1 Standards 7, 8 Review Rhombus Properties Review Isosceles Trapezoid Properties Review Trapezoid...

1

Standards 7, 8Review Rhombus Properties

Review Isosceles Trapezoid Properties

Review Trapezoid Properties

Finding the Area of a Triangle

PROBLEM 1 PROBLEM 2

Finding the Area of a Trapezoid

PROBLEM 3 PROBLEM 4

PROBLEM 5

Finding the Area of a Rhombus

PROBLEM 7 PROBLEM 8

PROBLEM 11

Standards 7, 8, 10

PROBLEM 6

PROBLEM 9 PROBLEM 10

END SHOWPRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

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

1. Two pairs of parallel sides.

2. All sides are congruent.

4. Diagonals bisect each other

6. Opposite angles are congruent and bisected

by diagonals.

7. Consecutive angles are supplementary.

m A m B+ = 180°

m B m C+ = 180°

m C m D+ = 180°

m D m A+ = 180°

A

D B

C

3. Diagonals are NOT congruent

5. Diagonals form a right angle

RHOMBUS

Standards 7, 8, 10



4

1. Exactly one pair of parallel sides.

2. One pair of congruent sides.

4. Diagonals DO NOT bisect each other

5. Base angles are congruent.

6. Opposite angles are supplementary.

m A m C+ = 180°

m B m D+ = 180°

AD

BC

3. Diagonals are congruent

ISOSCELES TRAPEZOID

7. Line connecting midpoints of congruent sides is called MEDIAN.

M

b1

b 2

b1

b 2and are bases of trapezoid

M is the medianMb

1 b 2

2=

+

M = b1 b 2+

12

Standards 7, 8, 10



5

Mb

1 b 2

2=

+

M = b1 b 2+

12

1. Exactly one pair of parallel sides.

TRAPEZOID

2. Line connecting midpoints of congruent sides is called MEDIAN.

b1


M is the median

M

b1

b 2

Standards 7, 8, 10



6

Area of a Triangle

A = bh12

h

b b

where:

b= base

h= height

h

Standards 7, 8, 10



7

20 cm

14 cm

Find the area:

A = bh12

b= 20 cm

h= 14 cm

A = (20cm)(14cm)12

This is a triangle so:

where:

A =(10 cm )(14cm)

A=140 cm2

Standards 7, 8, 10



8

Find the lenght of the base of the triangle below if its area is 50 cm 2

7 cm

A = bh12

The area is:

b= ?

h= 7 cm

where:

A= 50 cm 2

50 cm = b(7cm)12

2

50 cm = b(7cm)12

2 (2)(2)

100 cm = b(7 cm)2

(7 cm) (7 cm)

b 14.3 cm

Substituting and solving for b:

Standards 7, 8, 10



9

A

B C

D

Area of Trapezoid

Area of Trapezoid = Area of + Area of ABC ADC

b1

b2

hh

Area of Trapezoid = 12

b1

h 12

b2

h+


b1

h b2

+

h

b1

b2

Can you define the Area in function of the Median? M = b1 b 2+

12

Click to find out…

Area of ABC =12

b1

h

Area of ADC =12

b2 h

Standards 7, 8, 10



10

h

b1

b2

Area of a Trapezoid


b1

h b2

+

b1

b2

MArea of Trapezoid = h M

Mb

1 b 2

2=

+

M = b1 b 2+

12

OR

where:

b1


M is the median

h is the height

Standards 7, 8, 10



11

h = 7

Find the area of the figure below:

30

50

7

This is a trapezoid so:

A = 12

b1

h b2

+

where:

b1=30

b2=50

A = 12

30 + 50(7)

= 72 (80)

= 5602

= 280 square units.

Substituting:

Standards 7, 8, 10



12

40 cm

10 cm

Find the length of the missing base of the trapezoid below if the area is 340 cm :2

The formula for the Area is:

A = 12

b1

h b2

+

where:

h = 10

b1=40

b2=?

A=340 cm2

We substitute and solve for :b2

340 = 12

(10) b2

+40

340 = (5) b2

+40

340 = (5)(40) b2

+ 5

340 = 200 b2

+ 5

-200 -200

140 = b25

5 5

b2 = 28 cm

Which means the trapezoid is:

28 cm

10 cm

40 cm

Standards 7, 8, 10



13

60

20

Find the Median, using both bases, for the following trapezoid if its area is 840 square units.

The formula for the Area is:

A = 12

b1

h b2

+

where:

h = 20

b1=?

b2=60

A= 840

We substitute and solve for :b1

840 = 12

(20) b1

+60

840 = (10) b1

+60

840 = (10)(60) b1

+ 10

840 = 600 b1

+ 10

-600 -600

240 = b110

10 10

b1 = 24

now the Median is:

Mb

1 b 2

2=

+

substituting values:

M2

=+24 60

M2

=84

M= 42 lineal units.

Standards 7, 8, 10

Is there a shorter way? Click to find out…



14

Area of Trapezoid = h M

840=20M20 20

M= 42

Standards 7, 8, 10If we don’t have to use the bases, then:



15

The area of an isosceles trapezoid is 88 square in. Its height is 3 in and its sides are 5 in. Find the lenght of the bases.

88 in2 5 in5 in3 in

Formula for the area:

A = 12

b1

h b2

+

b1 88 = 1

2( 3) b

2+

b1

b1176 = b

2+3

b1

176 = b2

+3 3 First Equation

Z

From the figure calculating Z

using Pythagorean Theorem:

5 = Z + 32 22

25 = Z + 92

-9 -9

16 = Z2

4 = Z

Z = 4

Z

b2

Now from the figure:

+ b1

b2 = Z + Z

b2 = 2Z + b

1

b2 = 2( ) + b

14

b2 = 8 + b

1 Second Equation

Solving both equations:

b1

176 = b2

+3 3

b2 = 8 + b

1

b1

176 = +3 3( )8 + b1

b1

176 = +3 24 + b1

3

b1

176 = +6 24

-24 -24

152 = 6b1

6 6

b1 = 25.3

Sustituting in Second Equation:

b2 = 8 + b

1

= 8 + 25.3

b2 = 33.3

Standards 7, 8, 10


2

16

The area of an isosceles trapezoid is 88 square in. Its height is 3 in and its sides are 5 in. Find the lenght of the bases.

88 in2 5 in5 in3 in

Z

From the figure calculating Z

using Pythagorean Theorem:

5 = Z + 32 22

25 = Z + 92

-9 -9

16 = Z2

4 = Z

Z = 4

Z

Now from the figure:

+ b1

b2 = Z + Z

b2 = 2Z + b

1

b2 = 2( ) + b

14

b2 = 8 + b

1

Standards 7, 8, 10

3

4

A = bh12

A = (4)(3)12

A= 6Now area of central rectangle is Area of trapezoid minus Area of triangles:

A rec = 88-2(6) = 76

Area of triangles:

b1A rec = (3)

b2

b1

76 = (3)b1

3 3

b1= 25.3

b2 = 8 + ( )25.3

b2 = 33.3

Alternate solution:

Area of central rectangle is also:

Which solution is easier for you?PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17

d2

d1

Area of a rhombus

A = 12

d1d

2

where:

d1 = small diagonal

d2 =large diagonal

Standards 7, 8, 10



18

Find the area of the figure below:

A =12

d1d

2

where:

d1 = small diagonal

d2 =large diagonal

30

15

Calculating value for diagonals:

d1 = 15 + 15

d2 = 30 +30

d2 =60

d1 = 30

This is a rhombus so: Then substituting values:

A =12

(30)(60)

A = 15(60)

A= 900

Standards 7, 8, 10



19

Giving that the figure below has an area of 300 square units, find the value for both diagonals:

A =12

d1d

2

where:

d1 = small diagonal

d2 =large diagonal

45 d2 = 45 + 45

d2 = 90

This is a rhombus so:

Then substituting values and solving for :d1

Calculating value for :d2

300 = 12

(90) d1

300 = 45d1

45 45

d1 6.7

A = 300

Standards 7, 8, 10



20

The area of a rhombus is 300 cm . If one of its diagonals is 1 more than twice the other, find the length of each diagonal.

2

d2

d1

A = 12

d1d

2

A= 300 cm2

300= 12

d1d

2

300= 12

d1d

2 22

600= d1d

2 First Equation

Second Equation

Solving both equations together:

600= d1d

2

600= d1

Standards 7, 8, 10

d2 =2d

1 + 1

d2 =2d

1 + 1

2 d1

+1

600= 2 d1

2d

1+

-600 -600

0= 2 d1

2d

1+ -600

Let’s solve this quadratic equation and then return to the problem:

=2d1 +1



21

Using the Quadratic Formula:

X=-b b - 4ac

2a

2+_

where: 0 = aX +bX +c2

Our equation is:

a= 2b= 1

c= -600

We substitute values:

2

21 1 -600

+ -

Standards 7, 8, 10

0= 2 d1

2d

1+ -600

-1 69.3 =

4

+d1

=-( ) ( ) - 4( )( )

2( )

2+_d

1

=-1 1 + 4800

4

+_d

1

-1 4801 =

4

+_d

1

-1 69.3 =

4

+_d

1

4

68.3 = =17d1

-1 69.3 =

4

-d

1

4

-70.3 = =-17.6d1

No sense in problem

17d1

-17.6d1

Let’s go back to problem!



22

The area of a rhombus is 300 cm . If one of its diagonals is 1 more than twice the other, find the length of each diagonal.

2

d1

A = 12

d1d

2

A= 300 cm2

300= 12

d1d

2

300= 12

d1d

2 22

600= d1d

2 First Equation

Second Equation

Solving both equations together:

600= d1d

2 Using second equation:

17

d2 35

Standards 7, 8, 10

d2 =2d

1 + 1

d2 =2d

1 + 1

600= d1 2 d

1+1

600= 2 d1

2d

1+

-600 -600

0= 2 d1

2d

1+ -600

17d1

=2d1 +1

d2 =2d

1 +1

d2 =2 +1

d1



23

The perimeter of a rhombus is 40 units and one of its diagonals is 5 units. Find the area of the rhombus.

Perimeter:

P=

X

X

X

+ X

X

+ X

X

+ X

P = 4X

40 = 4X4 4

X= 10

=105

To find the area we apply the formula:

A =12

d1d

2

d1=

From the figure:

d1= 5

10 = Z + 2.52 22

100 = Z + 6.252

-6.25 -6.25

93.75 = Z2

Z 9.7

From the figure:

d2 =2Z

d2 =2( )9.7

A =12

( )( ) 5 19.4

A 48.5 Square units

d2 19.4

d2 19.4

Using Area Formula:

Standards 7, 8, 10

We need to find the second diagonal using the Pythagorean Theorem; we know that so let’s find Z.

d2 =2Z

2.5

Z


24

Calculate the non-shaded area of the figure below:

50

3030 20

25

25A1

A2

Total Area = A1 A2-

is a trapezoid:A1

A = 12

b1

h b2

+1

where:b

1 =50

b2 = 30+20+30

=80

h= 25+25=50

So the trapezoid is:

A = 12

(50) +1

50 80

A = (25)1

130

A =1

3250

A = 12

(20)2

25

A2 is a triangle:

A = bh122

b= 20

h= 25

Calculating Total Area:

A =2

250

Total Area = 3250 - 250

Total Area = 3000

The Total non-shaded Area is 3000 square units

Standards 7, 8, 10

A =(10)2

25

80

50


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1 Standards 7, 8 Review Rhombus Properties Review Isosceles Trapezoid Properties Review Trapezoid...

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