Math Problem Statement

Solution

To solve the given problems in the image, let's break down the tasks:

  1. Derive an equation based on the areas in Figure 1:

    • In Figure 1, a large square is formed using smaller rectangles and squares. Let’s denote the side length of the large square as aa and bb, where aa is the length of the rectangle and bb is the width of the rectangle.
    • The area of the large square is (a+b)2(a+b)^2.
    • The large square is divided into four regions: two rectangles and two smaller squares.
      • The area of each smaller square is a2a^2 and b2b^2.
      • The area of each rectangle is abab.

    Therefore, the equation based on the areas is: (a+b)2=a2+b2+2ab(a+b)^2 = a^2 + b^2 + 2ab

  2. Calculate the area of the shaded region in Figure 2 given ab=3a - b = 3 and a2+b2=15a^2 + b^2 = 15:

    • First, find the values of aa and bb:

      • We have the equations: ab=3a - b = 3 a2+b2=15a^2 + b^2 = 15

      • Solve these equations simultaneously. From the first equation, express aa in terms of bb: a=b+3a = b + 3

      • Substitute this into the second equation: (b+3)2+b2=15(b+3)^2 + b^2 = 15 b2+6b+9+b2=15b^2 + 6b + 9 + b^2 = 15 2b2+6b+9=152b^2 + 6b + 9 = 15 2b2+6b6=02b^2 + 6b - 6 = 0 b2+3b3=0b^2 + 3b - 3 = 0

      • Solve this quadratic equation using the quadratic formula: b=3±3241(3)21b = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} b=3±9+122b = \frac{-3 \pm \sqrt{9 + 12}}{2} b=3±212b = \frac{-3 \pm \sqrt{21}}{2}

        Therefore, b=3+212b = \frac{-3 + \sqrt{21}}{2} (taking the positive root as length must be positive).

      • Now, find aa: a=b+3=3+212+3=3+212a = b + 3 = \frac{-3 + \sqrt{21}}{2} + 3 = \frac{3 + \sqrt{21}}{2}

    • Calculate the area of the shaded region in Figure 2:

      • The shaded region is a right-angled triangle.
      • The base of the triangle is aa and the height is bb.

      Therefore, the area of the shaded region is: Area=12×a×b=12×(3+212)×(3+212)\text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times \left(\frac{3 + \sqrt{21}}{2}\right) \times \left(\frac{-3 + \sqrt{21}}{2}\right)

      Simplify the expression: Area=12×(3+21)(3+21)4\text{Area} = \frac{1}{2} \times \frac{(3 + \sqrt{21})(-3 + \sqrt{21})}{4} Area=12×9+214\text{Area} = \frac{1}{2} \times \frac{-9 + 21}{4} Area=12×124\text{Area} = \frac{1}{2} \times \frac{12}{4} Area=12×3=32=1.5\text{Area} = \frac{1}{2} \times 3 = \frac{3}{2} = 1.5

Thus, the area of the shaded region in Figure 2 is 1.51.5 square units.

Do you want details or have any questions? Here are some more questions you might want to ask:

  1. Can you provide more detailed steps for solving quadratic equations?
  2. How can I verify the solution to the area problem?
  3. Are there any alternative methods to solve this area problem?
  4. Can you explain more about the properties of the shapes involved?
  5. How would the problem change if aa and bb were different values?

Tip: Always verify your algebraic solutions by substituting back into the original equations to ensure consistency.

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Math Problem Analysis

Mathematical Concepts

Geometry
Quadratic Equations
Algebra

Formulas

Area of a square: side^2
Area of a rectangle: length × width
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

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Suitable Grade Level

Grades 10-12