Transcript Standards 8, 10, 11 Classifying Solids Classifying Pyramids Surface Area of Pyramids Volume of a Right Pyramid Reviewing Perimeters PROBLEM 1 PROBLEM 2 PRESENTATION CREATED BY SIMON PEREZ.

Standards 8, 10, 11
Classifying Solids
Classifying Pyramids
Surface Area of Pyramids
Volume of a Right Pyramid
Reviewing Perimeters
PROBLEM 1
PROBLEM 2
1
PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved
END SHOW
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.
Standard 11:
Students determine how changes in dimensions affect the perimeter, area, and volume of common
geomegtric figures and solids.
Estándar 11:
Los estudiantes determinan cambios en dimensiones que afectan perímetro, área, y volumen de figuras
geométricas comunes y sólidos.
2
Standards 8, 10, 11
SOLIDS
PYRAMID
PRISM
CYLINDER
CONE
SPHERE
3
TRIANGULAR
RECTANGULAR
PENTAGONAL
PYRAMID
PYRAMID
PYRAMID
Standards 8, 10, 11
HEXAGONAL
OCTAGONAL
PYRAMID
PYRAMID
4
SURFACE AREA IN PYRAMIDS
l
l
h
x
x
x
x
x
L=
1
2
xl
L=
1
2
l x+x+x+x+x+x
xl
+
1
2
xl +
The perimeter of the BASE is:
P= x + x + x + x + x + x
LATERAL AREA IS:
L=
1
2
lP
or
x
x
x
x
Calculating Lateral Area:
1
2
l
l
l
x
x
x
+
l
l
L=
1
2
Pl
1
2
xl
+
1
2
xl +
1
2
xl
TOTAL SURFACE AREA:
T=
1
2
Pl + B
P= perimeter of base
B= Area of base polygon
l= slant height
5
h= height
Standards 8, 10, 11
Standards 8, 10, 11
VOLUME OF A PYRAMID:
l
h
x
V=
1
3
Bh
where:
B= Area of the base
h= height
6
Standards 8, 10, 11
REVIEWING PERIMETERS
X
L
X
W
W
Y
Y
P= L+W +L +W
P=L +L +W +W
P = 2L + 2W
X
X
X
X
L
X
P = Y +Y +X
P= 2Y + X
X
X
P = X +X +X +X +X +X +X +X
P = 8X
X
X
X
P = 6X
P = 5X
P =4X
7
Standards 8, 10, 11
Find the lateral area and the surface area and volume of a right pyramid whose
slant height is 9 in and whose height is 7 in. Its base is an equilateral triangle whose
side is 10 in. Round your answers to the nearest tenth.
LATERAL AREA:
9 in =l
L=
L=
7 in =h
1
2
1
2
( 30 in )( 9 in )
= 25 3
in2
TOTAL SURFACE AREA:
T=
2
10 in
10 in
( 10 ) 5 3
=5 5 3
L = 135 in
10 in
1
2
Pl
L =(15 in )(9 in)
Base perimeter:
B=
1
2
Pl + B
T = 135 in 2 + 25 3
Base Area:
in2
T = 135 in2 + 43.3 in2
2
10 in
T = 178.3 in
30°
5 3
P = 3(10 in)
60°
P = 30 in
10
5
VOLUME:
1
V = 3 Bh
1 (
25 3
V=
3
V 101 in3
in2 )( 7 in )
8
Standards 8, 10, 11
Find the lateral area, the surface area and volume of a right pyramid with a
height of 26 ft whose base is a regular hexagon with side of 6 ft. Round your
answers to the nearest tenth.
LATERAL AREA:
Perimeter:
1
L = 2 Pl
P = 6( 6 feet )
1
L
=
P = 36 feet
2 ( 36 ft )( 26.5ft)
l
1
L =(18ft )(26.5 ft)
B = 2 Pa
26 ft =h
L = 477 ft 2
B= 1 36
3 3
TOTAL SURFACE AREA:
2
1
= 18 3 3
T = 2 Pl + B
6 ft
3 3
2
B= 54 3 feet
T = 477 ft 2 + 93.5 ft 2
Calculating base area:
2
2
B 93.5 feet
T 570.5 ft
we need to find the slant height,
60° using the Pythagorean Theorem: VOLUME:
1
60° 60°
2
2
Bh
V
=
2
l = 26 + ( 3 3 )
3
2
2
2
1 (
30°
2
l
=
676
+
27
)( 26 ft )
93.5
ft
V=
a
3
3 3
3
3
2
60°
l = 703
(9)(3)
3
9
6
V
810.
ft
27
l
26.5 ft
3