Today we began class by correcting yesturdays homework from page 596 numbers 26-30 + 32-34. After correcting it we went straight to learn new material, and there is a QUIZ on WEDNESDAY about areas of right prisms, cylinders, pyramids and cones. Today we learned about pyramids and cones. They are the following:

Right +Regular (base is a regular polygon) Pyramids:

a pyramid has: base, vertex, lateral face, base edge, laterla edge, side of base, altitude, and slant.

Miss Saatkamp showed a pyramid with a right triangle inside where h was the altitude (height) of the pyramid, a was the apothem of the base, and l was the height of of a lateral face.

Total SUrface Area: T= L + B

L=lateral Area

B= area of base

To find the lateral area we can use the following diagram

l= slant height

s = side of base

To find the area of one triangle you (l*s)/2 and to get the area of all the triangles (lateral are) you multiply it by six, since the solid has six lateral faces. If you multiply the side of the base six times it is immediatley the perimeter. In the diagram above, the red line represents the perimenter of the base. SO therefore the second formula can be used to find the lateral area. perimeter times slant height divided by two.

Right Cones:

a right cone has: vertex, base (circle), altitude, and slant height.

For cones, ms. Saatkamp also showed a cone with a right triangle inside, were h was altitude, l was slant height, but the third side, instead of being the apothem is the radius of the circle.

Total Surface Area: T= L + B

where L = lateral area

B = area of the base ()

Lateral Area: L=

This Formula works because

This was what we learned today and the homework on this that we were assigned for tomorrow is on p. 604 #14-29 ALL