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
$12^2 + 5^2 = x^2$
$144 + 25 = x^2$
$sqrt{169} = sqrt{x^2}$
$sqrt{13} = x$ (note: likely a typo, should be $sqrt{169}=13$)
2)
right triangle with legs 6 ft and 8 ft, hypotenuse x
3)
right triangle with legs 10.3 ft and 5 ft, hypotenuse x
$10.3^2 + 5^2 = x^2$
$106.09 + 25 = x^2$ (note: $10.3^2=106.09$)
$sqrt{131.09} = sqrt{x^2}$
$11.49 = x$ (rounded to 11.5?)
4)
right triangle with legs 9.4 m and 8.6 m, hypotenuse x

Explanation:

Problem 1:

Step1: Apply Pythagorean theorem

For a right triangle, \( a^2 + b^2 = c^2 \), where \( a = 12 \), \( b = 5 \), \( c = x \). So \( 12^2 + 5^2 = x^2 \)
\( 144 + 25 = x^2 \)

Step2: Simplify and solve for \( x \)

\( 169 = x^2 \), take square root: \( \sqrt{169} = \sqrt{x^2} \)
\( 13 = x \)

Step1: Apply Pythagorean theorem

\( a = 6 \), \( b = 8 \), \( c = x \). So \( 6^2 + 8^2 = x^2 \)
\( 36 + 64 = x^2 \)

Step2: Simplify and solve for \( x \)

\( 100 = x^2 \), take square root: \( \sqrt{100} = \sqrt{x^2} \)
\( 10 = x \)

Step1: Apply Pythagorean theorem

\( a = 10.3 \), \( b = 5 \), \( c = x \). So \( 10.3^2 + 5^2 = x^2 \)
\( 106.09 + 25 = x^2 \)

Step2: Simplify and solve for \( x \)

\( 131.09 = x^2 \), take square root: \( \sqrt{131.09} = \sqrt{x^2} \)
\( 11.45 \approx x \) (rounded to nearest tenth: \( 11.5 \))

Answer:

\( 13 \) m

Problem 2: