} = \sqrt{455} \approx 21.3$$
Step7: Solve Q6 (find leg)
Rearrange Pythagorean theorem:
$$x = \sqrt{16^2 - 5^2} = \sqrt{256 - 25} = \sqrt{231} \approx 15.2$$
Step8: Solve Q7 (find total base)
First find partial base for left triangle:
$$x_1 = \sqrt{14^2 - 12^2} = \sqrt{196 - 144} = \sqrt{52} \approx 7.2$$
Find partial base for right triangle:
$$x_2 = \sqrt{29^2 - 12^2} = \sqrt{841 - 144} = \sqrt{697} \approx 26.4$$
Sum partial bases:
$$x = 7.2 + 26.4 = 33.6$$
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Filled Notes:
- Used to find the missing side of a right triangle.
- Sides $\boldsymbol{a}$ and $\boldsymbol{b}$ are called legs.
- Side $\boldsymbol{c}$ is called the hypotenuse.
- For any right triangle: $a^2 + b^2 = c^2$
Problem Solutions:
- $15.3$
- $34.8$
- $5.7$
- $23.8$
- $21.3$
- $15.2$
- $33.6$