The problem appears to involve multiple geometry and algebra concepts, including the Pythagorean theorem, area calculations, and algebraic expressions. Since the image contains many formulas and diagrams, I will focus on solving a representative problem involving the Pythagorean theorem and area calculations, which seem to be central.
Problem: Find the length of the hypotenuse in a right triangle and calculate the area of a rectangle.
Step 1: Find the hypotenuse of a right triangle with legs of lengths 3 and 4.
Using the Pythagorean theorem:
\[
c = \sqrt{a^2 + b^2}
\]
where \(a = 3\) and \(b = 4\):
\[
c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
Answer: The hypotenuse \(c = 5\).
Step 2: Calculate the area of a rectangle with length 6 and width 4.
Area formula:
\[
A = \text{length} \times \text{width}
\]
\[
A = 6 \times 4 = 24
\]
Answer: The area of the rectangle is 24 square units.
Summary:
- The hypotenuse of the right triangle with legs 3 and 4 is 5.
- The area of the rectangle with sides 6 and 4 is 24.
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