Question

7.1 practice problems
directions: find the missing side of each triangle. round your answers to the nearest tenth if necessary.
1)
triangle with legs 12 m and 5 m, hypotenuse x
$12^2 + 5^2 = x^2$
$144 + 25 = x^2$
$sqrt{169} = sqrt{x^2}$
$sqrt{13} = x$
2)
right triangle with legs 6 ft and 8 ft, hypotenuse x

Explanation:

Problem 1:

Step1: Identify the triangle type (right triangle)

We have a right triangle with legs \( 12 \, \text{m} \) and \( 5 \, \text{m} \), and hypotenuse \( x \). Use the Pythagorean theorem: \( a^2 + b^2 = c^2 \), where \( a = 12 \), \( b = 5 \), \( c = x \).
\( 12^2 + 5^2 = x^2 \)

Step2: Calculate the squares

\( 144 + 25 = x^2 \)
\( 169 = x^2 \)

Step3: Solve for \( x \)

Take the square root of both sides: \( \sqrt{169} = \sqrt{x^2} \)
\( x = 13 \) (since length can't be negative)

Step1: Identify the triangle type (right triangle)

We have a right triangle with legs \( 6 \, \text{ft} \) and \( 8 \, \text{ft} \), and hypotenuse \( x \). Use the Pythagorean theorem: \( a^2 + b^2 = c^2 \), where \( a = 6 \), \( b = 8 \), \( c = x \).
\( 6^2 + 8^2 = x^2 \)

Step2: Calculate the squares

\( 36 + 64 = x^2 \)
\( 100 = x^2 \)

Step3: Solve for \( x \)

Take the square root of both sides: \( \sqrt{100} = \sqrt{x^2} \)
\( x = 10 \) (since length can't be negative)

Answer:

\( x = 13 \, \text{m} \)

Problem 2: