8 th Grade Math Chart Brisa Alcorta 2 nd Period
Pi The ratio of the circumference of a circle to its diameter. Approximate value: 3.14
Perimeter Perimeter of a square=4s
Word Problem What is the perimeter of this square? 23 inches 92 inches
Diameter The distance from one side of the circle to the opposite. Twice the size of the radius.
Word Problem What is the diameter? Radius= 4.6 Diameter= 9.2
Area Area of a rectangle = L*H
Word Problem If 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 Problem If the diameter of a circle is 8.9 then what is it’s circumference? 3.14 *
Perimeter The complete distance around a shape Perimeter of a Rectangle= 2l+2w
Word Problem If 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
Radius The distance from the center of a circle to its outer edge. Also equals half of the diameter.
Word Problem If a circle has a Diameter of 83 cm. what is its Radius? 83/2=41.5 Radius=41.5
Area Area for a circle= r ^2
Area Area of a Trapezoid= ½(b1+b2)
Area Triangle=1/2bh
Word Problem What 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 Pyramid V=1/3Bh
Volume of a Prism V=Bh
Volume Volume of a cone= 1/3 Bh
Volume Sphere= 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.
P P represents the perimeter of the base
Surface Area Prism(lateral) S=Ph
Word Problem If the perimeter of the base of a prism is 8.6 inches and the height is 4.6 what is the lateral surface area? inches
Surface Area Prism(total) S=Ph+2B
Word Problem S=Ph+2B
Surface Area Pyramid (lateral) S= ½ Pl
Word Problem
Surface Area Pyramid (total) S= ½ Pl+B
Surface Area Cylinder- S=2(pi)rh
Word Problem Find the surface area of the cylinder S=2(pi)rh 2.5 inches 7.3 inches inches