# 1 Standards 7, 8, 10 Area of a Regular Polygon PROBLEM 1 PROBLEM 2 PROBLEM 3 Review of Perimeter of a Circle Area of a Circle PROBLEM 4a PROBLEM 4b PROBLEM.

## Presentation on theme: "1 Standards 7, 8, 10 Area of a Regular Polygon PROBLEM 1 PROBLEM 2 PROBLEM 3 Review of Perimeter of a Circle Area of a Circle PROBLEM 4a PROBLEM 4b PROBLEM."— Presentation transcript:

1 Standards 7, 8, 10 Area of a Regular Polygon PROBLEM 1 PROBLEM 2 PROBLEM 3 Review of Perimeter of a Circle Area of a Circle PROBLEM 4a PROBLEM 4b PROBLEM 5 APPLICATION PROBLEM 6 APPLICATION END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 Standard 7: Students prove and use theorems involving the properties of parallel lines cut by a transversal, the properties of quadrilaterals, and the properties of circles. Estándar 7: Los estudiantes prueban y usan teoremas involucrando las propiedades de líneas paralelas cortadas por una transversal, las propiedades de cuadriláteros, y las propiedades de círculos. Standard 8: Students know, derive, and solve problems involving perimeter, circumference, area, volume, lateral area, and surface area of common geometric figures. Estándar 8: Los estudiantes saben, derivan, y resuelven problemas involucrando perímetros, circunferencia, área, volumen, área lateral, y superficie de área de figuras geométricas comunes. Standard 10: Students compute areas of polygons including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. Estándar 10: Los estudiantes calculan áreas de polígonos incluyendo rectángulos, triángulos escalenos, triángulos equiláteros, rombos, paralelogramos, y trapezoides. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 Calculating the area of a regular polygon with side x: a A= x a xaxa 1 2 x a xaxa 1 2 + x a xaxa 1 2 + x a xaxa 1 2 + x a xaxa 1 2 + x a xaxa 1 2 + a 1 2 x + x + x + x + x + x The total Area of the polygon is: x Standards 7, 8, 10 apothem PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 Calculating the area of a regular polygon with side x: a x x x x x x a a a a a a A= xaxa 1 2 xaxa 1 2 + xaxa 1 2 + xaxa 1 2 + xaxa 1 2 + xaxa 1 2 + a 1 2 x + x + x + x + x + x P= x x x x + x x + x x + x x + x x + The perimeter of the polygon is: Substituting in Area Formula: A = aP 1 2 or A = Pa 1 2 The total Area of the polygon is: Standards 7, 8, 10 A = Pa 1 2 Area of a Regular Polygon: If a regular polygon has an area of A square units, a perimeter of P units, and an apothem of a units, then: A = Pa 1 2 Área de un Polígono Regular: Si un póligono regular tiene un área de A unidades cuadradas, un perimétro de P unidades, y un apotema de a unidades, entonces: apothem PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 Calculate the area of a square with apothem of 16 cm. 16 45° 16 + 32 and the Perimeter: A = Pa 1 2 6. As a Regular Polygon the square has an area: where: a = 16 cm P = 32 + 32 + 32 + 32 P = 128 cm Substituting in the Area Formula: A = (128cm)(16cm) 1 2 A =(64 cm )(16cm) A=1024cm 2 1. A square has four right angles. 2. Let’s draw a circle circumscribing the square and a diameter to the circle. 3. The diameter is a diagonal and the radius forms a 45°-45°-90° triangle. 4. Both legs are congruent in the right triangle. 5. The apothem is also Perpedicular Bisector, so: One leg is half the square’s side. Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 Calculate the area of a regular hexagon with perimeter of 120 feet. P= x x x + x x x + x x + x x + x x + The perimeter: P = 6x 120 = 6x 6 x= 20 We can divide in 6 triangles that are equiangular because the vertex sharing a common point in all of them is 360°/6= 60° and the other two angles are (180°-60°)/2= 60°, each. We use this information to calculate the apothem a: a a 20 30° 60° 10 3 3 S 2S S 60° 30° a=a= remember Then using the Area Formula for a Regular Polygon: A = Pa 1 2 P= 120 a =10 3 10 3 120 A= 1 2 600 3 feet 2 or A 1039feet 2 60° Standards 7, 8, 10 10 3 60= 30° 60° 20 10 3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 What would be the Area for the regular octagon below if the sides measure 8? 1. Let’s divide in 8 congruent isosceles triangles the total area. 1 2 3 4 5 6 7 8 2. Let’s take one of the triangles to calculate the apothem. 8 8 3. The vertex angle for each triangle would be: 4. Let’s obtain the base angles: 180°-45° 2 =67.5° 360° 8 =45° 45° 67.5° 5. Let’s draw the altitude and median which is the apothem a. a a 4 67.5° 6. Let’s use trigonometry to find “a” Tan67.5°= 4 a (4) Tan 67.5°=(4) a a = 4 ( ) 2.4142 a 9.7 Tan 67.5°= 4 a Standards 7, 8, 10 P= 8 8 + 8 8 + 8 8 + 8 + 8 8 + 8 8 8 8 8 + 8 8 + P= 64 8. Substituting in the Area Formula: A = ( )( ) 1 2 A =(32 )(9.7) A= 310.4 u 2 7. Calculating the perimeter: 649.7 A = Pa 1 2 apothem 9.7 8 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 D D=Diameter rr r= Radius D= r r + D=2r or D=2r 2 r= D 2 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 Circumference of a Circle: Consider that a circle has a circumference of C units and a radius of r units then: C=2 r where: =3.1416 Q if r= D 2 then C=2 rand C=2D 2 C= D Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 Perimeter or circumference of the circle. Area of the circle. C=2 r C= D or A = r 2 or Standards 7, 8, 10 A = D2D2 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 Find the area and the circumference of a circle with radius 50 cm. C=2 r 1. Finding the circumference: r= 50 cm C=2 ( ) 50 cm C=100 cm or 2. Finding the Area: A = r 2 r= 50 cm A = ( ) 2 50 cm or A=2500 cm 2 A 7854 cm 2 Standards 7, 8, 10 C 314.2 cm PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 Find the area and the circumference of a circle with radius 70 cm. C=2 r 1. Finding the circumference: r= 70 cm C=2 ( ) 70 cm C=140 cm or 2. Finding the Area: A = r 2 r= 70 cm A = ( ) 2 70 cm or A=4900 cm 2 A 15394 cm 2 Standards 7, 8, 10 C 440 cm PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13 John is planning to repaint the facade of the house below. How many square feet of paint does he need? 1. Let’s brake into areas the house, windows and door. 2. Let’s get the measures that we need to calculate the area for each one of them. 12’ 5’ 6’ 2’ 2.5’ 2’ 1.5’ 1.2’ 7’ 5’ 6’ 2’ 18’ 6’ 2’ 12’ 14’ 1.2’ 1.5’ 2.5’ 2’ 7’ 3. Let’s lable the Areas as A1, A2,A3, A4, A5, and A6. A1 A2 A3 A4 A5 A6 4. To calculate the Total Area, let’s calculate the individual areas. Click to continue… Standards 7, 8, 10

14 Standards 7, 8, 10 5’ 18’ 12’ 14’ 1.2’ 1.5’ 2’ 2.5’ 7’ 2’ A1= ( )( ) 1 2 A1 =(9’ )(5’) A1= 45 feet 2 18’ 5’ A1 = bh 1 2 A2 =L W where: W= 12’ L= 14’ A2 = (14’)(12’) A2 = 168 feet 2 A3 = r 2 r= 1.2’/2 A3 = ( ) 2.6 feet A3=.36 feet 2 A3 1.13 feet 2 d=1.2’ r=.6’ a 1.5’ 30° 60°.75 3 30° 60°.75’ a=.75 3 a 1.3 1.5’ 1.3’ Perimeter: P= 6( ) 1.5’ P= 9’ A4= Pa 1 2 1 2 = ( )( ) 9’ 1.3’ A4 5.85 feet 2 A5 =L W where: W= 2’ L= 2.5’ A5 = ( 2.5’)(2’) A5 = 5 feet 2 A6 =L W where: W= 7’ L= 2’ A6 = (7’)(2’) A6 = 14 feet 2 Click to continue… 360° 6 = 60°

15 John is planning to repaint the facade of the house below. How many square feet of paint does he need? 5. Now, knowing the area for each figure in the façade, let’s obtain the TOTAL AREA: A1= 45 feet 2 A3 1.13 feet 2 A4 5.85 feet 2 A5 = 5 feet 2 A6 = 14 feet 2 A2 = 168 feet 2 TOTAL AREA = A1 + A2 – A3 – A4 - 2A5 – A6 = 45 168 + 1.3 - 5.85 - 2(5) - 14 - TOTAL AREA 182 feet 2 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16 14’ 6’ 3’ 2’ 3’ 2’ 1.3’ 1’ 8’ Alice is planning to repaint the facade of the house below. How many square feet of paint does she need? 1.3’ 1’ 3’ 2’ 3’ 8’ A3 A4 A5 A6 Click to continue… 6’ 3’ 15’ 6’ 19’ 14’ A1 A2 1. Let’s brake into areas the house, windows and door. 2. Let’s get the measures that we need to calculate the area for each one of them. 3. Let’s lable the Areas as A1, A2,A3, A4, A5, and A6. 4. To calculate the Total Area, let’s calculate the individual areas. Standards 7, 8, 10 6’ 3’ 2’

17 Standards 7, 8, 10 6’ 19’ 14’ 15’ 1.3’ 2’ 3’ 8’ 3’ A1= ( )( ) 1 2 A1 =(9.5’ )(6’) A1= 57 feet 2 19’ 6’ A1 = bh 1 2 A2 =L W where: W= 14’ L= 15’ A2 = (15’)(14’) A2 = 210 feet 2 A3 = r 2 r= 1.3’/2 A3 = ( ) 2.65’ A3=.42 feet 2 A3 1.3 feet 2 d=1.3’ r=.65’ Perimeter: P= 5( ) 1’ P= 5’ A4= Pa 1 2 1 2 = ( )( ) 5’.7’ A4 1.8 feet 2 A5 =L W where: W= 2’ L= 3’ A5 = ( 3’)(2’) A5 = 6 feet 2 A6 =L W where: W= 8’ L= 3’ A6 = (8’)(3’) A6 = 24 feet 2 1’ a.5’ 360° 5 = 72° 72° 2 =36° Tan 36°=.5 a aTan36°=.5 (a) Tan36° a=.7.7’ 1’ a.5’ 36°

18 Alice is planning to repaint the facade of the house below. How many square feet of paint does she need? A1= 57 feet 2 A3 1.3 feet 2 A4 1.8 feet 2 A5 = 6 feet 2 A6 = 24 feet 2 A2 = 210 feet 2 5. Now, knowing the area for each figure in the façade, let’s obtain the TOTAL AREA: TOTAL AREA = A1 + A2 – 2A3 – A4 - 2A5 – A6 = 57 210 + 2(1.3) - 1.8 - 2(6) - 24 - TOTAL AREA 227 feet 2 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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