1. Surface Area of Pyramids Volume of a Right Pyramid
Standards 8, 10, 11
Classifying Solids
Classifying Pyramids
Surface Area of Pyramids
Volume of a Right Pyramid
Reviewing Perimeters
PROBLEM 1
PROBLEM 2
END SHOW
PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved

2. 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.

3. SOLIDS
Standards 8, 10, 11
PRISM
PYRAMID
CYLINDER
CONE
SPHERE

4. PYRAMID
TRIANGULAR
PYRAMID
RECTANGULAR
PYRAMID
PENTAGONAL
Standards 8, 10, 11
PYRAMID
HEXAGONAL
PYRAMID
OCTAGONAL

5. SURFACE AREA IN PYRAMIDSxl 1 2 + x l xl 1 2 x l xl 1 2 + l x xl 1 2 + x l h xl 1 2 + x l xl 1 2 + x l l x x x + x + x + x +
Calculating Lateral Area: x + L= L= l 1 2
x + x + x + x + x + x
TOTAL SURFACE AREA:
The perimeter of the BASE is:
T = Pl 1 2
+ B P=
LATERAL AREA IS:
P= perimeter of base
B= Area of base polygon
L = lP 1 2
L = Pl 1 2 or
l= slant height
h= height
Standards 8, 10, 11

6. V = B h Standards 8, 10, 11 VOLUME OF A PYRAMID: l h x 1 3 where:
B= Area of the base
h= height

7. Standards 8, 10, 11
REVIEWING PERIMETERS X L X +X X +X Y Y +Y W
+ W W
+ W X +X +X X X +X +X X +X X L
+ L X +X
P =
P =
P =
P = L
+ L
+ W
P= 2Y + X
P = 8X
P = 2L
+ 2W X X X
P =4X
P = 6X
P = 5X

8. 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.
B = ( ) 1 2 l
9 in = 10 5 3 h
7 in =
LATERAL AREA:
= 5 5 3
L = Pl 1 2
L = ( )( ) 1 2
= 25 3 in 2
30 in
9 in
L =(15 in )(9 in)
TOTAL SURFACE AREA:
T = Pl 1 2
+ B
L = 135 in 2
10 in
T = 135 in
+ 25 3 in 2
Base Area:
Base perimeter:
T = 135 in in 2
10 in
T = in 2
30°
60°
VOLUME: 5 3
V = B h 1 3
P = 3( )
10 in
V = 1 3
( )( )
P = 30 in 25 3 in 2
7 in 10 5
V in 3

9. 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:
L = Pl 1 2 h
26 ft = l
P = 6( )
6 feet
L = ( )( ) 1 2
P = 36 feet
36 ft
26.5ft
B = Pa 1 2
L =(18ft )(26.5 ft)
L = ft 2 B= 1 2 36 3
TOTAL SURFACE AREA: 3 18 =
T = Pl 1 2
+ B
6 ft 3 B= 54 3
feet 2
93.5 ft 2
T = ft +
Calculating base area:
B feet 2
T ft 2
we need to find the slant height, using the Pythagorean Theorem:
60°
VOLUME:
60°
V = B h 1 3 l
= ( ) 2 3 a
l = 2 3 2
V = 1 3
( )( )
30°
93.5 ft 2
26 ft 3
60°
l = 703 2
(9)(3) 6
V ft 3 27 3
l ft