How to Find the Area of Overlapping Squares?

Finding the area of overlapping squares can be a bit tricky, but with a clear step-by-step approach, it becomes manageable. Let’s break it down.

Step-by-Step Process

  1. Identify the Coordinates
    First, identify the coordinates of the vertices of both squares. Let’s assume we have two squares, Square A and Square B. Each square has four vertices, and we need their coordinates in the form of $(x, y)$

  1. Determine the Overlapping Region
    Next, determine the overlapping region. This involves finding the intersecting area of the two squares. To do this, you need to find the maximum of the minimum $x$ and $y$ coordinates and the minimum of the maximum $x$ and $y$ coordinates. This will give you the coordinates of the overlapping rectangle.

    For example, if Square A has vertices at $(x1, y1)$, $(x2, y2)$, $(x3, y3)$, and $(x4, y4)$, and Square B has vertices at $(x5, y5)$, $(x6, y6)$, $(x7, y7)$, and $(x8, y8)$, the overlapping region can be found by:

    $text{left} = text{max}(x1, x5)$

    $text{right} = text{min}(x2, x6)$

    $text{top} = text{min}(y1, y5)$

    $text{bottom} = text{max}(y2, y6)$

  1. Calculate the Width and Height of the Overlapping Region
    Now, calculate the width and height of the overlapping region using the coordinates found in

$text{width} = text{right} – text{left}$

$text{height} = text{top} – text{bottom}$

  1. Calculate the Area of the Overlapping Region
    Finally, the area of the overlapping region is simply the product of the width and height:

    $text{Area} = text{width} times text{height}$

Example

Let’s consider an example to illustrate this process.

Suppose Square A has vertices at $(1, 4)$, $(4, 4)$, $(4, 1)$, and $(1, 1)$, and Square B has vertices at $(2, 5)$, $(5, 5)$, $(5, 2)$, and $(2, 2)$. The overlapping region can be found as follows:

$text{left} = text{max}(1, 2) = 2$

$text{right} = text{min}(4, 5) = 4$

$text{top} = text{min}(4, 5) = 4$

$text{bottom} = text{max}(1, 2) = 2$

The width and height of the overlapping region are:

$text{width} = 4 – 2 = 2$

$text{height} = 4 – 2 = 2$

Therefore, the area of the overlapping region is:

$text{Area} = 2 times 2 = 4 text{ square units}$

Conclusion

Understanding how to find the area of overlapping squares involves identifying the coordinates of the vertices, determining the overlapping region, and then calculating the width, height, and area of that region. With practice, this process can become second nature.

Citations

  1. 1. Khan Academy – Area of Shapes
  2. 2. Math is Fun – Geometry
  3. 3. Wolfram Alpha – Geometry

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ