Surface Area / Volume TEST

TEST Tomorrow! 

Test Specs:
7 Surface Area problems
5 Volume problems
Questions will be worth 8 or 9 points depending on the degree of difficulty.

You get points for including the formula on each problem.  This will need to be the specific fromula, not the generalized “B” for base. 
For example:
The formula for Surface Area of a trapezoidal pyramid would be  SA = 1/2 P l + 1/2 (b1 + b2) h

You will get to use your graphic organizer!!
You do NOT have to memorize the formulas. 🙂

I have posted the answers to the optional review assignment on the Homework page of my web site AND the powerpoint presentations with problems worked out step-by-step.

I won’t be able to check the GeoBlog tonight…..Bad news, my home computer is sick and in the hospital.  I’m still at school now….Jordyn’s 1-Act Play…Bravo!……Sorry about no late night posting!!  🙁

I WILL be in my classroom at 7:00 tomorrow morning if anyone has any problems / questions.

Mrs. C

Volume of Pyramids and Cones


V= volume      B= area of the base    h= height      r= radius

Volume of a Pyramid- V=1/3Bh

Volume of a Cone- V=1/3πr2h

Example 1


V= 1/3 (8×8)(6.5)


V=138.67 or 138 2/3 m3

Example 2

This is a 30-60-90 triangle so in order to find the height we multiply the long leg by √3 the sh.leg.


V=1/3π r2h

V= 1/3π(92)(9√3)

V= 1/3π(81)(9√3)

V= 243π√3 m3

By Amanda C



Volume of a Prism and a Cylinder

First, V=volume, B=base area, H=prism height, r=radius, and h=cylinder height


The volume of a three-dimensional figure represents how many cubed units can fit into the figure. To find the volume of a prism, multiply the base area times the prism height( V=BH ).

Example 1:V=BH

 The base area would be base times height since the base is a parrallelogram. So, 4 x 5 equals 20. Next, the height is 6. So, now our equation is V=20 x 6. Then, the volume is 120cm cubed. 






To find the volume of a cylinder, multiply pi times the radius squared times the height( V=Pi x r squared x h ).  

Example 2:V=Pi x r squared x

First, the diameter is 13, so the radius is 6.5, and 6.5 squared is 42.25. Next, the cylinder height is 6. So, Pi x 42.25 x 6 is 253.5Pi ft cubed.



 By Matt N

Surface Area of Prism


SA=surface area

B=area of base

P=perimeter of base

H=height of prism


Ex. 1




SA=360 units^2




SA=416 units^2


~posted by: EliseG~

Area (First 4 Formulas)

Perimeter=sum of all the sides

Area for a parallelogram= base * height


Ex. 1


Before finding area you must find height. Since this “has” a 30-60-90 triangle(just move the line over to where one side meets a vertex) then:

leg=sq.r.3 * short leg

h= sq.r.3 * 5



A= b * h

  =5 * 5(sq.r. 3)

 =25(sq.r. 3)ft. squared

Ex. 2

Determine type of quadrilateral and area.

T(0,0) V(2,6) X(6,6) Y(4,0)

slope TV=3                slope XY=3

slope VX=0                slope YT=0

therefore it is a parallelogram because opposite sides are parallel.

Graph to determine area.

b=TY= sq.r.[(0-4)sq. + (0-0)sq.)]=sq.r. 16= 4

h=from graph=6

A=b*h= 4*6=24


Area of a Triangle


Inside a rectangle, there are two congruent triangles if it is cut out and rearranged.

If the Area of a rectangle= b * h then to find one of the triangles



Area of a Trapezoid


 A= 1/2 (b1+b2)*h



Area of a Rhombus


A=1/2 (d1*d2)


~posted by karap~

Circle III Test

TEST Tomorrow! 

Test Specs:
Circle Graph  20 pts
Probability  12 pts
Arc Length  12 pts
Area of Sector  12 pts
Equation  24 pts

Check your answers to the Review Sheetfrom the Homework page at….

I’ll check the GeoBlog @ 8:30 and 10:00 tonight.
So, post your questions/problems and I’ll respond.

I have Math Make-up tomorrow morning, so I’ll post the answers outside the door.  You can work together and I ‘ll any answer questions during class before the test.

Mrs. C

Equation of A Circle

(can’t figure out how to do the ‘square root’ sign, so I am going to use'{ }’.)

(also, ^ means ‘squared’)


Deriving Formula

ex. radius=4, center= (3,2)

4= {(x-3)^+(y-2)^}






Hope this helps clarify any questions.



Segment and Arc Theorems – 5 and 6

radius - tangent.JPG
J is perpendicular to K
A radius drawn to a point of tangency is perpendicular to the tangent.  K is the radius and J is the tangent.
tangents from same pt.JPG
Mk is congruent to ML
Tangents from the same ext. pt. are congruent.
These theorems should be pretty easy if you don’t think too hard!
-Katherine H.

segments and arcs in a circle

s chords.JPG


** If 2 chords intersect, then the product of their segments are equal


2 secants

Relationship- FHxFG=FJxFI

** if 2 secants are drawn from the same pt the ext. section times the secant equals the ext. section times the secant

tangent and secant

Relationship- BExBD=BC sqr

** If a tangent and a secant are drawn from the same pt the the secant times the ext. segment equals the tangent sqr

~Stephanie~ good luck on the test tomorrow everyone~