Post on 15-Jan-2016
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
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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.
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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
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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
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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
b 2and are bases of trapezoid
M is the median
M
b1
b 2
Standards 7, 8, 10
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6
Area of a Triangle
A = bh12
h
b b
where:
b= base
h= height
h
Standards 7, 8, 10
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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
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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
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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+
Area of Trapezoid = 12
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
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10
h
b1
b2
Area of a Trapezoid
Area of Trapezoid = 12
b1
h b2
+
b1
b2
MArea of Trapezoid = h M
Mb
1 b 2
2=
+
M = b1 b 2+
12
OR
where:
b1
b 2and are bases of trapezoid
M is the median
h is the height
Standards 7, 8, 10
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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
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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
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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…
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14
Area of Trapezoid = h M
840=20M20 20
M= 42
Standards 7, 8, 10If we don’t have to use the bases, then:
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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
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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:
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17
d2
d1
Area of a rhombus
A = 12
d1d
2
where:
d1 = small diagonal
d2 =large diagonal
Standards 7, 8, 10
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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
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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
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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
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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!
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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
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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
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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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