How to Determine the Area of a Figure?

Determining the area of a figure is a fundamental concept in geometry. The area measures the surface covered by a shape and is usually expressed in square units. Let’s explore how to find the area of various common geometric shapes.

1. Area of a Rectangle

A rectangle is a four-sided figure with opposite sides equal and all angles right angles. The formula to find the area of a rectangle is straightforward:

$A = l times w$

where $A$ is the area, $l$ is the length, and $w$ is the width. For example, if a rectangle has a length of 5 units and a width of 3 units, its area would be:

$A = 5 times 3 = 15 text{ square units}$

2. Area of a Square

A square is a special type of rectangle where all four sides are equal. The formula for the area of a square is:

$A = s^2$

where $s$ is the length of one side. For instance, if each side of a square is 4 units, the area would be:

$A = 4^2 = 16 text{ square units}$

3. Area of a Triangle

The area of a triangle can be calculated using the formula:

$A = frac{1}{2} times b times h$

where $b$ is the base and $h$ is the height. For example, if the base of a triangle is 6 units and the height is 4 units, the area would be:

$A = frac{1}{2} times 6 times 4 = 12 text{ square units}$

4. Area of a Parallelogram

A parallelogram has opposite sides that are equal and parallel. The formula for the area of a parallelogram is similar to that of a rectangle:

$A = b times h$

where $b$ is the base and $h$ is the height. If a parallelogram has a base of 8 units and a height of 5 units, its area would be:

$A = 8 times 5 = 40 text{ square units}$

5. Area of a Trapezoid

A trapezoid has one pair of parallel sides. The formula to find its area is:

$A = frac{1}{2} times (b_1 + b_2) times h$

where $b_1$ and $b_2$ are the lengths of the parallel sides, and $h$ is the height. For instance, if $b_1$ is 7 units, $b_2$ is 5 units, and the height is 4 units, the area would be:

$A = frac{1}{2} times (7 + 5) times 4 = 24 text{ square units}$

6. Area of a Circle

The area of a circle is determined using the formula:

$A = pi r^2$

where $r$ is the radius. For example, if the radius of a circle is 3 units, the area would be:

$A = pi times 3^2 = 9pi text{ square units}$

7. Area of a Rhombus

A rhombus is a type of parallelogram where all sides are equal in length. The area can be calculated using the formula:

$A = frac{1}{2} times d_1 times d_2$

where $d_1$ and $d_2$ are the lengths of the diagonals. For instance, if $d_1$ is 6 units and $d_2$ is 8 units, the area would be:

$A = frac{1}{2} times 6 times 8 = 24 text{ square units}$

8. Area of a Regular Polygon

A regular polygon has all sides and angles equal. The general formula for the area of a regular polygon is:

$A = frac{1}{4} n s^2 frac{1}{tan(frac{pi}{n})}$

where $n$ is the number of sides and $s$ is the length of one side. For example, for a regular pentagon (5 sides) with each side 3 units long, the area would be:

$A = frac{1}{4} times 5 times 3^2 times frac{1}{tan(frac{pi}{5})}$

Conclusion

Understanding how to determine the area of various geometric figures is essential for solving many real-world problems, from architecture to everyday tasks. By mastering these formulas, you can easily calculate the area of any shape you encounter.

3. CK-12 Foundation – Area4. Wikipedia – Area

Citations

  1. 1. Khan Academy – Area
  2. 2. Math is Fun – Area

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