Find the area of each trapezoid, rhombus, or
kite.
1.
SOLUTION:
ANSWER:
2.
SOLUTION:
ANSWER:
3.
SOLUTION:
ANSWER:
4.SHORT RESPONSE Suki is doing fashion design
at 4-H Club. Her first project is to make a simple A-
line skirt. How much fabric will she need according
to the design at the right?
SOLUTION:
The area A of a trapezoid is A = h(b1 + b2).
She will need · 2 or squarefeetoffabric.
ANSWER:
ALGEBRA Find x.
5.
SOLUTION:
ANSWER:
8 cm
6.
SOLUTION:
ANSWER:
6.6 in.
7.
SOLUTION:
ANSWER:
6.3 ft
CCSSSTRUCTUREFind the area of each
trapezoid, rhombus, or kite.
8.
SOLUTION:
h = 13, b1 = 18 and b2 = 24
ANSWER:
9.
SOLUTION:
h = 23, b1 = 22 and b2 = 37
ANSWER:
10.
SOLUTION:
d1 = 22 and d2=24
ANSWER:
11.
SOLUTION:
d1 = 16 in. and d2=17in.
ANSWER:
12.
SOLUTION:
d1 = 7 cm. and d2 = 6 + 9 = 15 cm
ANSWER:
13.
SOLUTION:
d1 = 11 ft. and d2=25ft
ANSWER:
MICROSCOPES Find the area of the identified
portion of each magnified image. Assume that
the identified portion is either a trapezoid,
rhombus, or kite. Measures are provided in
microns.
14.human skin
Refer to the image on Page 777.
SOLUTION:
ANSWER:
26 square microns
15.heartleaf plant
RefertotheimageonPage777.
SOLUTION:
ANSWER:
24.5 square microns
16.eye of a fly
RefertotheimageonPage777.
SOLUTION:
Thefigureconsistsoftwotrapezoids.
ANSWER:
9.9 square microns
17.JOBS Jimmy works on his neighbors’yards after
school to earn extra money to buy a car. He is going
to plant grass seed in Mr. Troyer’s yard. What is the
area of the yard?
SOLUTION:
ANSWER:
ALGEBRA Find each missing length.
18.One diagonal of a kite is twice as long as the other
diagonal. If the area of the kite is 240 square inches,
what are the lengths of the diagonals?
SOLUTION:
The area A of a kite is one half the product of the
lengths of its diagonals, d1 and d2.
Therefore, the diagonals are of length 15.5 in. and
31.0 in.
ANSWER:
15.5 in., 31.0 in.
19.The area of a rhombus is 168 square centimeters. If
one diagonal is three times as long as the other, what
are the lengths of the diagonals?
SOLUTION:
The area A of a rhombus is one half the product of
the lengths of its diagonals, d1 and d2.
Therefore, the diagonals are of length 10.6 cm. and
31.7 cm.
ANSWER:
10.6 cm, 31.7 cm
20.A trapezoid has base lengths of 12 and 14 feet with
an area of 322 square feet. What is the height of the
trapezoid?
SOLUTION:
The area A of a trapezoid is one half the product of
the height h and the sum of the lengths of its bases,
b1 and b2.
.
ANSWER:
24.8 ft
21.A trapezoid has a height of 8 meters, a base length of
12 meters, and an area of 64 square meters. What is
the length of the other base?
SOLUTION:
The area A of a trapezoid is one half the product of
the height h and the sum of the lengths of its bases,
b1 and b2.
ANSWER:
4 m
22.HONORS Estella has been asked to join an honor
society at school. Before the first meeting, new
members are asked to sand and stain the front side
of a piece of wood in the shape of an isosceles
trapezoid. What is the surface area that Allison will
need to sand and stain?
SOLUTION:
The required area is the difference between the
larger trapezoid and the smaller trapezoid.
The area A of a trapezoid is one half the product of
the height h and the sum of the lengths of its bases,
b1 and b2.
ANSWER:
For each figure, provide a justification showing
that .
23.
SOLUTION:
Set the area of the kite equal to the sum of the areas
of the two triangles with bases d1 and d2.
ANSWER:
The area of andtheareaof
. Therefore, the area of
, and the area of .
The area of kite FGHJ is equal to the area of
the area of . After
simplification, the area of kite FGHJ is equal to
.
24.
SOLUTION:
Set the area of the kite equal to the sum of the areas
of the two triangles with bases d1 and d2.
ZWX = ZYX by SSS
ANSWER:
The area of and the area of
. Therefore, the area of
, and the area of .
The area of rhombus WXYZ is equal to the area of
the area of .
After simplification, the area of rhombus WXYZ is
equal to .
25.CRAFTS Ashanti is in a kite competition. The
yellow, red, orange, green, and blue pieces of her kite
design shown are all congruent rhombi.
a. How much fabric of each color does she need to
buy?
b. Competition rules require that the total area of
each kite be no greater than 200 square inches. Does
Ashanti’s kite meet this requirement? Explain.
SOLUTION:
The area of the yellow rhombus is or24
in2. Since the yellow, red , orange, green, and blue
pieces area all congruent rhombi, each have an area
of 24 in2.
The area of the purple kite-shaped piece is
or20in2.
The total area of the entire kite is 24(5) + 20 or 140
in2, which is less than the maximum 200 in2 allowed.
Therefore, her kite meets this requirement.
ANSWER:
a. each of yellow, red, orange, green, and
blue; of purple
b. Yes; her kite has an area of , which is less
than 200 in2.
CCSS SENSE-MAKING Find the area of each
quadrilateral with the given vertices.
26.A(–8, 6), B(–5, 8), C(–2, 6), and D(–5, 0)
SOLUTION:
Graph the quadrilateral.
The quadrilateral is a kite. The area A of a kite is one
half the product of the lengths of its diagonals, d1 and
d2. The lengths of the diagonals are 6 units and 8
units. Therefore, the area of the kite is
ANSWER:
24 sq. units
27.W(3, 0), X(0, 3), Y(–3, 0), and Z(0, –3)
SOLUTION:
Graph the quadrilateral.
The quadrilateral is a rhombus. The area A of a
rhombus is one half the product of the lengths of its
diagonals, d1 and d2.
The lengths of the diagonals are 6 units each.
Therefore, the area of the kite is
ANSWER:
18 sq. units
28.METALS When magnified in very powerful
microscopes, some metals are composed of grains
that have various polygonal shapes.
a. What is the area of figure 1 if the grain has a
height of 4 microns and bases with lengths of 5 and 6
microns?
b. If figure 2 has perpendicular diagonal lengths of
3.8 microns and 4.9 microns, what is the area of the
grain?
SOLUTION:
a. Figure1isatrapezoid.
The area of figure 1 is 22 square microns.
b. Figure2isarhombus.
The area of figure 2 is about 9.3 square microns.
ANSWER:
a. 22 square microns
b. 9.3 square microns
29.PROOF The figure at the right is a trapezoid that
consists of two congruent right triangles and an
isosceles triangle. In 1876, James A. Garfield, the
20th president of the United States, discovered a
proof of the Pythagorean Theorem using this
diagram.Provethat .
SOLUTION:
Following is an algebraic proof.
The trapezoid is on its side, so the bases are x and y
and the height is x + y. Set the area of the trapezoid
equal to the sum of the areas of the triangles and
simplify.
ANSWER:
The area of a trapezoid is So,
.
The area of . The area
of . Set the
area of the trapezoid equal to the combined areas of
the triangles to get . Multiply by
2 on each side: .
When simplified, .
DIMENSIONAL ANALYSIS Find the
perimeter and area of each figure in feet. Round
to the nearest tenth, if necessary.
30.
SOLUTION:
Both diagonals are perpendicular bisectors, so the
figure is a rhombus and all four triangles are
congruent. All of the sides are 12 feet, so the
perimeter is 48 feet.
Use trigonometry to find the lengths of the diagonals.
Now find the area.
ANSWER:
48 ft; 129.4 ft2
31.
SOLUTION:
Use the 30-60-90 triangle to find the dimensions of
the isosceles trapezoid. The base of the triangle is 0.5
(12 – 8) = 2.
Don't forget to use dimensional analysis to convert
the units to feet.
ANSWER:
2.3 ft; 0.24 ft2
32.
SOLUTION:
The figure is a kite because one of the diagonals is a
perpendicular bisector. Find the perimeter. Convert
the units to feet.
Use the 45-45-90 triangle to find the lengths of the
congruent parts of the diagonals.
d1 = 3 + 3 = 6
Use the Pythagorean theorem to find the other piece
of d2.
Now find the area of the kite.
ANSWER:
67.6 ft; 267.8 ft2
33.MULTIPLE REPRESENTATIONS In this
problem, you will investigate perimeters of kites.
a. GEOMETRIC Draw a kite like the one shown if
x = 2.
b. GEOMETRIC Repeat the process in part a for
three x-values between 2 and 10 and for an x-value
of 10.
c. TABULAR Measure and record in a table the
perimeter of each kite, along with the x-value.
d. GRAPHICAL Graph the perimeter versus the x-
value using the data from your table.
e. ANALYTICAL Make a conjecture about the
value of x that will minimize the perimeter of the kite.
Whatisthesignificanceofthisvalue?
SOLUTION:
a. When x = 2, the long diagonal will be in pieces of
length 2 and 10.
b. When x = 4, the long diagonal will be in pieces of
length 4 and 8.
When x = 6, the long diagonal will be in pieces of
length 6 and 6.
When x = 8, the long diagonal will be in pieces of
length 8 and 4.
When x = 10, the long diagonal will be in pieces of
length 10 and 2.
c. UsethePythagoreanTheoremtofirstfindthe
lengths of the sides of the kite and then find its
perimeter.
For x = 2: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm each
and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and
10 cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of
the kite is cm.
For x = 4: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 4
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and 8
cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of the
kite is cm.
For x = 6: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 6
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and 6
cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of the
kite is cm.
For x = 8: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 8
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and 4
cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of the
kite is cm.
For x = 10: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 10
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm each.
Thus, the length of two legs =
and the length of the other two sides =
. So, the perimeter of the kite is
cm.
d.
e. Sample answer: Based on the graph, the perimeter
will be minimized when x = 6. This value is
significant because when x = 6, the figure is a
rhombus.
x P
2 cm 26.1
4 cm 25.4
6 cm 25.3
8 cm 25.4
10 cm 26.1
ANSWER:
a.
b.
c.
d.
e. Sample answer: Based on the graph, the perimeter
will be minimized when x = 6. This value is
significant because when x = 6, the figure is a
rhombus.
x P
2 cm 26.1
4 cm 25.4
6 cm 25.3
8 cm 25.4
10 cm 26.1
34.CCSS CRITIQUE Antonio and Madeline want to
draw a trapezoid that has a height of 4 units and an
area of 18 square units. Antonio says that only one
trapezoid will meet the criteria. Madeline disagrees
and thinks that she can draw several different
trapezoids with a height of 4 units and an area of 18
square units. Is either of them correct? Explain your
reasoning.
SOLUTION:
There is more than one trapezoid with a height of 4
units and an area of 18 square units. The sum of the
bases of the trapezoid has to be 9, so one possibility
is a trapezoid with bases of 4 and 5 units and a height
of 4 units. Another is a trapezoid with bases of 3 and
6 units and a height of 4 units. Therefore, Madeline is
correct.
ANSWER:
Madeline; sample answer: There is more than one
trapezoid with a height of 4 units and an area of 18
square units. The sum of the bases of the trapezoid
has to be 9, so one possibility is a trapezoid with
bases of 4 and 5 units and a height of 4 units.
Another is a trapezoid with bases of 3 and 6 units
and a height of 4 units.
35.CHALLENGE Find x in parallelogram ABCD.
SOLUTION:
Let the two segments of the side bea and 15 –
a.
We have two variables, x and a. They are a part of
two right triangles, so we can use the Pythagorean
Theorem for each triangle and then combine the
equations.
We now have x2 on the left side of each equation.
Use substitution.
Use the value of a to find the value of x.
ANSWER:
7.2
36.OPEN ENDED Draw a kite and a rhombus with an
area of 6 square inches. Label and justify your
drawings.
SOLUTION:
Sample answer: Since the area formula for both a
rhombus and a kite is one half the product of the
lengths of the two diagonals, if the area is 6 square
inches, the product of the two diagonals must be 12. I
used 3 and 4 inches for the diagonals of the rhombus
and 2 and 6 inches for the diagonals of the kite.
Both diagonals must be perpendicular bisectors for
the rhombus and one diagonal must be a
perpendicular bisector for the kite.
ANSWER:
Sample answer: Since the area formula for both a
rhombus and a kite is one half the product of the
lengths of the two diagonals, if the area is 6 square
inches, the product of the two diagonals must be 12. I
used 3 and 4 inches for the diagonals of the rhombus
and 2 and 6 inches for the diagonals of the kite.
37.REASONING If the areas of two rhombi are equal,
are the perimeters sometimes, always, or never
equal? Explain.
SOLUTION:
If the areas are equal, it means that the products of
the diagonals are equal. The only time that the
perimeters will be equal is when the diagonals are
also equal, or when the two rhombi are congruent.
Therefore, the statement is sometimes true.
For example,
When the diagonals are 6 and 8, the area is 24 and
the perimeter is 26.
When the diagonals are 12 and 4, the area is 24 and
the perimeter is 32.
ANSWER:
Sometimes; sample answer: If the areas are equal, it
means that the products of the diagonals are equal.
The only time that the perimeters will be equal is
when the diagonals are also equal, or when the two
rhombi are congruent.
38.WRITING IN MATH How can you use
trigonometry to find the area of a figure?
SOLUTION:
Trigonometry can be used with known angles and
sides to find unknown sides. For example;
The height of the parallelogram can be found using
the sine function:
And the area can be found as usual:
ANSWER:
Sample answer: You can use trigonometry and
known angle and side measures to find unknown
triangular measures that are required to calculate the
area.
39.The lengths of the bases of an isosceles trapezoid are
shown below.
If the perimeter is 74 meters, what is its area?
A
B
C
D
SOLUTION:
The lengths of the legs of an isosceles trapezoid are
congruent. The perimeter is 74 and the sum of the
lengths of the bases is 54 m, so the sides are each 10
m.
We can draw dashed lines for the height to produce
two congruent triangles
The bases of the triangles are 8 because the middle
figure is a rectangle and the entire base is 35.
Use the Pythagorean Theorem to find the height h.
Now find the area of the trapezoid.
ANSWER:
A
40.SHORT RESPONSE One diagonal of a rhombus is
three times as long as the other diagonal. If the area
of the rhombus is 54 square millimeters, what are the
lengths of the diagonals?
SOLUTION:
Therefore, the diagonals are of length 6 mm. and 18
mm.
ANSWER:
6 mm, 18 mm
41.ALGEBRA What is the effect on the graph of the
equation whentheequationischangedtoy
= –2x?
F The graph is moved 1 unit down.
G The graph is moved 1 unit up.
H Thegraphisrotated45°abouttheorigin.
J Thegraphisrotated90°abouttheorigin.
SOLUTION:
The product of the slopes of the two lines is –1. So,
thelinesareperpendiculartoeachother.
When two lines are perpendicular, the first line is
rotated 90°abouttheorigintogetthesecondline.
Therefore, the correct choice is J.
ANSWER:
J
42.A regular hexagon is divided into 6 congruent
triangles. If the perimeter of the hexagon is 48
centimeters, what is the height of each triangle?
A 4 cm
B cm
C cm
D 8 cm
E cm
SOLUTION:
The perimeter of the hexagon is 48 cm, so the length
of each side of the hexagon is 8 cm. The height of
each triangle is equal to the apothem of the hexagon.
Each interior angle of a regular hexagon measures
So, each diagonal will make an
angleof60°withthesides.
An apothem bisects the side of the polygon.
The triangle with the sides of one apothem, half the
side of the polygon to which the apothem is drawn,
and the line joining the center and the vertex at
endpoints of the side form a triangle.
Let h be the height of the triangle. Use the tangent
ratio to find h.
The height of each triangle is .
ANSWER:
B
COORDINATE GEOMETRY Find the area of
each figure.
43. JKL with J(–4, 3), K(–9, –1), and L(–4, –4)
SOLUTION:
Plot the points and draw the triangle.
The length of base JL is 7 units. The corresponding
height is from the line x = –9 to x = –4, so the height
is5units.
ANSWER:
44. RSTV with R(–5, 7), S(2, 7), T(0, 2), and V(–7, 2)
SOLUTION:
Plot the points.
The parallelogram has base length of 7 units.
The height is from y = 2 to y = 7, so the height is 5
units. The area of the parallelogram is 7(5) or 35
units2.
ANSWER:
45.WEATHER Meteorologists track severe storms
using Doppler radar. A polar grid is used to measure
distances as the storms progress. If the center of the
radar screen is the origin and each ring is 10 miles
farther from the center, what is the equation of the
fourth ring?
Refer to the image on Page 780.
SOLUTION:
The standard form of the equation of a circle with
center at (h, k) and radius r is (x – h)2 + (y – k)2 =
r2.
(h, k)=(0,0)
The fourth ring will be 40 miles away from the origin,
so r=40.
Therefore, the equation is (x – 0)2 + (y – 0)2 = 402 or
x2 + y2 = 1600.
ANSWER:
Find x and y.
46.
SOLUTION:
Thisisa30°-60°-90°triangle.
Sideoppositethe30°=a
Sideoppositethe60°=
Hypotenuse = 2a
For this triangle:
Sideoppositethe30°=8
Sideoppositethe60°=
Hypotenuse = 16
Therefore, x = and y = 12
ANSWER:
47.
SOLUTION:
Thisisa30°-60°-90°triangle.
Sideoppositethe30°=a
Sideoppositethe60°=
Hypotenuse = 2a
For this triangle:
Sideoppositethe30°=9
Sideoppositethe60°=
Hypotenuse = 18
Therefore, x = 9 and y =
ANSWER:
Use the Venn diagram to determine whether
each statement is always, sometimes, or never
true.
48.A parallelogram is a square.
SOLUTION:
When a figure is a parallelogram, it can be anything
inside the yellow sphere of the Venn diagram. The
figure is a square only when it is inside the purple
intersection in the middle of the diagram. Therefore,
a figure that is a parallelogram is sometimes also a
square.
ANSWER:
sometimes
49.A square is a rhombus.
SOLUTION:
When a figure is a square, it is inside the purple
intersection in the middle of the diagram. This region
is within the region of a rhombus, so when a figure is
a square, it is alwaysarhombusaswell.
ANSWER:
always
50.A rectangle is a parallelogram.
SOLUTION:
When a figure is a rectangle, it can be anything inside
the blue sphere of the Venn diagram. This region is
within the region of a parallelogram, so any figure
that is a rectangle is alwaysalsoaparallelogram.
ANSWER:
always
51.A rhombus is a rectangle but not a square.
SOLUTION:
When a figure is a rhombus, it can be anything inside
the red sphere of the Venn diagram. The figure is a
rectangle when it is inside the blue sphere. The
intersection of these two spheres indicate when a
figure is both a rhombus and a rectangle. However,
everything within this region is also a square, so a
figure can never be both a rhombus and a rectangle
while not also being a square.
ANSWER:
never
52.A rhombus is a square.
SOLUTION:
When a figure is a rhombus, it can be anything inside
the red sphere of the Venn diagram. When a figure
is within the purple region, it is a square. However, a
figure within the red region but not within the purple
region is a rhombus that is not a square.
Therefore, a figure can sometimes be both a
rhombus and a square.
ANSWER:
sometimes
Find the circumference and area of each figure.
Round to the nearest tenth.
53.
SOLUTION:
ANSWER:
54.
SOLUTION:
ANSWER:
55.
SOLUTION:
ANSWER:
Find the area of each trapezoid, rhombus, or
kite.
1.
SOLUTION:
ANSWER:
2.
SOLUTION:
ANSWER:
3.
SOLUTION:
ANSWER:
4.SHORT RESPONSE Suki is doing fashion design
at 4-H Club. Her first project is to make a simple A-
line skirt. How much fabric will she need according
to the design at the right?
SOLUTION:
The area A of a trapezoid is A = h(b1 + b2).
She will need · 2 or squarefeetoffabric.
ANSWER:
ALGEBRA Find x.
5.
SOLUTION:
ANSWER:
8 cm
6.
SOLUTION:
ANSWER:
6.6 in.
7.
SOLUTION:
ANSWER:
6.3 ft
CCSSSTRUCTUREFind the area of each
trapezoid, rhombus, or kite.
8.
SOLUTION:
h = 13, b1 = 18 and b2 = 24
ANSWER:
9.
SOLUTION:
h = 23, b1 = 22 and b2 = 37
ANSWER:
10.
SOLUTION:
d1 = 22 and d2=24
ANSWER:
11.
SOLUTION:
d1 = 16 in. and d2=17in.
ANSWER:
12.
SOLUTION:
d1 = 7 cm. and d2 = 6 + 9 = 15 cm
ANSWER:
13.
SOLUTION:
d1 = 11 ft. and d2=25ft
ANSWER:
MICROSCOPES Find the area of the identified
portion of each magnified image. Assume that
the identified portion is either a trapezoid,
rhombus, or kite. Measures are provided in
microns.
14.human skin
Refer to the image on Page 777.
SOLUTION:
ANSWER:
26 square microns
15.heartleaf plant
RefertotheimageonPage777.
SOLUTION:
ANSWER:
24.5 square microns
16.eye of a fly
RefertotheimageonPage777.
SOLUTION:
Thefigureconsistsoftwotrapezoids.
ANSWER:
9.9 square microns
17.JOBS Jimmy works on his neighbors’yards after
school to earn extra money to buy a car. He is going
to plant grass seed in Mr. Troyer’s yard. What is the
area of the yard?
SOLUTION:
ANSWER:
ALGEBRA Find each missing length.
18.One diagonal of a kite is twice as long as the other
diagonal. If the area of the kite is 240 square inches,
what are the lengths of the diagonals?
SOLUTION:
The area A of a kite is one half the product of the
lengths of its diagonals, d1 and d2.
Therefore, the diagonals are of length 15.5 in. and
31.0 in.
ANSWER:
15.5 in., 31.0 in.
19.The area of a rhombus is 168 square centimeters. If
one diagonal is three times as long as the other, what
are the lengths of the diagonals?
SOLUTION:
The area A of a rhombus is one half the product of
the lengths of its diagonals, d1 and d2.
Therefore, the diagonals are of length 10.6 cm. and
31.7 cm.
ANSWER:
10.6 cm, 31.7 cm
20.A trapezoid has base lengths of 12 and 14 feet with
an area of 322 square feet. What is the height of the
trapezoid?
SOLUTION:
The area A of a trapezoid is one half the product of
the height h and the sum of the lengths of its bases,
b1 and b2.
.
ANSWER:
24.8 ft
21.A trapezoid has a height of 8 meters, a base length of
12 meters, and an area of 64 square meters. What is
the length of the other base?
SOLUTION:
The area A of a trapezoid is one half the product of
the height h and the sum of the lengths of its bases,
b1 and b2.
ANSWER:
4 m
22.HONORS Estella has been asked to join an honor
society at school. Before the first meeting, new
members are asked to sand and stain the front side
of a piece of wood in the shape of an isosceles
trapezoid. What is the surface area that Allison will
need to sand and stain?
SOLUTION:
The required area is the difference between the
larger trapezoid and the smaller trapezoid.
The area A of a trapezoid is one half the product of
the height h and the sum of the lengths of its bases,
b1 and b2.
ANSWER:
For each figure, provide a justification showing
that .
23.
SOLUTION:
Set the area of the kite equal to the sum of the areas
of the two triangles with bases d1 and d2.
ANSWER:
The area of andtheareaof
. Therefore, the area of
, and the area of .
The area of kite FGHJ is equal to the area of
the area of . After
simplification, the area of kite FGHJ is equal to
.
24.
SOLUTION:
Set the area of the kite equal to the sum of the areas
of the two triangles with bases d1 and d2.
ZWX = ZYX by SSS
ANSWER:
The area of and the area of
. Therefore, the area of
, and the area of .
The area of rhombus WXYZ is equal to the area of
the area of .
After simplification, the area of rhombus WXYZ is
equal to .
25.CRAFTS Ashanti is in a kite competition. The
yellow, red, orange, green, and blue pieces of her kite
design shown are all congruent rhombi.
a. How much fabric of each color does she need to
buy?
b. Competition rules require that the total area of
each kite be no greater than 200 square inches. Does
Ashanti’s kite meet this requirement? Explain.
SOLUTION:
The area of the yellow rhombus is or24
in2. Since the yellow, red , orange, green, and blue
pieces area all congruent rhombi, each have an area
of 24 in2.
The area of the purple kite-shaped piece is
or20in2.
The total area of the entire kite is 24(5) + 20 or 140
in2, which is less than the maximum 200 in2 allowed.
Therefore, her kite meets this requirement.
ANSWER:
a. each of yellow, red, orange, green, and
blue; of purple
b. Yes; her kite has an area of , which is less
than 200 in2.
CCSS SENSE-MAKING Find the area of each
quadrilateral with the given vertices.
26.A(–8, 6), B(–5, 8), C(–2, 6), and D(–5, 0)
SOLUTION:
Graph the quadrilateral.
The quadrilateral is a kite. The area A of a kite is one
half the product of the lengths of its diagonals, d1 and
d2. The lengths of the diagonals are 6 units and 8
units. Therefore, the area of the kite is
ANSWER:
24 sq. units
27.W(3, 0), X(0, 3), Y(–3, 0), and Z(0, –3)
SOLUTION:
Graph the quadrilateral.
The quadrilateral is a rhombus. The area A of a
rhombus is one half the product of the lengths of its
diagonals, d1 and d2.
The lengths of the diagonals are 6 units each.
Therefore, the area of the kite is
ANSWER:
18 sq. units
28.METALS When magnified in very powerful
microscopes, some metals are composed of grains
that have various polygonal shapes.
a. What is the area of figure 1 if the grain has a
height of 4 microns and bases with lengths of 5 and 6
microns?
b. If figure 2 has perpendicular diagonal lengths of
3.8 microns and 4.9 microns, what is the area of the
grain?
SOLUTION:
a. Figure1isatrapezoid.
The area of figure 1 is 22 square microns.
b. Figure2isarhombus.
The area of figure 2 is about 9.3 square microns.
ANSWER:
a. 22 square microns
b. 9.3 square microns
29.PROOF The figure at the right is a trapezoid that
consists of two congruent right triangles and an
isosceles triangle. In 1876, James A. Garfield, the
20th president of the United States, discovered a
proof of the Pythagorean Theorem using this
diagram.Provethat .
SOLUTION:
Following is an algebraic proof.
The trapezoid is on its side, so the bases are x and y
and the height is x + y. Set the area of the trapezoid
equal to the sum of the areas of the triangles and
simplify.
ANSWER:
The area of a trapezoid is So,
.
The area of . The area
of . Set the
area of the trapezoid equal to the combined areas of
the triangles to get . Multiply by
2 on each side: .
When simplified, .
DIMENSIONAL ANALYSIS Find the
perimeter and area of each figure in feet. Round
to the nearest tenth, if necessary.
30.
SOLUTION:
Both diagonals are perpendicular bisectors, so the
figure is a rhombus and all four triangles are
congruent. All of the sides are 12 feet, so the
perimeter is 48 feet.
Use trigonometry to find the lengths of the diagonals.
Now find the area.
ANSWER:
48 ft; 129.4 ft2
31.
SOLUTION:
Use the 30-60-90 triangle to find the dimensions of
the isosceles trapezoid. The base of the triangle is 0.5
(12 – 8) = 2.
Don't forget to use dimensional analysis to convert
the units to feet.
ANSWER:
2.3 ft; 0.24 ft2
32.
SOLUTION:
The figure is a kite because one of the diagonals is a
perpendicular bisector. Find the perimeter. Convert
the units to feet.
Use the 45-45-90 triangle to find the lengths of the
congruent parts of the diagonals.
d1 = 3 + 3 = 6
Use the Pythagorean theorem to find the other piece
of d2.
Now find the area of the kite.
ANSWER:
67.6 ft; 267.8 ft2
33.MULTIPLE REPRESENTATIONS In this
problem, you will investigate perimeters of kites.
a. GEOMETRIC Draw a kite like the one shown if
x = 2.
b. GEOMETRIC Repeat the process in part a for
three x-values between 2 and 10 and for an x-value
of 10.
c. TABULAR Measure and record in a table the
perimeter of each kite, along with the x-value.
d. GRAPHICAL Graph the perimeter versus the x-
value using the data from your table.
e. ANALYTICAL Make a conjecture about the
value of x that will minimize the perimeter of the kite.
Whatisthesignificanceofthisvalue?
SOLUTION:
a. When x = 2, the long diagonal will be in pieces of
length 2 and 10.
b. When x = 4, the long diagonal will be in pieces of
length 4 and 8.
When x = 6, the long diagonal will be in pieces of
length 6 and 6.
When x = 8, the long diagonal will be in pieces of
length 8 and 4.
When x = 10, the long diagonal will be in pieces of
length 10 and 2.
c. UsethePythagoreanTheoremtofirstfindthe
lengths of the sides of the kite and then find its
perimeter.
For x = 2: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm each
and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and
10 cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of
the kite is cm.
For x = 4: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 4
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and 8
cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of the
kite is cm.
For x = 6: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 6
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and 6
cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of the
kite is cm.
For x = 8: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 8
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm and 4
cm. Thus, the length of two legs =
and the length of the other two
sides = . So, the perimeter of the
kite is cm.
For x = 10: Two sides of the kite are the length of the
hypotenuse of right triangle with legs of 2 cm and 10
cm and the other two sides are the length of the
hypotenuse of a right triangle with legs of 2 cm each.
Thus, the length of two legs =
and the length of the other two sides =
. So, the perimeter of the kite is
cm.
d.
e. Sample answer: Based on the graph, the perimeter
will be minimized when x = 6. This value is
significant because when x = 6, the figure is a
rhombus.
x P
2 cm 26.1
4 cm 25.4
6 cm 25.3
8 cm 25.4
10 cm 26.1
ANSWER:
a.
b.
c.
d.
e. Sample answer: Based on the graph, the perimeter
will be minimized when x = 6. This value is
significant because when x = 6, the figure is a
rhombus.
x P
2 cm 26.1
4 cm 25.4
6 cm 25.3
8 cm 25.4
10 cm 26.1
34.CCSS CRITIQUE Antonio and Madeline want to
draw a trapezoid that has a height of 4 units and an
area of 18 square units. Antonio says that only one
trapezoid will meet the criteria. Madeline disagrees
and thinks that she can draw several different
trapezoids with a height of 4 units and an area of 18
square units. Is either of them correct? Explain your
reasoning.
SOLUTION:
There is more than one trapezoid with a height of 4
units and an area of 18 square units. The sum of the
bases of the trapezoid has to be 9, so one possibility
is a trapezoid with bases of 4 and 5 units and a height
of 4 units. Another is a trapezoid with bases of 3 and
6 units and a height of 4 units. Therefore, Madeline is
correct.
ANSWER:
Madeline; sample answer: There is more than one
trapezoid with a height of 4 units and an area of 18
square units. The sum of the bases of the trapezoid
has to be 9, so one possibility is a trapezoid with
bases of 4 and 5 units and a height of 4 units.
Another is a trapezoid with bases of 3 and 6 units
and a height of 4 units.
35.CHALLENGE Find x in parallelogram ABCD.
SOLUTION:
Let the two segments of the side bea and 15 –
a.
We have two variables, x and a. They are a part of
two right triangles, so we can use the Pythagorean
Theorem for each triangle and then combine the
equations.
We now have x2 on the left side of each equation.
Use substitution.
Use the value of a to find the value of x.
ANSWER:
7.2
36.OPEN ENDED Draw a kite and a rhombus with an
area of 6 square inches. Label and justify your
drawings.
SOLUTION:
Sample answer: Since the area formula for both a
rhombus and a kite is one half the product of the
lengths of the two diagonals, if the area is 6 square
inches, the product of the two diagonals must be 12. I
used 3 and 4 inches for the diagonals of the rhombus
and 2 and 6 inches for the diagonals of the kite.
Both diagonals must be perpendicular bisectors for
the rhombus and one diagonal must be a
perpendicular bisector for the kite.
ANSWER:
Sample answer: Since the area formula for both a
rhombus and a kite is one half the product of the
lengths of the two diagonals, if the area is 6 square
inches, the product of the two diagonals must be 12. I
used 3 and 4 inches for the diagonals of the rhombus
and 2 and 6 inches for the diagonals of the kite.
37.REASONING If the areas of two rhombi are equal,
are the perimeters sometimes, always, or never
equal? Explain.
SOLUTION:
If the areas are equal, it means that the products of
the diagonals are equal. The only time that the
perimeters will be equal is when the diagonals are
also equal, or when the two rhombi are congruent.
Therefore, the statement is sometimes true.
For example,
When the diagonals are 6 and 8, the area is 24 and
the perimeter is 26.
When the diagonals are 12 and 4, the area is 24 and
the perimeter is 32.
ANSWER:
Sometimes; sample answer: If the areas are equal, it
means that the products of the diagonals are equal.
The only time that the perimeters will be equal is
when the diagonals are also equal, or when the two
rhombi are congruent.
38.WRITING IN MATH How can you use
trigonometry to find the area of a figure?
SOLUTION:
Trigonometry can be used with known angles and
sides to find unknown sides. For example;
The height of the parallelogram can be found using
the sine function:
And the area can be found as usual:
ANSWER:
Sample answer: You can use trigonometry and
known angle and side measures to find unknown
triangular measures that are required to calculate the
area.
39.The lengths of the bases of an isosceles trapezoid are
shown below.
If the perimeter is 74 meters, what is its area?
A
B
C
D
SOLUTION:
The lengths of the legs of an isosceles trapezoid are
congruent. The perimeter is 74 and the sum of the
lengths of the bases is 54 m, so the sides are each 10
m.
We can draw dashed lines for the height to produce
two congruent triangles
The bases of the triangles are 8 because the middle
figure is a rectangle and the entire base is 35.
Use the Pythagorean Theorem to find the height h.
Now find the area of the trapezoid.
ANSWER:
A
40.SHORT RESPONSE One diagonal of a rhombus is
three times as long as the other diagonal. If the area
of the rhombus is 54 square millimeters, what are the
lengths of the diagonals?
SOLUTION:
Therefore, the diagonals are of length 6 mm. and 18
mm.
ANSWER:
6 mm, 18 mm
41.ALGEBRA What is the effect on the graph of the
equation whentheequationischangedtoy
= –2x?
F The graph is moved 1 unit down.
G The graph is moved 1 unit up.
H Thegraphisrotated45°abouttheorigin.
J Thegraphisrotated90°abouttheorigin.
SOLUTION:
The product of the slopes of the two lines is –1. So,
thelinesareperpendiculartoeachother.
When two lines are perpendicular, the first line is
rotated 90°abouttheorigintogetthesecondline.
Therefore, the correct choice is J.
ANSWER:
J
42.A regular hexagon is divided into 6 congruent
triangles. If the perimeter of the hexagon is 48
centimeters, what is the height of each triangle?
A 4 cm
B cm
C cm
D 8 cm
E cm
SOLUTION:
The perimeter of the hexagon is 48 cm, so the length
of each side of the hexagon is 8 cm. The height of
each triangle is equal to the apothem of the hexagon.
Each interior angle of a regular hexagon measures
So, each diagonal will make an
angleof60°withthesides.
An apothem bisects the side of the polygon.
The triangle with the sides of one apothem, half the
side of the polygon to which the apothem is drawn,
and the line joining the center and the vertex at
endpoints of the side form a triangle.
Let h be the height of the triangle. Use the tangent
ratio to find h.
The height of each triangle is .
ANSWER:
B
COORDINATE GEOMETRY Find the area of
each figure.
43. JKL with J(–4, 3), K(–9, –1), and L(–4, –4)
SOLUTION:
Plot the points and draw the triangle.
The length of base JL is 7 units. The corresponding
height is from the line x = –9 to x = –4, so the height
is5units.
ANSWER:
44. RSTV with R(–5, 7), S(2, 7), T(0, 2), and V(–7, 2)
SOLUTION:
Plot the points.
The parallelogram has base length of 7 units.
The height is from y = 2 to y = 7, so the height is 5
units. The area of the parallelogram is 7(5) or 35
units2.
ANSWER:
45.WEATHER Meteorologists track severe storms
using Doppler radar. A polar grid is used to measure
distances as the storms progress. If the center of the
radar screen is the origin and each ring is 10 miles
farther from the center, what is the equation of the
fourth ring?
Refer to the image on Page 780.
SOLUTION:
The standard form of the equation of a circle with
center at (h, k) and radius r is (x – h)2 + (y – k)2 =
r2.
(h, k)=(0,0)
The fourth ring will be 40 miles away from the origin,
so r=40.
Therefore, the equation is (x – 0)2 + (y – 0)2 = 402 or
x2 + y2 = 1600.
ANSWER:
Find x and y.
46.
SOLUTION:
Thisisa30°-60°-90°triangle.
Sideoppositethe30°=a
Sideoppositethe60°=
Hypotenuse = 2a
For this triangle:
Sideoppositethe30°=8
Sideoppositethe60°=
Hypotenuse = 16
Therefore, x = and y = 12
ANSWER:
47.
SOLUTION:
Thisisa30°-60°-90°triangle.
Sideoppositethe30°=a
Sideoppositethe60°=
Hypotenuse = 2a
For this triangle:
Sideoppositethe30°=9
Sideoppositethe60°=
Hypotenuse = 18
Therefore, x = 9 and y =
ANSWER:
Use the Venn diagram to determine whether
each statement is always, sometimes, or never
true.
48.A parallelogram is a square.
SOLUTION:
When a figure is a parallelogram, it can be anything
inside the yellow sphere of the Venn diagram. The
figure is a square only when it is inside the purple
intersection in the middle of the diagram. Therefore,
a figure that is a parallelogram is sometimes also a
square.
ANSWER:
sometimes
49.A square is a rhombus.
SOLUTION:
When a figure is a square, it is inside the purple
intersection in the middle of the diagram. This region
is within the region of a rhombus, so when a figure is
a square, it is alwaysarhombusaswell.
ANSWER:
always
50.A rectangle is a parallelogram.
SOLUTION:
When a figure is a rectangle, it can be anything inside
the blue sphere of the Venn diagram. This region is
within the region of a parallelogram, so any figure
that is a rectangle is alwaysalsoaparallelogram.
ANSWER:
always
51.A rhombus is a rectangle but not a square.
SOLUTION:
When a figure is a rhombus, it can be anything inside
the red sphere of the Venn diagram. The figure is a
rectangle when it is inside the blue sphere. The
intersection of these two spheres indicate when a
figure is both a rhombus and a rectangle. However,
everything within this region is also a square, so a
figure can never be both a rhombus and a rectangle
while not also being a square.
ANSWER:
never
52.A rhombus is a square.
SOLUTION:
When a figure is a rhombus, it can be anything inside
the red sphere of the Venn diagram. When a figure
is within the purple region, it is a square. However, a
figure within the red region but not within the purple
region is a rhombus that is not a square.
Therefore, a figure can sometimes be both a
rhombus and a square.
ANSWER:
sometimes
Find the circumference and area of each figure.
Round to the nearest tenth.
53.
SOLUTION:
ANSWER:
54.
SOLUTION:
ANSWER:
55.
SOLUTION:
ANSWER:
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11-2 Areas of Trapezoids, Rhombi, and Kites
