9.7 Rectangles PDF
9.7 Rectangles PDF
9.7 Rectangles PDF
Quadratics - Rectangles
An application of solving quadratic equations comes from the formula for the area
of a rectangle. The area of a rectangle can be calculated by multiplying the width
by the length. To solve problems with rectangles we will first draw a picture to
represent the problem and use the picture to set up our equation.
Example 1.
The length of a rectangle is 3 more than the width. If the area is 40 square
inches, what are the dimensions?
The above rectangle problem is very simple as there is only one rectangle
involved. When we compare two rectangles, we may have to get a bit more cre-
ative.
Example 2.
If each side of a square is increased by 6, the area is multiplied by 16. Find the
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side of the original square.
Example 3.
2
6x + 21 = 33 Subtract 21 from both sides
− 21 − 21
6x = 12 Divide both sides by 6
6 6
x=2 x is the width of the original
(2) + 4 = 6 x + 4 is the length. Substitute 2 to find
2 ft by 6ft Our Solution
From one rectangle we can find two equations. Perimeter is found by adding all
the sides of a polygon together. A rectangle has two widths and two lengths, both
the same size. So we can use the equation P = 2l + 2w (twice the length plus
twice the width).
Example 4.
The area of a rectangle is 168 cm2. The perimeter of the same rectangle is 52 cm.
What are the dimensions of the rectangle?
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Both are the same rectangle, variables switched!
12 cm by 14cm Our Solution
World View Note: Indian mathematical records from the 9th century demon-
strate that their civilization had worked extensivly in geometry creating religious
alters of various shapes including rectangles.
Another type of rectangle problem is what we will call a “frame problem”. The
idea behind a frame problem is that a rectangle, such as a photograph, is centered
inside another rectangle, such as a frame. In these cases it will be important to
rememember that the frame extends on all sides of the rectangle. This is shown in
the following example.
Example 5.
An 8 in by 12 in picture has a frame of uniform width around it. The area of the
frame is equal to the area of the picture. What is the width of the frame?
8
Draw picture, picture if 8 by 10
12 12 + 2x
If frame has width x, on both sides, we add 2x
8 + 2x
8 · 12 = 96 Area of the picture, length times width
2 · 96 = 192 Frame is the same as the picture. Total area is double this.
(12 + 2x)(8 + 2x) = 192 Area of everything, length times width
96 + 24x + 16x + 4x2 = 192 FOIL
4x2 + 40x + 96 = 192 Combine like terms
− 192 − 192 Make equation equal to zero by subtracting 192
2
4x + 40x − 96 = 0 Factor out GCF of 4
4(x2 + 10x − 24) = 0 Factor trinomial
4(x − 2)(x + 12) = 0 Set each factor equal to zero
x − 2 = 0 or x + 12 = 0 Solve each equation
+2+2 − 12 − 12
x = 2 or − 12 Can ′t have negative frame width.
2 inches Our Solution
Example 6.
A farmer has a field that is 400 rods by 200 rods. He is mowing the field in a
spiral pattern, starting from the outside and working in towards the center. After
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an hour of work, 72% of the field is left uncut. What is the size of the ring cut
around the outside?
400 − 2x
Draw picture, outside is 200 by 400
200 − 2x 200 If frame has width x on both sides,
subtract 2x from each side to get center
400
400 · 200 = 80000 Area of entire field, length times width
80000 · (0.72) = 57600 Area of center, multiply by 28% as decimal
(400 − 2x)(200 − 2x) = 57600 Area of center, length times width
80000 − 800x − 400x + 4x2 = 57600 FOIL
4x2 − 1200x + 80000 = 57600 Combine like terms
− 57600 − 57600 Make equation equal zero
4x2 − 1200x + 22400 = 0 Factor out GCF of 4
4(x2 − 300x + 5600) = 0 Factor trinomial
4(x − 280)(x − 20) = 0 Set each factor equal to zero
x − 280 = 0 or x − 20 = 0 Solve each equation
+ 280 + 280 + 20 + 20
x = 280 or 20 The field is only 200 rods wide,
Can ′t cut 280 off two sides!
20 rods Our Solution
For each of the frame problems above we could have also completed the square or
use the quadratic formula to solve the trinomials. Remember that completing the
square or the quadratic formula always will work when solving, however, factoring
only works if we can factor the trinomial.
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
5
should the dimensions of the rectangular bed be?
3) A rectangular lot is 20 yards longer than it is wide and its area is 2400 square
yards. Find the dimensions of the lot.
5) The length of a rectangular lot is 4 rods greater than its width, and its area is
60 square rods. Find the dimensions of the lot.
8) A room is one yard longer than it is wide. At 75c per sq. yd. a covering for
the floor costs S31.50. Find the dimensions of the floor.
9) The area of a rectangle is 48 ft2 and its perimeter is 32 ft. Find its length and
width.
11) A mirror 14 inches by 15 inches has a frame of uniform width. If the area of
the frame equals that of the mirror, what is the width of the frame.
12) A lawn is 60 ft by 80 ft. How wide a strip must be cut around it when
mowing the grass to have cut half of it.
13) A grass plot 9 yards long and 6 yards wide has a path of uniform width
around it. If the area of the path is equal to the area of the plot, determine
the width of the path.
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16) A picture 10 inches long by 8 inches wide has a frame whose area is one half
the area of the picture. What are the outside dimensions of the frame?
17) A rectangular wheat field is 80 rods long by 60 rods wide. A strip of uniform
width is cut around the field, so that half the grain is left standing in the form
of a rectangular plot. How wide is the strip that is cut?
19) A rectangular field 225 ft by 120 ft has a ring of uniform width cut around
the outside edge. The ring leaves 65% of the field uncut in the center. What
is the width of the ring?
20) One Saturday morning George goes out to cut his lot that is 100 ft by 120 ft.
He starts cutting around the outside boundary spiraling around towards the
center. By noon he has cut 60% of the lawn. What is the width of the ring
that he has cut?
22) A farmer has a field 180 ft by 240 ft. He wants to increase the area of the
field by 50% by cultivating a band of uniform width around the outside. How
wide a band should he cultivate?
23) The farmer in the previous problem has a neighber who has a field 325 ft by
420 ft. His neighbor wants to increase the size of his field by 20% by
cultivating a band of uniform width around the outside of his lot. How wide a
band should his neighbor cultivate?
24) A third farmer has a field that is 500 ft by 550 ft. He wants to increase his
field by 20%. How wide a ring should he cultivate around the outside of his
field?
25) Donna has a garden that is 30 ft by 36 ft. She wants to increase the size of
the garden by 40%. How wide a ring around the outside should she cultivate?
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
7
9.7
Answers - Rectangles
1) 6 m x 10 m 10) 1.54 in 19) 15 ft
2) 5 11) 3 in 20) 20 ft
3) 40 yd x 60 yd 12) 10 ft
21) 1.25 in
4) 10 ft x 18 ft 13) 1.5 yd
22) 23.16 ft
5) 6 x 10 14) 6 m x 8 m
23) 17.5 ft
6) 20 ft x 35 ft 15) 7 x 9
24) 25 ft
7) 6” x 6” 16) 1 in
8) 6 yd x 7 yd 17) 10 rods 25) 3 ft
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)