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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