Question

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topic:
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main ideas/questions | notes/examples
pythagorean theorem
converse
given a triangle with sides $a$, $b$, and $c$:

• if _______________________, then the triangle is _____________.
• if _______________________, then the triangle is _____________.
• if _______________________, then the triangle is _____________.

always keep \$c^2$\ on the left side to avoid confusion when classifying!
examples
directions: first, determine if the three side lengths could form a triangle. (recall from earlier, the sum of the two smaller sides must be greater than the third side). if yes, classify the triangle further as acute, right, or obtuse.

1. 3, 7, 9

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 20, 21, 29

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 4, 11, 16

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 17, 17, 22

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 18, 24, 30

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 8, 15, 23

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 31, 35, 39

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse

1. 11, 19, 28

$square$ not a $delta$
$square$ acute
$square$ right
$square$ obtuse
© gina wilson (all things algebra®, llc), 2014-2018

Explanation:

Step1: Fill Pythagorean Converse Rules

For a triangle with sides \(a, b, c\) (where \(c\) is the longest side):

• If \(c^2 = a^2 + b^2\), then the triangle is right.
• If \(c^2 < a^2 + b^2\), then the triangle is acute.
• If \(c^2 > a^2 + b^2\), then the triangle is obtuse.

Step2: Verify triangle validity first

For 3 side lengths to form a triangle: sum of two smaller sides > longest side. Then use the converse rules.

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Example 1: 3, 7, 9

Step1: Check triangle validity

\(3 + 7 = 10 > 9\), valid triangle.

Step2: Apply converse rule

\(c=9\), compute \(9^2 = 81\), \(3^2 + 7^2 = 9 + 49 = 58\).
Since \(81 > 58\), \(c^2 > a^2 + b^2\).

Classification: Obtuse

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Example 2: 20, 21, 29

Step1: Check triangle validity

\(20 + 21 = 41 > 29\), valid triangle.

Step2: Apply converse rule

\(c=29\), compute \(29^2 = 841\), \(20^2 + 21^2 = 400 + 441 = 841\).
Since \(841 = 841\), \(c^2 = a^2 + b^2\).

Classification: Right

---

Example 3: 4, 11, 16

Step1: Check triangle validity

\(4 + 11 = 15 < 16\), does not satisfy triangle inequality.

Classification: Not a Δ

---

Example 4: 17, 17, 22

Step1: Check triangle validity

\(17 + 17 = 34 > 22\), valid triangle.

Step2: Apply converse rule

\(c=22\), compute \(22^2 = 484\), \(17^2 + 17^2 = 289 + 289 = 578\).
Since \(484 < 578\), \(c^2 < a^2 + b^2\).

Classification: Acute

---

Example 5: 18, 24, 30

Step1: Check triangle validity

\(18 + 24 = 42 > 30\), valid triangle.

Step2: Apply converse rule

\(c=30\), compute \(30^2 = 900\), \(18^2 + 24^2 = 324 + 576 = 900\).
Since \(900 = 900\), \(c^2 = a^2 + b^2\).

Classification: Right

---

Example 6: 8, 15, 23

Step1: Check triangle validity

\(8 + 15 = 23\), does not satisfy sum > longest side.

Classification: Not a Δ

---

Example 7: 31, 35, 39

Step1: Check triangle validity

\(31 + 35 = 66 > 39\), valid triangle.

Step2: Apply converse rule

\(c=39\), compute \(39^2 = 1521\), \(31^2 + 35^2 = 961 + 1225 = 2186\).
Since \(1521 < 2186\), \(c^2 < a^2 + b^2\).

Classification: Acute

---

Example 8: 11, 19, 28

Step1: Check triangle validity

\(11 + 19 = 30 > 28\), valid triangle.

Step2: Apply converse rule

\(c=28\), compute \(28^2 = 784\), \(11^2 + 19^2 = 121 + 361 = 482\).
Since \(784 > 482\), \(c^2 > a^2 + b^2\).

Classification: Obtuse

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Answer:

Pythagorean Converse Fill-Ins:

1. If $\boldsymbol{c^2 = a^2 + b^2}$, then the triangle is $\boldsymbol{\text{right}}$.
2. If $\boldsymbol{c^2 < a^2 + b^2}$, then the triangle is $\boldsymbol{\text{acute}}$.
3. If $\boldsymbol{c^2 > a^2 + b^2}$, then the triangle is $\boldsymbol{\text{obtuse}}$.

Example Classifications:

1. Obtuse
2. Right
3. Not a Δ
4. Acute
5. Right
6. Not a Δ
7. Acute
8. Obtuse