Areas of Polygons – Blue Problems
1. The lengths of the sides of an 8 cm x 8 cm square are increased by 2 cm
each. The area of the square has been increased by what percent? Express
your answer to the nearest whole percent.
2. For x > 0, the perimeter of the triangle with vertices (0, 0), (x, 0) and (x, 5)
is 30 units. What is the value of x?
3. The area of trapezoids ABEC, ABFD and ABFC are 133, 140 and 161 square
feet, respectively. Quadrilateral ABED is a rectangle. The length of
segment DE is 8 feet. In feet, what is the length of segment CF?
C
A
B
D
E
F
2mI
I
4. The area of a rectangular patio is 360 square meters. Surrounding the patio
on all four sides is a flower border 2 meters wide as shown in the diagram.
The border has an area of 184 square meters. What is the number of
meters
in
the
perimeter of
2m
flower
border
the patio?
PATIO
I
I 2m
2m
5. A square blanket measuring x feet by x feet was folded in half, folded in
half again and finally folded in half one last time. After these three
successive folds, without ever unfolding, the blanket covers an area of 8
square feet. What is the value of x?
6.
A rectangle has dimensions 20 inches by 30 inches. If one side is increased
by 30% and the other side is decreased by 20%, then what is the largest
possible number of square inches in the area of the new rectangle?
7. School portraits come in two sizes, 5” x 7” and 3” x 5”. The area of the
smaller portrait is what percent of the area of the larger portrait? Express
your answer to the nearest whole number.
8. A rectangular corral is to be built using 160 feet of fence. One side of the
corral will be part of a straight 100-foot wall of an adjacent building. What
is the maximum number of square feet possible for the area of the corral?
9. The area of ABC is 27 square feet. What is the number of square feet in
the area of the shaded triangle? Express your answer as a decimal to the
nearest tenth.
A
4
6
D
8
10
C
10. Challenge. You form a rhombus by putting two equilateral triangles with
side length 2n together, as shown. Write an expression for the area of the
rhombus in terms of n. Explain your reasoning.
30c
2n
11. Change in Area. The length of a rectangle is increased by 20 percent, and
its width is decreased by 10 percent. By what percent does the area of the
rectangle change? What if, instead, we decreased the length by 20 percent
and increased the width by 10 percent?
12. Rectangle Building. Bobby was using 1260 wooden square tiles, one inch on a
side, to build rectangles. How many different perimeters can he get with
rectangles that he can build that use all of the tiles?
13. Squares Galore. Find the area of the rectangle in the margin. Everything
that looks like a square is a square, and the smallest square has an area of 1.
14. Tiling, tiling. A rectangular tile measures 2 inches x 3 inches. What is the
least number of tiles that are needed to completely cover a square region
two feet on a side?
15. Partition a Square. A square measuring 10 centimeters on a side is divided
by vertical and horizontal line segments into rectangles with areas of 12, 18,
28, and 42 square centimeters. Where should the vertical and horizontal
lines be located? Draw a diagram.
16. Each small square in the figure to the right has an
area of 6 cm 2 . What is the number of square
centimeters in the area of the shaded region?
The isosceles trapezoid shown has area 24 cm 2 . How many
centimeters are in the length of the shorter base?
17.
8 cm
4 cm
18. A rectangle has integer side lengths and its area is equal to 24 square units.
The length of each side of the rectangle is increased by one unit. What is
the largest possible number of square units in the area of the new
rectangle?
19. Office Tile Choice. I need help in determining which tiles I should
purchase to cover the floor in my office.
I have found two patterns that I particularly like. The first pattern has
light green triangles on a dark green background. These 3-inch by 3-inch
tiles come boxed in packs of 75 at a cost of $55.21 per pack.
The second pattern has a dark green diamond on a light green background.
These 9-inch tiles come 12 tiles to a pack and cost $75.89 per package.
My office measures fifteen and a half feet by twelve and a quarter feet.
That means that some of the tiles will need to be cut. When a tile is cut,
only one portion of that tile may be used. There will be some wasted tile
pieces, but the pattern on the floor will match.
Since I can’t decide which pattern I prefer, my husband suggested we go
with less expensive tile.
a. Which tiles are less expensive?
In your explanation be sure to include the following information:
•
•
•
•
The number of 3-inch tiles needed to cover the floor.
The total cost of the 3-inch tiles.
The number of 9-inch tiles to cover the floor.
The total cost of the 9-inch tiles.
b. To have the flooring installed, the cost is $175 plus seventeen cents per
installed tile. There is an additional charge of $2.95 for each tile that
must be cut. After considering the installation costs, will my flooring
choice need to change? What is the total cost of the less expensive
flooring, including installation?
20. The school administration has found that students like to take a shortcut
across a grassy field that is 35 feet x 40 feet. They have decided to pave a
4 foot wide walkway along the shortcut the students take.
Here is a picture of the field and the proposed walkway.
35 feet
4 feet
WALKWAY
40 feet
30 feet
Determine the total area of the walkway. Descibe your strategy and show
all your calculations.
21. To solve the four-square puzzle at the right, you must reason logically about
the way squares fit together.
y
5
8
x
a. Use the given measurements to find
b. Find
c.
x.
y.
How can you find the area of the four-square puzzle?
Now apply this method to the figure below, which consists of ten squares
fitted together to form a rectangle. The length of the sides of two of the
squares are given. Write the dimensions of each of the remaining squares in
the given square. Write the dimensions and area of the rectangle in the
space below the figure.
12x12
7x7
d. Dimensions of rectangle: _____ Area: _____
22. A rectangular pool measuring 6 feet by 12 feet is surrounded by a walkway.
The width of the walkway is the same on all four sides of the pool. If the
total area of the walkway and pool is 520 square feet, what is the number
of feet in the width of the walkway?
23. Area to Length. The ares of three faces of a rectangular solid are 54, 72,
and 108 square units. Find the whole number measures of the edges.
Areas of Polygons – Blue Solutions
1. 56
2. We don’t know the value of x, but we can still graph
this triangle; we just don’t know how far the
(x, 5)
triangle actually stretches from left to right. From
c
the coordinates we do know it is a right triangle.
5
(0,
0)
The horizontal side is x units and the vertical side
x
is 5 units. Because this is a right triangle, we can
(x, 0)
determine the length of the hypotenuse, c, in terms
of x, with the help of the Pythagorean Theorem: c 2 = x 2 + 5 2 so c =
x 2 + 25 . The perimeter is 30 units, therefore:
x+5+
x 2 + 25 = 30;
x 2 + 25 = 25 – x; x 2 - 25 = 625 – 50x + x 2 ; 50x =
600;
x = 12. Notice this triangle has the rare quality that the value of its
perimeter is equal to the value of its area.
3. If we add the areas of trapezoids ABEC and ABFD, we will have all of the
area of trapezoid ABFC plus the area of rectangle ABED, which was
included twice. This means that 133 + 140 – 161 = 112 square feet must be
the area of rectangle ABED. Dividing 112 square feet by the length of base
DE, which is 8 feet, we get 14 feet, which is the height of the rectangle and
all the trapezoids. We know that the area of trapezoid ABFC is 161 square
feet and we know that AB measures 8 feet since DE = 8. To find the length
of base CF, let CF = b and use the area formula of trapezoid to solve for
the length of b as follows:
161 = 1 x 14 x (8 + b); 161 = 7 x (8 + b); 23 = 8 + b; 15 = b.
2
The length of segment CF is 15 feet.
4. At every corner of the patio, there is a 2 meter by 2 meter square portion
of flower bed that does not directly abut the perimeter of the patio. This
amounts to 4 square meters of flower bed at each of four corners for a
total of 16 square meters. If we subtract 16 square meters from the 184
square meters of flower bed, we have 168 square meters of flower bed
remaining. This is exactly double the perimeter of the patio, since the
border is 2 meters wide. The perimeter of the patio is thus 168 ' 2 = 84
meters.
5. If 8 square feet is half of half of half of the area of the blanket, then the
area of the blanket is 8 x 2 x 2 x 2 = 64 square feet. The side length x,
must be square root of 64, which is 8 feet.
6. Whether we compute (0.8 x 20) x (1.3 x 20) or (1.3 x 20) x (0.8 x 30) the
result is 624 square inches.
7. The area of the smaller portrait is 3 x 5 = 15 in 2 , and the area of the larger
portrait is 5 x 7 = 35 in 2 . The smaller portrait is 15 = 3 = 43% of the
7
35
larger one.
8. The 160 feet of fence must be used for three sides of the corral. It makes
sense that the side parallel to the barn is longer than the other two sides,
because the side of the barn acts as the fourth side of the corral but uses
no fencing. Trial and error shows that the maximum area for the corral
occurs when the side parallel to the barn is twice the length of the shorter
two sides. That is, use one-half the fencing for this side, and divide the
other half between the other two sides. That gives dimensions 40 x 80, and
the area is 3200 ft 2.
9.
ABC has area 27 and base 18. Label the unnamed vertices of the shaded
triangle D and E. Draw the altitudes for both triangles, and label the points
of intersection with their bases F and G. In
ABC, since the area is 27,
and area is half the height times the base, then 27 = 1 (18)(AG), so AG = 3
2
A
units.
4
6
B
E
8 F D
3
G
C
The height of
BED is unknown. The two triangles are not similar, so we
can’t compare their sides. However
BEF and
BAG are similar. Their
corresponding sides and heights will be in the same ratio, and EF can be
found. Compare the ratio of AG to EF.
10 = 3 Æ EF = 1.8.
EF
6
Since the altitude of
BED is 1.8 and the base is 8, the area is 1 (8)(1.8) =
2
7.2 square feet.
10. 11. Change in Area.
Eight percent increase in the first case and 12 percent decrease in the
second case. Let L represent the length and W represent the width, so LW
represents the area. If L is increased by 20 percent, it becomes 1.2L. If W
is decreased by 10 percent, it becomes 0.9W. The area of the new
rectangle is (1.2L)(0.9W) = 1.08 LW, which is an 8 percent increase over the
original LW. In the second case, we have a new area of (0.8L)(1.1W) = 0.88
LW, which accounts for a 12 percent decrease from LW.
12. Rectangle Building.
Eighteen different perimeters. List all factor pairs, using prime
h h
factorization of 1260 = 2 2 3 2 (5)(7). The list is the following: (1, 1260), (2,
630), (3, 420), (4, 315), (5, 252), (6, 210), (7, 180), (9, 140), (10, 126), (12,
105), (14, 90), (15, 84), (18, 70), (20, 63), (21, 60), (28, 45), (30, 42), (35,
36). Corresponding to the pairs are the perimeters: 2552, 1264, 846, 638,
514, 432, 374, 298, 272, 234, 208, 198, 176, 166, 162, 146, 144, 142.
13. Squares Galore.
1870 square units. The areas of squares in increasing sequence are given by
Fibonacchi sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, …. The rectangle is 34 x 55.
14. Tiling, tiling.
Ninety-six tiles. The square region is 24 inches x 24 inches, and its area is
576 square inches. Each small tile has an area of 6 square inches, so there
must be at least 576 = 96 tiles. Note that the tiles can be put in a 12 x 8
6
array to cover the square.
15. Partition a Square.
Divide the length or width into 6 cm and 4 cm and the other dimension into
3 cm and 7 cm. The there will be a 3 x 6 rectangle, a 3 x 4 rectangle, a 7x 6
rectangle, and a 4 x 7 rectangle. (See margin.)
3 cm
7 cm
4 cm 12cm 2
28 cm 2
6 cm 18cm 2
42 cm 2
16. Sometimes the easiest way to find the answer to a problem is to find the
answer to a different, but related, problem first. In this case, it may be
easier to find the areas of the non-shaded regions and subtract that from
the entire region. These non-shaded regions will help us to answer the
question about the shaded region.
The area of the entire figure is 3 x 5 x 6 = 90 cm 2 . (Remember there are
just 15 square regions, but each of those has an area of 6 cm 2 .)
The area of the three non-shaded regions are:
the right triangle at the top: 0.5 x 5 x 2 x
I.
6 = 30 cm 2
the right triangle at the bottom, on the
II.
III
right:
0.5 x 1 x 3 x 6 = 9 cm 2
III. the region at the bottom, on the left
(one square + one half-square + one right triangle):
6 + 3 + 0.5 x 1 x 2 x 6 = 15 cm 2
I
II
Therefore, the area of the shaded region is 90 – 30 – 9 – 15 = 36 cm 2
17. The area of a trapezoid can be found by averaging the bases and multiplying
by the height. As a formula, A = 1 (b 1 + b 2 )h. In this case, b 1 = 4 cm, h = 8
2
cm, and A = 24 cm 2 , so 24 = 1 (4 + b 2 )(8). Solving by b 2 gives an answer of 2
2
cm.
18. The largest increase will come from the thinnest original rectangle. A
rectangle measuring 1 unit by 24 units, increased to 2 units by 25 units will
have an area of 50 square units.
19. Office Tile Choice.
a. The first pattern (3” x 3” tiles) is the cheapest to buy.
First I converted the room size from feet to inches. There are 12
inches in 1 foot.
15 1 ’ = 186” x 12 1 ’ = 147”
2
4
Second, take the room size in inches and figure out how many tiles will
be needed:
186” divided by 3” = 62 tiles
147” divided by 3” = 49 tiles
Take 62 tiles x 49 tiles = 3038 tiles total needed.
Third, take the total tiles, 3038 and divide by the pack size of 75 = 40
full boxes and 38 loose tiles. Since the tiles are sold in full boxes, we
must purchase 41.
Fourth, take 41 total boxes x the price per box, $55.21 = $2263. 61
Total cost for the 3” x 3” tiles is $2263.61
Now, let’s figure out the 9” x 9” tile job
First, take the room size in inches and figure out how many tiles will be
needed:
186” divided by 9” = 20.6 tiles which is 21 tiles
147” divided by 9” = 16.3 tiles which is 17 tiles
Second, take 21 tiles x 17 tiles = 357 tiles needed for job
Third, take the total tiles needed, 357, and divide into pack size of 12,
= 29 full ctns and 9 loose tiles. Since they do not sell loose tiles, we
must buy 30 full ctns.
Fourth, take the 30 full ctns and multiply by case cost, $75.89 =
$2276.70
The 3” tiles job will cost $2263.61 for supplies and the 9” tile job will
cost $2276.70 for supplies. Therefore, the 3” tiles are cheaper.
b. The installation cost is a flat $175 per tile job. It will cost 17 cents per
tile to install and $2.95 per tile to be cut.
First, on the 3” tile job – there were 3038 tiles needed at 17 cents
each = $516.46. No tiles were needed to be cut on this job so the total
installation cost is $175.00 plus $516.46 = $691.46 plus $2263.61 for
the cost of the tiles = $2955.07 for the total job.
Second, on the 9” tile job – there were 357 tiles needed at 17 cents
each = $60.69. There were 2 tiles needed to be cut at $2.95 each =
$5.90. Total installation cost is $175.00 plus $60.69 plus $5.90 =
$241.59 plus $2276.70 for the cost of the tiles = $2518.29 for the
total job.
Bonus answer is the 9” tile becomes the cheapest with the installation
charges accrued.
20. The hypotenuse of the two large right triangles, each having sides of 30 ft
and 40 ft, is 50 ft. The small triangles formed at each end of the walkway
are 3-4-5 right triangles, so substracting 3 ft from each of the 50 ft sides
gives the dimension of the rectangle as 47 x 4, or 188 square feet. Each
small triangle is 6 square feet, so the sum of the component parts of the
walkway is 200 square feet.
It is possible to arrive at a solution without using the Pythagorean theorem.
The total area of the 35’ x 40’ rectangle is 1400 square feet. The area of
each of the right triangles may be found by calculating 1 bh, or 1 (30)(40),
2
2
for a total of 1200 square feet. Subtracting this from the whole area
leaves a walkway of 200 square feet.
Still another approach is to treat the walkway as a long, skinny
parallelogram with a base of 5 feet and a height of 40 feet, giving total
area of 200 square feet.
4
'
35
5
'
4
50
WALKWAY
40'
4
30
'
21. a. 13
b. 21
c. Add the area of the squares together.
d.
5
44 x 44
60 x 60
16x16
19x19
45 x 45
28 x 28
12x12
7x7
26 x 26
33 x 33
105 x 105; 11,025 units 2
22. Call the width of the walkway x. Then, the width of the pool and walk
together is 6 + 2 x, and the length of the walk and pool together is 12 + 2 x.
Algebra could then be used to solve this, but a guess-and-check approach
might be faster.
Because 12 x 6 = 72, and that’s significantly less than 520, a first guess of
5 feet for the width of the walkway is reasonable. The width of the pool
and walk is then 6 + 2(5) = 16 feet, and the length of the pool and walk is 12
+ 2(5) = 22 feet, giving an area of 16 x 22 = 352 ft 2. That’s still low.
Revise the guess so the width of the walk is 7 feet. Then the total width is
20 feet, the total length is 26 feet, and the area is 520 ft 2.
23. Area to Length.
The edges measure 6, 9, and 12. Let the dimensions be L, W, H and LW =
108, LH = 54, and WH = 72. Then, LW = 108 = 2(54) = 2LH. So, W = 2H.
Substituting, we get 2HH = 72 and then H = 6. Substituting H = 6 into LH =
54, we get L = 9. Substituting L = 9 into LW = 108, we get W = 12. This can
also be done by trial and error, looking for common factors.