Area Notes

Overview

• The transcript covers various geometric formulas related to area, perimeter, and applications of the Pythagorean Theorem.

Pythagorean Theorem

• Used to find the relationship between the sides of a right triangle.

• Formula: $a^2 + b^2 = c^2$, where:

  • $c$ is the hypotenuse

  • $a$ and $b$ are the lengths of the other two sides

Area Formulas

• General Area Formula for a Rectangle:

  • $A = lw$

  • Where $l$ is the length and $w$ is the width.

• Area of a Square:

  • $A = s^2$

  • Where $s$ is the side length.

• Area of a Triangle:

  • $A = rac{1}{2}bh$

  • Where $b$ is the base and $h$ is the height.

• Area of a Parallelogram:

  • $A = bh$

  • Where $b$ is the base and $h$ is the height.

• Area of a Trapezoid:

  • $A = rac{1}{2}(b_1 + b_2)h$

  • Where $b_1$ and $b_2$ are the lengths of the two bases and $h$ is the height.

• Area of a Circle:

  • $A = rac{1}{4} imes πd^2$ or equivalently $A = πr^2$

  • Where $d$ is the diameter and $r$ is the radius.

Problem Examples

1. Area of a Square:

   • Given side length: 8.5 ft

   • Calculation:
     $A = (8.5)^2 = 72.25 ext{ ft}^2$

2. Area of a Parallelogram:

   • Given base = 12.9 in, height = 7.1 in

   • Calculation:
     $A = (12.9)(7.1) = 91.59 ext{ in}^2$

3. Area of a Trapezoid:

   • Given bases = 19 cm and 27 cm, height = 16 cm

   • Calculation:
     $A = rac{1}{2}(19 + 27)(16) = 368 ext{ cm}^2$

4. Area of a Triangle:

   • Given base = 29.5 in, height = 14 in

   • Calculation:
     $A = rac{1}{2}(29.5)(14) = 206.5 ext{ in}^2$

5. Area of a Half Circle:

   • Given diameter = 17 ft

   • Calculation:
     $A = rac{1}{2}(r^2π) = rac{1}{2}(8.5^2π) = 453.96 ext{ ft}^2$

Application Problems

1. Finding Height of Triangle:

   • Area = 45.5 cm², base = 14 cm

   • Formula: $A = rac{1}{2}bh$

   • Calculation:

     • $45.5 = rac{1}{2}(14)(h)$

     • $45.5 = 7h$

     • $h = rac{45.5}{7} = 6.5 cm$

2. Finding Base of Trapezoid:

   • Area = 390 km², bases = 23 km and 29 km

   • Formula: $A = rac{1}{2}(b_1 + b_2)h$

   • Calculation:

     • $390 = rac{1}{2}(23 + 29)(h)$

     • $390 = 26h$

     • $h = rac{390}{26} = 15 km$

3. Finding Diameter of Circle:

   • Area = 380.13 m²

   • Formula: $A = πr^2$

   • Calculation:

     • $380.13 = πr^2$

     • $r^2 = rac{380.13}{π}$

     • r = ext{sqrt}igg{(} rac{380.13}{ ext{π}}igg{)}

     • Therefore, diameter $d = 2 imes 11 = 22 m$

4. Finding Side of Square with Area:

   • Area = 361 square feet

   • Calculation:

     • Using $A = s^2$

     • $361 = s^2$

     • $s = 19$

     • Perimeter $P = 4s = 4 imes 19 = 76 ft$

Application Calculation for Mr. Payton's Deck

• Deck dimensions = 14 ft x 17.2 ft

• Calculate area:

  • Area = $14 imes 17.2 = 240.8 ext{ ft}^2$

• Since two coats are required, total area to be painted = $240.8 imes 2 = 481.6 ext{ ft}^2$

• Paint coverage = 72 square feet per can

• Number of cans needed:

  • $ext{Cans} = rac{481.6}{72} ext{ (round up)} <br>ightarrow 7 cans$

Conclusion

• These calculations and formulas provide a comprehensive understanding of basic geometric properties and their applications in real-world scenarios. Efficiently calculating areas and dimensions can be critical in different contexts such as construction, landscaping, and home improvement projects.