Question

1. solve for the unknown variable for a right triangle:

1. solve for the unknown variable for a right triangle:

1. solve for the unknown variables for an isosceles triangle :

(images show right/isosceles triangles with labeled sides: 1. a=5in, b=4in, c=; 2. c=16in, b=12in, a=; 3. a=15in, b=20in, c=__)

Explanation:

Problem 1

Step1: Recall Pythagorean theorem

For a right triangle, \(a^2 + b^2 = c^2\), where \(a = 5\) in, \(b = 4\) in.

Step2: Substitute values into formula

\(c^2 = 5^2 + 4^2 = 25 + 16 = 41\)

Step3: Solve for \(c\)

\(c = \sqrt{41} \approx 6.40\) in (or keep as \(\sqrt{41}\) in)

Step1: Recall Pythagorean theorem

\(a^2 + b^2 = c^2\), so \(a^2 = c^2 - b^2\), with \(c = 16\) in, \(b = 12\) in.

Step2: Substitute values

\(a^2 = 16^2 - 12^2 = 256 - 144 = 112\)

Step3: Solve for \(a\)

\(a = \sqrt{112} = 4\sqrt{7} \approx 10.58\) in (or keep as \(4\sqrt{7}\) in)

Step1: Identify right triangle sides

Height \(a = 15\) in, base half \(= 10\) in, hypotenuse \(c\).

Step2: Apply Pythagorean theorem

\(c^2 = 15^2 + 10^2 = 225 + 100 = 325\)

Step3: Solve for \(c\)

\(c = \sqrt{325} = 5\sqrt{13} \approx 18.03\) in (or keep as \(5\sqrt{13}\) in)

Answer:

\(c = \sqrt{41}\) in (or approximately \(6.40\) in)

Problem 2