8th Grade Math ChartBrisa Alcorta 2nd Period

PiThe ratio of the circumference of

a circle to its diameter. Approximate value: 3.14

PerimeterPerimeter of a square=4s

Word ProblemWhat is the perimeter of this

square?23 inches

92 inches

DiameterThe distance from one side of the

circle to the opposite. Twice the size of the radius.

Word ProblemWhat is the diameter?

Radius= 4.6

Diameter= 9.2

AreaArea of a rectangle = L*H

Word ProblemIf one side of a rectangle is 4.5

and the other is 2.9, what is the area?

4.5*2.9=13.05

Circumference

Formula: r^2

Word ProblemIf the diameter of a circle is 8.9

then what is it’s circumference?

3.14* 8.927.946

PerimeterThe complete distance around a

shapePerimeter of a Rectangle= 2l+2w

Word ProblemIf one side of a rectangle

measures 7.9 inches, and the other measures 3.6 inches what is the perimeter of the square?

7.9(2)+3.6(2)=38.8

RadiusThe distance from the center of a

circle to its outer edge. Also equals half of the diameter.

Word ProblemIf a circle has a Diameter of 83

cm. what is its Radius?

83/2=41.5Radius=41.5

Area

Area for a circle= r^2

AreaArea of a Trapezoid= ½(b1+b2)

AreaTriangle=1/2bh

Word ProblemWhat is the area of a triangle

with a base that measures 5.6 and a height of 9.5?

Volume of a Cylinder V==Bh

Volume of a PyramidV=1/3Bh

Volume of a Prism V=Bh

VolumeVolume of a cone= 1/3 Bh

VolumeSphere= V= 4/3 r^3

Surface Area A measure of the number of

square units needed to cover the faces or surfaces of a figure.

PP represents the perimeter of the

base

Surface AreaPrism(lateral) S=Ph

Word ProblemIf the perimeter of the base of a

prism is 8.6 inches and the height is 4.6 what is the lateral surface area? 39.56

inches

Surface AreaPrism(total) S=Ph+2B

Word ProblemS=Ph+2B

Surface AreaPyramid (lateral) S= ½ Pl

Word Problem

Surface AreaPyramid (total) S= ½ Pl+B

Surface AreaCylinder- S=2(pi)rh

Word ProblemFind the surface area of the cylinder

S=2(pi)rh2.5 inches

7.3 inches

114.61 inches