Transcript 1. Find the discounted price of a tent with a price of $89 and a discount of 15%.
Slide 1
1. Find the discounted price of a tent with a price
of $89 and a discount of 15%. $75.65
2. Find the final price of a pair of hiking boots
with a price of $78, a discount of 10%, and a
tax of 6%. $74.41
3. On September 1, a stock sold for $46 per share
and on October 1 it sold for $48.30 per share.
What was the percent of change in the price of
the stock? 5% increase
Slide 2
Math 8H
Problem Solving Day 3
Rectangle/Frame Problems
Algebra 1
Glencoe McGraw-Hill
JoAnn Evans
Slide 3
Let Statements:
Let x = width of the pool
Let x + 4 = length of the pool
Let x + 6 = total length
Let x + 2 = total width
1 + x + 1
A rectangular swimming pool is 4 meters longer than
it is wide. The pool is surrounded by a cement
sidewalk that is 1 meter wide. The area of the
sidewalk is 32 m2. Find the dimensions of the pool.
x
x + 4
x + 4
x
1 + (x + 4) + 1
Verbal Sentence:
Area of pool + area of sidewalk = total area
x (x + 4) + 32
= (x + 6) (x + 2)
Slide 4
Equation:
x (x + 4) + 32 = (x + 6) (x + 2)
x2 + 4x + 32 = x2 + 2x + 6x + 12
Solution:
4x + 32
-4x
32
-12
20
5
= 8x + 12
-4x
= 4x + 12
-12
= 4x
= x
The pool is 5 meters wide and 9 meters long.
Slide 5
A painting is 10 cm longer than it is wide. It is
mounted in a frame that is 1.5 cm wide. The
area of the frame is 339 cm2. Find the
dimensions of the painting.
x
1.5 + (x+10) + 1.5
x + 10
Let Statements:
Let x = width of the painting
Let x + 10 = length of the painting
Let x + 3 = width of framed painting
Let x + 13 = length of framed painting
1.5 + x + 1.5
Verbal Sentence:
Area of painting + area of frame = total area
x ( x + 10) +
339
= (x + 3) (x + 13)
Slide 6
Equation:
x ( x + 10) +
339 = (x + 3) (x + 13)
x2 + 10x + 339 = x2 + 13x + 3x + 39
Solution:
10x + 339
-10x
339
-39
300
50
= 16x + 39
-10x
= 6x + 39
-39
= 6x
= x
The painting is 50 cm wide and 60 cm long.
Slide 7
A rectangle is 4 meters longer than it is wide. If
the length and width are both increased by 5
meters, the area is increased by 115 m2. Find the
original dimensions.
Let
Let
Let
Let
Let
Statements:
x = original width of rectangle
x + 4 = original length of rectangle
x + 5 = new rectangle width
x + 9 = new rectangle length
Verbal Sentence:
original area + 115
x
x + 4
x + 5
(x + 4) + 5
= new larger rectangle area
x (x + 4) + 115 = (x + 5) (x + 9)
Slide 8
Equation:
x (x + 4) + 115 = (x + 5) (x + 9)
x2 + 4x + 115 = x2 + 9x + 5x + 45
Solution:
4x + 115
-4x
115
-45
70
7
= 14x + 45
-4x
= 10x + 45
-45
= 10x
= x
The original dimensions were 7 m wide and 11 m long.
Slide 9
A Mini Cooper and a truck heading for San Pedro on the
same freeway left the Redwood Middle School parking
lot at the same time. The Mini Cooper, which drove 20
mi/h faster than the truck, arrived in San Pedro after
two hours. The truck arrived in San Pedro one hour later
than the car. Find the rate of the car.
RMS
distance = distance
San Pedro
Let Statements:
Let x = rate of Mini Cooper
Let x - 20 = rate of truck
Verbal Sentence:
distance Mini Cooper drives = distance truck drives
Slide 10
Equation:
Mini’s distance = Truck’s distance
rt
= rt
x(2)
=
(x – 20)3
2x = 3x – 60
-3x -3x
-x = -60
x = 60
Solution:
The car’s rate was 60 mi/h.
1. Find the discounted price of a tent with a price
of $89 and a discount of 15%. $75.65
2. Find the final price of a pair of hiking boots
with a price of $78, a discount of 10%, and a
tax of 6%. $74.41
3. On September 1, a stock sold for $46 per share
and on October 1 it sold for $48.30 per share.
What was the percent of change in the price of
the stock? 5% increase
Slide 2
Math 8H
Problem Solving Day 3
Rectangle/Frame Problems
Algebra 1
Glencoe McGraw-Hill
JoAnn Evans
Slide 3
Let Statements:
Let x = width of the pool
Let x + 4 = length of the pool
Let x + 6 = total length
Let x + 2 = total width
1 + x + 1
A rectangular swimming pool is 4 meters longer than
it is wide. The pool is surrounded by a cement
sidewalk that is 1 meter wide. The area of the
sidewalk is 32 m2. Find the dimensions of the pool.
x
x + 4
x + 4
x
1 + (x + 4) + 1
Verbal Sentence:
Area of pool + area of sidewalk = total area
x (x + 4) + 32
= (x + 6) (x + 2)
Slide 4
Equation:
x (x + 4) + 32 = (x + 6) (x + 2)
x2 + 4x + 32 = x2 + 2x + 6x + 12
Solution:
4x + 32
-4x
32
-12
20
5
= 8x + 12
-4x
= 4x + 12
-12
= 4x
= x
The pool is 5 meters wide and 9 meters long.
Slide 5
A painting is 10 cm longer than it is wide. It is
mounted in a frame that is 1.5 cm wide. The
area of the frame is 339 cm2. Find the
dimensions of the painting.
x
1.5 + (x+10) + 1.5
x + 10
Let Statements:
Let x = width of the painting
Let x + 10 = length of the painting
Let x + 3 = width of framed painting
Let x + 13 = length of framed painting
1.5 + x + 1.5
Verbal Sentence:
Area of painting + area of frame = total area
x ( x + 10) +
339
= (x + 3) (x + 13)
Slide 6
Equation:
x ( x + 10) +
339 = (x + 3) (x + 13)
x2 + 10x + 339 = x2 + 13x + 3x + 39
Solution:
10x + 339
-10x
339
-39
300
50
= 16x + 39
-10x
= 6x + 39
-39
= 6x
= x
The painting is 50 cm wide and 60 cm long.
Slide 7
A rectangle is 4 meters longer than it is wide. If
the length and width are both increased by 5
meters, the area is increased by 115 m2. Find the
original dimensions.
Let
Let
Let
Let
Let
Statements:
x = original width of rectangle
x + 4 = original length of rectangle
x + 5 = new rectangle width
x + 9 = new rectangle length
Verbal Sentence:
original area + 115
x
x + 4
x + 5
(x + 4) + 5
= new larger rectangle area
x (x + 4) + 115 = (x + 5) (x + 9)
Slide 8
Equation:
x (x + 4) + 115 = (x + 5) (x + 9)
x2 + 4x + 115 = x2 + 9x + 5x + 45
Solution:
4x + 115
-4x
115
-45
70
7
= 14x + 45
-4x
= 10x + 45
-45
= 10x
= x
The original dimensions were 7 m wide and 11 m long.
Slide 9
A Mini Cooper and a truck heading for San Pedro on the
same freeway left the Redwood Middle School parking
lot at the same time. The Mini Cooper, which drove 20
mi/h faster than the truck, arrived in San Pedro after
two hours. The truck arrived in San Pedro one hour later
than the car. Find the rate of the car.
RMS
distance = distance
San Pedro
Let Statements:
Let x = rate of Mini Cooper
Let x - 20 = rate of truck
Verbal Sentence:
distance Mini Cooper drives = distance truck drives
Slide 10
Equation:
Mini’s distance = Truck’s distance
rt
= rt
x(2)
=
(x – 20)3
2x = 3x – 60
-3x -3x
-x = -60
x = 60
Solution:
The car’s rate was 60 mi/h.