Question

x = \sqrt{11881} = 109
given the side lengths, determine whether the triangle is (1) acute, (2) obtuse, or (3) right.

1. 20, 21, and 29

20^2 + 21^2 = 400 + 441 = 841
29^2 = 841
right

1. 15, 19, and 24
2. 11, 23, and 26

find the value of x.

1. a = x, b = 45, c = 53

diagram of a right triangle with legs a, b and hypotenuse c

Explanation:

Let's solve these problems one by one. We'll use the Pythagorean theorem and its converse to determine the type of triangle (right, acute, obtuse) and find the value of \( x \).

Problem 8: Determine the type of triangle with sides 20, 21, and 29

To determine if a triangle is right, acute, or obtuse, we use the converse of the Pythagorean theorem. For a triangle with side lengths \( a \), \( b \), and \( c \) (where \( c \) is the longest side), we check:

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

Step 1: Identify the longest side

The longest side is 29, so \( c = 29 \), \( a = 20 \), and \( b = 21 \).

Step 2: Calculate \( a^2 + b^2 \) and \( c^2 \)

\[
a^2 + b^2 = 20^2 + 21^2 = 400 + 441 = 841
\]
\[
c^2 = 29^2 = 841
\]

Step 3: Compare \( a^2 + b^2 \) and \( c^2 \)

Since \( 20^2 + 21^2 = 29^2 \) (both equal 841), the triangle is right.

Problem 9: Determine the type of triangle with sides 15, 19, and 24

Step 1: Identify the longest side

The longest side is 24, so \( c = 24 \), \( a = 15 \), and \( b = 19 \).

Step 2: Calculate \( a^2 + b^2 \) and \( c^2 \)

\[
a^2 + b^2 = 15^2 + 19^2 = 225 + 361 = 586
\]
\[
c^2 = 24^2 = 576
\]

Step 3: Compare \( a^2 + b^2 \) and \( c^2 \)

Since \( 15^2 + 19^2 = 586 > 576 = 24^2 \), the triangle is acute.

Problem 10: Determine the type of triangle with sides 11, 23, and 26

Step 1: Identify the longest side

The longest side is 26, so \( c = 26 \), \( a = 11 \), and \( b = 23 \).

Step 2: Calculate \( a^2 + b^2 \) and \( c^2 \)

\[
a^2 + b^2 = 11^2 + 23^2 = 121 + 529 = 650
\]
\[
c^2 = 26^2 = 676
\]

Step 3: Compare \( a^2 + b^2 \) and \( c^2 \)

Since \( 11^2 + 23^2 = 650 < 676 = 26^2 \), the triangle is obtuse.

Problem 11: Find the value of \( x \) where \( a = x \), \( b = 45 \), \( c = 53 \) (assuming it's a right triangle with \( c \) as the hypotenuse)

We use the Pythagorean theorem: \( a^2 + b^2 = c^2 \) (if \( c \) is the hypotenuse) or \( b^2 + c^2 = a^2 \) (if \( a \) is the hypotenuse) or \( a^2 + c^2 = b^2 \) (if \( b \) is the hypotenuse). Since \( c = 53 \) is longer than \( b = 45 \), \( c \) is likely the hypotenuse.

Step 1: Apply the Pythagorean theorem

\[
a^2 + b^2 = c^2
\]
Substitute \( a = x \), \( b = 45 \), \( c = 53 \):
\[
x^2 + 45^2 = 53^2
\]

Step 2: Solve for \( x^2 \)

\[
x^2 = 53^2 - 45^2
\]
Calculate \( 53^2 \) and \( 45^2 \):
\[
53^2 = 2809, \quad 45^2 = 2025
\]
\[
x^2 = 2809 - 2025 = 784
\]

Step 3: Solve for \( x \)

\[
x = \sqrt{784} = 28
\]

Final Answers

1. The triangle with sides 20, 21, and 29 is a right triangle.
2. The triangle with sides 15, 19, and 24 is an acute triangle.
3. The triangle with sides 11, 23, and 26 is an obtuse triangle.
4. The value of \( x \) is \( \boxed{28} \).

Answer:

Let's solve these problems one by one. We'll use the Pythagorean theorem and its converse to determine the type of triangle (right, acute, obtuse) and find the value of \( x \).

Problem 8: Determine the type of triangle with sides 20, 21, and 29

To determine if a triangle is right, acute, or obtuse, we use the converse of the Pythagorean theorem. For a triangle with side lengths \( a \), \( b \), and \( c \) (where \( c \) is the longest side), we check:

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

Step 1: Identify the longest side

The longest side is 29, so \( c = 29 \), \( a = 20 \), and \( b = 21 \).

Step 2: Calculate \( a^2 + b^2 \) and \( c^2 \)

\[
a^2 + b^2 = 20^2 + 21^2 = 400 + 441 = 841
\]
\[
c^2 = 29^2 = 841
\]

Step 3: Compare \( a^2 + b^2 \) and \( c^2 \)

Since \( 20^2 + 21^2 = 29^2 \) (both equal 841), the triangle is right.

Problem 9: Determine the type of triangle with sides 15, 19, and 24

Step 1: Identify the longest side

The longest side is 24, so \( c = 24 \), \( a = 15 \), and \( b = 19 \).

Step 2: Calculate \( a^2 + b^2 \) and \( c^2 \)

\[
a^2 + b^2 = 15^2 + 19^2 = 225 + 361 = 586
\]
\[
c^2 = 24^2 = 576
\]

Step 3: Compare \( a^2 + b^2 \) and \( c^2 \)

Since \( 15^2 + 19^2 = 586 > 576 = 24^2 \), the triangle is acute.

Problem 10: Determine the type of triangle with sides 11, 23, and 26

Step 1: Identify the longest side

The longest side is 26, so \( c = 26 \), \( a = 11 \), and \( b = 23 \).

Step 2: Calculate \( a^2 + b^2 \) and \( c^2 \)

\[
a^2 + b^2 = 11^2 + 23^2 = 121 + 529 = 650
\]
\[
c^2 = 26^2 = 676
\]

Step 3: Compare \( a^2 + b^2 \) and \( c^2 \)

Since \( 11^2 + 23^2 = 650 < 676 = 26^2 \), the triangle is obtuse.

Problem 11: Find the value of \( x \) where \( a = x \), \( b = 45 \), \( c = 53 \) (assuming it's a right triangle with \( c \) as the hypotenuse)

We use the Pythagorean theorem: \( a^2 + b^2 = c^2 \) (if \( c \) is the hypotenuse) or \( b^2 + c^2 = a^2 \) (if \( a \) is the hypotenuse) or \( a^2 + c^2 = b^2 \) (if \( b \) is the hypotenuse). Since \( c = 53 \) is longer than \( b = 45 \), \( c \) is likely the hypotenuse.

Step 1: Apply the Pythagorean theorem

\[
a^2 + b^2 = c^2
\]
Substitute \( a = x \), \( b = 45 \), \( c = 53 \):
\[
x^2 + 45^2 = 53^2
\]

Step 2: Solve for \( x^2 \)

\[
x^2 = 53^2 - 45^2
\]
Calculate \( 53^2 \) and \( 45^2 \):
\[
53^2 = 2809, \quad 45^2 = 2025
\]
\[
x^2 = 2809 - 2025 = 784
\]

Step 3: Solve for \( x \)

\[
x = \sqrt{784} = 28
\]

Final Answers

1. The triangle with sides 20, 21, and 29 is a right triangle.
2. The triangle with sides 15, 19, and 24 is an acute triangle.
3. The triangle with sides 11, 23, and 26 is an obtuse triangle.
4. The value of \( x \) is \( \boxed{28} \).