2nd degree word problem, areas and volumes

1. 2nd-Degree-Equation Word Problems

2. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems

3. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems

4. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.

5. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x.

6. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).

7. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96

8. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0.

9. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0.
(3x + 16)(x – 6) = 0.

10. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0.
(3x + 16)(x – 6) = 0.
x = –16/3 or x = 6

11. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)
= -16 + 64
= 48 ft
So it reaches the height of 48 feet.

12. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
= 64 (ft)
Therefore, the maximum height is 64 feet.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

13. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.

14. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.

15. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.

16. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
1 in
1 in
1 m
1 m
1 mi
1 mi

17. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
1 in
1 in
If each side of a square is 1 unit, then we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
1 m
1 m
1 mi
1 mi

18. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
1 in
1 in
If each side of a square is 1 unit, then we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 m
1 m
1 mi
1 mi

19. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
1 in
1 in
1 in2
If each side of a square is 1 unit, then we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 m
1 m
1 mi
1 mi
1 square-inch

20. Area
If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.
The loop forms a
perimeter or border
that encloses an area, i.e. a surface in the plane.
The word “area” also denotes the amount of surface
enclosed which is defined below.
1 in
1 in
1 in2
If each side of a square is 1 unit, then we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.
Hence the areas of the following squares are:
1 m
1 m
1 mi
1 mi
1 m2 1 mi2
1 square-inch 1 square-meter 1 square-mile

21. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW

22. If L and W are in a given unit, then A is in unit2.
1 in
1 in
1 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW

23. If L and W are in a given unit, then A is in unit2.
So if L = 2 inch W = 3 inch then A = 2*3 = 6 in2.
1 in
1 in
1 in2
2 in
3 in
6 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW

24. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
R

25. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
R

26. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
R
There are two basic approaches.

27. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
There are two basic approaches.
R
12
8
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
4
R
12
4

28. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
There are two basic approaches.
R
12
8
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
4
Hence the area of R is
R
144 – 32
= 112 m2
12
4

29. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
There are two basic approaches.
Il. We may dissect R into two
rectangles I and II as shown.
R
12
8
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
4
12
12
4 4
Hence the area of R is
R
144 – 32
= 112 m2
12
4
8
I II

30. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
There are two basic approaches.
Il. We may dissect R into two
rectangles I and II as shown.
R
12
8
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
4
12
12
4 4
Hence the area of R is
R
144 – 32
= 112 m2
12
4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.

31. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
There are two basic approaches.
Il. We may dissect R into two
rectangles I and II as shown.
R
12
8
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
4
12
12
4 4
Hence the area of R is
R
144 – 32
= 112 m2
12
4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.

32. Example C. Find the area of the following
shape R. Assume the unit is meter.
Area
12
12
44
There are two basic approaches.
Il. We may dissect R into two
rectangles I and II as shown.
R
12
8
I. We may view R as a
12 x 12 = 144 m2 square
with a 4 x 8 = 32 m2
corner removed.
4
12
12
4 4
Hence the area of R is
R
144 – 32
= 112 m2
12
4
8
I II
Area of I is 12 x 8 = 96,
area of II is 4 x 4 = 16.
The area of R is the sum of the
two or 96 + 16 = 112 m2.
12
12
4 4
8
iii
iv
(We may also cut R
into iii and iv as shown

33. b. Find the area of the following shape R.
Area
2 ft
Let’s cut R into three rectangles
as shown.

34. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
2 ftI
II
III

35. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
III

36. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,

37. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.

38. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.

39. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
6 ft
4 ft
25 ft
20 ft

40. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
6 ft
4 ft
25 ft
The area of the larger strip is 25 x 6 = 150 ft2
20 ft

41. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
6 ft
4 ft
25 ft
The area of the larger strip is 25 x 6 = 150 ft2
and the area of the smaller strip is 20 x 4 = 80 ft2.
20 ft

42. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
6 ft
4 ft
25 ft
The area of the larger strip is 25 x 6 = 150 ft2
and the area of the smaller strip is 20 x 4 = 80 ft2.
Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2)
rectangular overlap twice.
20 ft

43. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
The area of the larger strip is 25 x 6 = 150 ft2
and the area of the smaller strip is 20 x 4 = 80 ft2.
Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2)
rectangular overlap twice. Hence the total area covered is
230 – 24 = 206 ft2.
6 ft
4 ft
25 ft
20 ft

44. Area
b. Find the area of the following shape R.
Let’s cut R into three rectangles
as shown.
Area of I is 2 x 2 = 4,
2 ftI
II
IIIarea of II is 2 x 6 = 12,
and area of III is 2 x 5 = 10.
Hence the total area is 4 + 12 + 10 = 16 ft2.
c. Two rectangular walk-ways crosses each
other perpendicularly as shown. The larger
one is 25 ft x 6 ft and the smaller one is
20 ft x 4 ft. What is the total area they cover?
The area of the larger strip is 25 x 6 = 150 ft2
and the area of the smaller strip is 20 x 4 = 80 ft2.
Their sum 150 + 80 = 230 includes the 4 ft x 6 ft (= 24 ft2)
rectangular overlap twice. Hence the total area covered is
230 – 24 = 206 ft2.
6 ft
4 ft
25 ft
20 ft

45. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
2nd-Degree-Equation Word Problems
Area formulas lead to 2nd degree equations.

46. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
2nd-Degree-Equation Word Problems
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

47. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
2nd-Degree-Equation Word Problems
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

48. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
2nd-Degree-Equation Word Problems
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

49. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
2nd-Degree-Equation Word Problems
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

50. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

51. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. H=height
B=base
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

52. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown, we
get a rectangle
H=height
B=base
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

53. Example D. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines. If we
move the shaded part as shown, we
get a rectangle. Hence the area A
of the parallelogram is A = BH.
H=height
B=base
x + 4
x A = LW
Area formulas lead to 2nd degree equations.

54. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
2nd-Degree-Equation Word Problems

55. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
2nd-Degree-Equation Word Problems
2x + 3
x

56. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
The area is (base)(height) so that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x

57. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
The area is (base)(height) so that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0

58. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
The area is (base)(height) so that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2 so x = 3 ft

59. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
The area is (base)(height) so that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown,
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2 so x = 3 ft

60. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
The area is (base)(height) so that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown,
taking another copy and place it above the original one,
we obtain a parallelogram.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2 so x = 3 ft

61. Example E.
The area of the parallelogram shown is 27 ft2. Find x.
The area is (base)(height) so that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown,
taking another copy and place it above the original one,
we obtain a parallelogram.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2 so x = 3 ft
Hence the area A of the triangle
satisfies 2A = HB or that
A = BH
2

62. Example F. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find its base and
height.
2nd-Degree-Equation Word Problems

63. Example F. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find its base and
height.
Let x = height, then the base = (2x – 3)
2nd-Degree-Equation Word Problems
2x– 3
x

64. Example F. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find its base and
height.
Let x = height, then the base = (2x – 3)
using the formula 2A = BH
2*10 = (2x – 3) x
2nd-Degree-Equation Word Problems
2x– 3
x

65. Example F. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find its base and
height.
Let x = height, then the base = (2x – 3)
using the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
2nd-Degree-Equation Word Problems
2x– 3
x

66. Example F. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find its base and
height.
Let x = height, then the base = (2x – 3)
using the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
Therefore the height is 4 in. and the base is 5 in.
2nd-Degree-Equation Word Problems
2x– 3
x

67. Example F. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find its base and
height.
Let x = height, then the base = (2x – 3)
using the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
Therefore the height is 4 in. and the base is 5 in.
2nd-Degree-Equation Word Problems
2x– 3
x
r
A = πr2 The area A of the circle with radius r
is given by A = πr2 (unit2)
where π ≈ 3.1415... So the circle with radius
r = 1 m covers an area A ≈ 3.14 m2.
Areas of Circles

68. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.

69. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
r = ?
50 m2
C = ?

70. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
The area is πr2 = A = 50 so r2 = 50/π
and the radius r = √50/π.r = ?
50 m2
C = ?

71. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
The area is πr2 = A = 50 so r2 = 50/π
and the radius r = √50/π.
Therefore its circumference is
C = 2π√50/π = 2√50π.
r = ?
50 m2
C = ?

72. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
The area is πr2 = A = 50 so r2 = 50/π
and the radius r = √50/π.
Therefore its circumference is
C = 2π√50/π = 2√50π.
The volume of an object measures the 3-dimensional space
the object occupies.
Volume
r = ?
50 m2
C = ?

73. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
The area is πr2 = A = 50 so r2 = 50/π
and the radius r = √50/π.
Therefore its circumference is
C = 2π√50/π = 2√50π.
The volume of an object measures the 3-dimensional space
the object occupies. The volume V of a box
With length = l, width = w, height = h
is given as V = l x w x h (unit3)
h
w
Volume
r = ?
50 m2
l
C = ?

74. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
The area is πr2 = A = 50 so r2 = 50/π
and the radius r = √50/π.
Therefore its circumference is
C = 2π√50/π = 2√50π.
The volume of an object measures the 3-dimensional space
the object occupies. The volume V of a box
With length = l, width = w, height = h
is given as V = l x w x h (unit3)
h
w
Volume
r = ?
50 m2
1m x 1m x 1m cube has V = 1 m3
1 m3
1 cm3
l
C = ?

75. Length and Perimeter
Example G. The area of a circle is 50 m2
find its radius and circumference.
The area is πr2 = A = 50 so r2 = 50/π
and the radius r = √50/π.
Therefore its circumference is
C = 2π√50/π = 2√50π.
The volume of an object measures the 3-dimensional space
the object occupies. The volume V of a box
With length = l, width = w, height = h
is given as V = l x w x h (unit3)
h
w
Volume
r = ?
50 m2
1m x 1m x 1m cube has V = 1 m3
2m x 2m x 2m cube has V = 8m3
and the s x s x s cube has
V = s3 (unit3)
1 m3
1,000,000 cm3
= 1 m3
l
C = ?
1 cm3

76. Length and Perimeter
Here are some important volume formulas.
l
h
w
V = l x w x h
r
h
Box Cylinder
V = πr2h/3
r
h
Cone
r
Sphere
V = 4πr3/3
V = πr2h

77. Length and Perimeter
Here are some important volume formulas.
l
h
w
V = l x w x h
r
h
Box Cylinder
V = πr2h/3
r
h
Cone
r
Sphere
V = 4πr3/3
V = πr2h
Example H. The circumference of a giant
chocolate ball is 10 ft. What is its volume?
10 ft

78. Length and Perimeter
Here are some important volume formulas.
l
h
w
V = l x w x h
r
h
Box Cylinder
V = πr2h/3
r
h
Cone
r
Sphere
V = 4πr3/3
V = πr2h
Example H. The circumference of a giant
chocolate ball is 10 ft. What is its volume?
The circumference C = 2πr = 10 so r = 5/π.
10 ft

79. Length and Perimeter
Here are some important volume formulas.
l
h
w
V = l x w x h
r
h
Box Cylinder
V = πr2h/3
r
h
Cone
r
Sphere
V = 4πr3/3
V = πr2h
Example H. The circumference of a giant
chocolate ball is 10 ft. What is its volume?
The circumference C = 2πr = 10 so r = 5/π.
Its volume is 4π(5/π)3/3 = 500/(3π2) ≈ 16.9ft3
10 ft

80. Area
The area of the parallelogram is
A = b x h where
b = base and h = height.
h
b
The area A of the rectangle is
A = h x w (unit2) where
w = width (units) and h = height (units)
h
w
h
b
The area of a triangle is
b x h
2
A = (b x h) ÷ 2 or A =
where b = base and h = height.
b
h
a The area of a trapezoid is
A = (a + b)h where
a, b = lengths of two parallel sides
h = height.

81. Exercise A. Use the formula h = –16t2 + vt for the following
problems.
2nd-Degree-Equation Word Problems
1. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it’s height to reach 48 ft?
2. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it’s height to reach 28 ft?
3. A stone is thrown upward at a speed of v = 96 ft/sec,
a. how long does it take for its height to reach 80 ft? Draw a
picture.
b. how long does it take for its height to reach the highest
point?
c. What is the maximum height it reached?
4. A stone is thrown upward at a speed of v = 128 ft/sec,
a. how long does it take for its height to reach 256 ft?
Draw a picture. How long does it take for its height to reach
the highest point and what is the maximum height it reached?

82. B. Find the areas of the following regions.
2nd-Degree-Equation Word Problems
5.
2 ft
2 ft6 ft
4ft
6.
2 ft
6 ft
6 ft
4ft
7. 3 ft
3 ft
3 ft
10 m
2 m3 m
8 m
12 m2 m
Given the following rectangles find the
following covered areas.
4 m
2 m
2 m
8.
10.
9.
11. 12.

83. C. Given the following area measurements, find x’s.
2nd-Degree-Equation Word Problems
13.
8 ft2
x + 2
x
14.
12 ft2
x
(x – 1)
15.
x + 2
16.
12 ft2 x
(x + 4)
17.
24 ft2
(3x – 1)
x
18.
15 ft2x
18 ft2 x
(4x + 1)

84. C. Given the following area measurements, find x’s.
2nd-Degree-Equation Word Problems
2x + 1
19.
x5cm2
20.
x
2x – 3
9cm2
2x + 1
21.
x
18km2
22.
(x + 3)
(5x + 3)
24km2
23. 24.
16km2
2
x
x + 1
35km2
2
x
2x – 1

85. Length and Perimeter
D. Find the following volumes.
5 m
10 m
3 m
7 m
2 m
5 ft
2 ft
25. 26. 27.
28. The circumference of the earth is approximately
40,000 km. Estimated its volume using this.
x
x
29. Given the following silo with
equal radius and height, and whose
circumference is 10 ft. Find its volume.
10 ft.
30. A silo with equal radius and height (as in 29)
has volume of 100 ft3 find its radius and circumference.