Algebra
Pythagorean Theorem
Distance & Length
Perimeter
Area
100
Solve for x
3x = 27
x=9
100
The formula of the Pythagorean Theorem
a^2+b^2=c^2
100
A=4 and B=12
Find the length of \overline(AB)
8
100
The perimeter of a square, 3 cm on each side.
12 cm
100
Area of a square, 3 cm on each side.
9 " cm"^2
200
Solve for x
Simplify!
6x = 3
x=1/2
200
a=3 cm and b=4 cm
Find c
a^2+b^2=c^2
5 cm
200
C=-3 and D=6
Find the length of \overline(CD)
9
200
The perimeter of a rectangle with length 8 m and width 10 m
36 m
200
Area of a rectangle:
base 20 m and height 5 m
A_("rectangle")=b cdot h
100 "m"^2
300
Solve for x
4x -5 = 27
x = 8
300
a=8 cm and b=15 cm
Find c
a^2+b^2=c^2
17 cm
300
E=-12 and F=-5
Find the length of \overline(EF)
7
300
The perimeter of a regular hexagon with sides of 3 m
18 m
300
Area of a trapezoid with base 12 cm and 10 cm. Height 5 cm
A_("trapezoid")=1/2(b_1+b_2)h
55 "cm"^2
400
Solve for x
(35 + x)/3 = 10
x=-5
400
b=12 cm and c=15 cm
Find a
a^2+b^2=c^2
9 cm
400
G=(8,5) and H=(16,11)
Find the length of \overline(GH)
D=sqrt([x_2-x_1]^2+[y_2-y_1]^2)
10
400
The perimeter of a parallelogram, base 7 cm,
height 4 cm and overhang 3 cm
P = s_1+s_2+s_3+s_4
24 cm
400
Area of a triangle, base 18m and height 24 m
A_("triangle")=1/2b\cdoth
216 "m"^2
500
Solve for x
10x - 14 = 3x + 70
x=12
500
a=10 cm and c=26 cm
Find b
a^2+b^2=c^2
24 cm
500
J=(-5,4) and K=(19,-6)
Find the length of \overline(JK)
D=sqrt([x_2-x_1]^2+[y_2-y_1]^2)
26
500
The perimeter of a right triangle with base 3 cm and height 4 cm.
a^2+b^2=c^2
12 cm
500
Area of a circle,
with radius 4.5 m
A_("circle")=pi cdot r^2
63.62 "m"^2