Math 152A
Steps for Solving Application Problems:
1. Read, understand the question
2. Assign a variable (What is it asking
for?)
3. Write an equation
4. Solve the equation
5. Check, does it make sense?
Chapter 5.8: Solving Applications by Factoring
Objectives:
• Finding unknown sides of rectangles
• Finding widths of borders
• Finding unknown sides of right triangles
• Projectiles
Finding Unknown Sides of Rectangles
Area of a rectangle formula: area = (length)(width)
Ex: The length of a rectangle is 7 feet more than twice the width. The rectangle has an area of 85
square feet. Find the length and width of the rectangle.
Let
Then
A,
9P'(^:
h<V^ =
Equation:
the width times the length
Answer:
5 feet = the width
17 feet = the length
area of the rectangle
)U^l^iT
<^ifw cr\
(3X+i
O
You try:
'^^"^''^ X*5
1. The width of a rectangle is 9 feet less than 5 times le length. The rectangle has an area of 248
square feet. Find the length and width of the rectangle.
Let
Then
= \€VXM>
fe^A-A.^
J2fc3__='jMV^Jr
Equation:
the width times the length
area of the rectangle
:.0
Finding Widths of Borders
Ex: A 2 yard by 2 yard rug was placed directly in the center of a square room, and did not cover
140 square yards. What are the dimensions of the room?
•
l + lX
\
X
Then
— X — • 2yds — X — •
2vds
9t^A-
^ knjh
Equation:
t
X
1
die
the area of the room
area of the rug
area not covered
Answer: 5 yards = the widtlfofthe border
12 yards = the width and length of the room
You try:
1. A rectangular garden measures 2 yards by 4 yards. The gardener wants to place a border of
bark of width x around the garden. He has enough bark to cover 72 square yards. How wide
of a border should be made so that the border uses all the bark?
2 yards
•4
N
4 vaids
the area of the garden and border
Answer: 3 yards = the width of the border
Equation:
area of the garden
area of the bark
Application Problems Involving Right Triangles
Ex: A ladder is leaning against a building. The distance of the ground between the bottom of the
ladder to the building is 4 feet less than the length of the ladder. Also, the ladder reaches up the
wall 2 feet less than the length of the ladder. Find the length of the ladder, distance of the ground
between the bottom of the ladder to the building, and distance the ladder reaches up the wall.
Let X - ^ e f
Then
Then
-jh^ \trkkc
(§)
x-"2
Equation:
(legr
(other leg)-
(hypotenuse)*
Wal
Answer:
>^ 1x3^
10 feet the length of the ladder
6 feet = distance of the ground between the bottom of the ladder to the building
8 feet = distance the ladder reaches up the wall
You try:
1. A tree is supported by ropes. One rope goes from the top of the tree to a point on the ground.
The height of the tree is 1 foot more than the distance between the base of the tree and the rope
anchored in the ground. The length of the rope is 1 foot less than twice the distance between
the base of the tree and the rope anchored in the ground. Find the distance between the base of
the tree and the rope anchored in the ground, the height o|"the tree, and the length of the rope.
Then^X-H
Then .^IX-T
= miM
A \<n
yr\ (j
Equation:
(other leg)
%
(hypolenuse)''
X*+X"-»-^x+\ '^X^-VX+l
5lA^..^^y=Mx!-^ix+y
-ax^ '3iK/\
X=3
Answer: 3 feet = the distance between the base of thelfeeana the rope anchored in the ground
4 feet = the height of the tree
5 feet = the length of the rope
Ex: The equation for the height of a ball thrown, from a 200 foot cUff, into the air at 40 feet per
second is h{t) = -I6t^ +40/^ + 200, where/z(0 is the height of the ball after / seconds,
a.) Calculate the time it takes for the ball to be 176 feet above the ground.
Equation:
helghi
200 ft
176 ft
equation
/=0
Answer:
3 seconds = time it takes to 176 feet
et above the ground
b.) Calculate the seconds it takes for the ball to hit the ground.
Equation:
O
^"t^-HtHt"
C ^
YC/'^L^'^^N ^
= ntf^-h^Dtt^CC
height
200 ft
Q
equation
t=7
0= (.^t"+^5t)+Mob-J^S)
Answer:
5 seconds = time it takes to hit the ground ^ _ -^^^^ tS) " 5 ( 2"t.
You try:
1. The equation for the height of a ball thrown, from a 160 footfeliff,TnToth#%if aP48 feet per
second is h(t) = -16?" +4& +160, where/?(0 is the height of the ball after / seconds,
a.) Calculate the time it takes for the ball to be 192 feet above the ground.
= ^^t^fct
Equation:
equation
0= t^-5t,^^.
Answer:
1 and 2 seconds = time it takes to be 192 feet above the ground
b.) Calculate the seconds it takes for the ball to hit the ground.
Equation:
160 ft
Answer:
f
= " / f t ^ ^^P^t
height
5 seconds = time it takes to hit the ground \
equation
W^O
)