(1)
Billy is hiking in Colorado. He walks eastward four miles, then turns 60
degrees northward and walks six miles. How far is he from his starting point? Express your
answer in simplest radical form.
(2)
The sides of triangle CAB are in the ratio of 2 : 3 : 4. Segment BD is the
angle bisector drawn to the shortest side, dividing it into segments AD and DC. What is
the length, in inches, of the longer subsegment of side AC if the length of side AC is 10
inches? Express your answer as a common fraction.
(3)
In right triangle JKL, angle J measures 60 degrees and angle K measures
30 degrees. When drawn, the angle bisectors of angles J and K intersect at a point M.
What is the measure of obtuse angle JMK?
K
M
J
L
(4)
Two cross sections of a right hexagonal pyramid are obtained by cutting the
pyramid
√ with planes parallel√to the hexagonal base. The areas of the cross sections are
216 3 square feet and 486 3 square feet. The two planes are 8 feet apart. How far from
the apex of the pyramid is the larger cross section?
(5)
Circle O has radius 8 units and circle P has radius 2 units. The circles are
externally tangent to each other at point Q. Segment T S is tangent to both circle O and
circle P at points T and S, respectively. What is the length of segment OS? Express your
answer in simplest radical form.
(6)
The vertices of a convex pentagon are (−1, −1), (−3, 4), (1, 7), (6, 5) and
(3, −1). What is the area of the pentagon?
(7)
Fido’s leash is tied to a stake at the center of his yard, which is in the shape
of a regular hexagon. His leash is exactly long enough to reach the midpoint of each side of
his yard. If the fraction of the area of Fido’s
yard that he is able to reach while on his leash
√
a
is expressed in simplest radical form as b π, what is the value of the product ab?
(8)
The region shown is bounded by the arcs of circles having radius 4 units,
having a central angle measure of 60 degrees and√intersecting at points of tangency.
The
√
area of the region can be expressed in the form a b + cπ square units, where b is a
radical in simplest form. What is the value of a + b + c?
(9)
In the circle with center O and diameters AC and BD, the angle AOD
measures 54 degrees. What is the measure, in degrees, of angle AOB?
A
B
D
O
C
(10)
If the diameter of a right cylindrical can with circular bases is increased by
25%, by what percent should the height be increased in order to double the volume of the
original can?
(11)
In rectangle ABCD, side AB measures 6 units and side BC measures 3
units, as shown. Points F and G are on side CD with segment DF measuring 1 unit and
segment GC measuring 2 units, and lines AF and BG intersect at E. What is the area of
triangle AEB?
E
D 1
2
F
C
G
3
A
6
B
(12)
Circle O has radius 10 units. Point P is on radius OQ and OP = 6 units.
How many different chords containing P, including the diameter, have integer lengths?
Q
O
P
(13)
√ A square and an equilateral triangle have equal perimeters. The area of the
triangle is 16 3 square centimeters. How long, in centimeters, is a diagonal of the square?
Express your answer in simplest radical form.
(14)
Two similar right triangles have areas of 6 square inches and 150 square
inches. The length of the hypotenuse of the smaller triangle is 5 inches. What is the sum
of the lengths of the legs of the larger triangle?
(15)
Three coplanar squares with sides of lengths two, four and six units,
respectively, are arranged side-by-side, as shown so that one side of each square lies on line
AB and a segment connects the bottom left corner of the smallest square to the upper
right corner of the largest square. What is the area of the shaded quadrilateral?
6
4
2
A
(16)
B
A quarter-circle of radius 3 units is drawn at each of the vertices of a square
with sides of 6 units.
The area of the shaded region can be expressed in the form a − bπ square units, where a
and b are both integers. What is the value of a + b?
(17)
What is the area enclosed by the graph of |x | + |2y | = 10 shown here?
y
x
(18)
The point A (3, 4) is reflected over the x -axis to B. Then B is reflected
over the line y = x to C. What is the area of triangle ABC?
(19)
In triangle ABC, AB = 12 units and AC = 9 units. Point D is on segment
BC so that BD : DC = 2 : 1. If AD = 6 units, what is the length of segment BC? Express
your answer in simplest radical form.
(20)
How many unit cubes would it take to construct the complete exterior of a
hollow cube with edges of 4 units and faces 1 unit thick?
(21)
Side AB of regular hexagon ABCDEF is extended past B to point X such
that AX = 3AB. Given that each side of the hexagon is 2 units long, what is the length of
segment F X? Express your answer in simplest radical form.
(22)
In trapezoid ABCD, the parallel sides AB and CD have lengths of 8 and 20
units, respectively, and the altitude is 12 units. Points E and F are the midpoints of sides
AD and BC, respectively. What is the area of quadrilateral EF CD in square units?
(23)
What is the volume, in cubic centimeters, of a right rectangular prism with
all integer edge lengths, and faces having areas of 30, 180 and 24 square centimeters?
(24)
If a fly is buzzing randomly around a room 8 ft long, 12 ft wide and 10 ft
high, what is the probability that, at any given time, the fly is within 6 feet of the ceiling?
Express your answer as a common fraction.
(25)
The points (1, 7), (13, 16) and (5, k), where k is an integer, are vertices of a
triangle. What is the sum of the values of k for which the area of the triangle is a
minimum?
(26)
In pentagon ABCDE, BC = CD = DE = 2 units, ∠E is a right angle and
◦
m∠B = m∠C = m∠D
√ = 135 . The length of segment AE can be expressed in simplest
radical form as a + 2 b units. What is the value of a + b?
(27)
The point (0, 0) is reflected across the vertical line x = 1. Its image is then
reflected across the line y = 2. What are the coordinates of the resulting point?
(28)
In triangle ABC, AB = AC and D is a point on AC so that BD bisects
angle ABC. If BD = BC, what is the measure, in degrees, of angle A?
(29)
In the figure shown, the ratio of BD to DC is 4 to 3. The area of △ABD is
24 square centimeters. What is the area of △ADC?
A
B
(30)
D C
When the diameter of a pizza increases by 2 inches, the area increases by
44%. What was the area, in square inches, of the original pizza? Express your answer in
terms of π.
Copyright MATHCOUNTS Inc. All rights reserved
Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Answer
√
2 19 miles
40
7 inches
135 degrees
24
√ feet
8 2
47 sq units
18
11
126 degrees
28 %
18 square units
8√chords
6 2 cm
35 inches
8 sq units
45
100 square units
28
√ square units
3 33 (units)
56
√
2 13 units
102 square units
360 cu cm
3/5
20
6
(2, 4)
36 degrees
18 sq cm
25pi square inches
Problem ID
14CC2
A5CC2
55103
05CC2
3A253
BC403
B1553
C34D3
4A303
35CC2
20403
AD253
3D103
03453
DC403
530D3
B2453
41553
0B253
C30D3
45CC2
4BCC2
D4CC2
B1453
D0503
BA253
3BCC2
B0403
C1CC2
CC3D3
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
√
ID: [14CC2]
(1) 2 19 miles
Suppose Billy starts at point A, turns at point B, and ends at point D, as shown below. If
Billy turns 60◦ northward and walks six miles, then we can draw a 30 − 60 − 90 triangle
whose hypotenuse is 6 miles (triangle BCD below).
D
6
A
4
B
C
It follows√
that Billy traveled 6/2 = 3 miles eastward during these 6 miles, and that he
miles northward during these 6 miles. In total, Billy traveled 4 + 3 = 7 miles
traveled 3 · 3 √
eastward and 3 q
3 miles northward. By the Pythagorean Theorem, the distance from his
√
√
√
√
starting point is (7)2 + (3 3)2 = 49 + 27 = 76 = 2 19 .
(2)
40
7
inches
ID: [A5CC2]
Without loss of generality, suppose that BA < BC. Since BD is the angle bisector of ∠B,
by the Angle Bisector Theorem, it follows that
AD
BA
3
=
= .
CD
BC
4
Thus, AD < CD, so CD is the longer subsegment of AC. Solving for AD, it follows that
AD = 3CD
4 . Also, we know that AD + CD = AC = 10, and substituting our previous value
40
7CD
inches.
for AD, we find that 3CD
4 + CD = 4 = 10 =⇒ CD =
7
(3) 135 degrees
ID: [55103]
Since JM bisects ∠J, we know that the measure of ∠KJM is 60/2 = 30 degrees.
Similarly, since MK bisects ∠K, we know that the measure of ∠JKM is 30/2 = 15
degrees. Finally, since the sum of the measures of the angles of a triangle always equals
180 degrees, we know that the sum of the measures of ∠JKM, ∠KJM, and ∠JMK equals
180 degrees. Thus, the measure of ∠JMK = 180 − 30 − 15 = 135 degrees.
(4) 24 feet
ID: [05CC2]
√
√3 = 4 . Since the ratio of the
The ratio of the areas of the cross sections is equal to 216
9
486 3
area of two similar figures is the square of the ratio of their correspondingq
sides, it follows
that the ratio of the corresponding sides of the cross-sections is equal to 49 = 23 .
Now consider the right triangles formed by the apex of the pyramid, the foot of the
altitude from the apex to the cross section, and a vertex of the hexagon. It follows that
these two right triangles will be similar, since they share an angle at the apex. The ratio of
their legs in the cross-section is 2/3, so it follows that the heights of the right triangles are
in the same ratio. Suppose that the larger cross section is h feet away from the apex; then
h − 23 h = 8, so h3 = 8 =⇒ h = 24 feet.
√
(5) 8 2
ID: [3A253]
We create a diagram with the given information from the problem:
T
S
O
Q P
We draw in radii OT and P S and connect O and P . Then we drop a perpendicular from
P to OT that intersects OT at R:
T
R
S
O
P
∠OT S and ∠P ST are right angles as tangents create right angles with radii at points of
tangency. RT SP is a rectangle, and △ORP is right. We use Pythagorean theorem on
△ORP√: we have OP =√8 + 2 = 10 and OR = 8 − 2 = 6, so
RP = OP 2 − OR2 = 102 − 62 = 8. Then T S = 8 as well.
T
8
8
S
O
P
Finally, OS is the hypotenuse of right triangle △OT S with OT = T S = 8. Hence
√
√
OS = 82 + 82 = 8 2 .
(6) 47 sq units
ID: [BC403]
Draw a rectangle with vertices (−3, 7), (−3, −1), (6, −1), (6, 7) about the pentagon, as
shown below:
A2
A1
A3
A4
The area of the pentagon is the difference between the area of the rectangle and the
four right triangles formed at the vertices of the rectangle. We find that
1
A1 = · 5 · 2 = 5,
2
1
A2 = · 4 · 3 = 6,
2
1
A3 = · 2 · 5 = 5,
2
1
A4 = · 3 · 6 = 9,
2
while the area of the entire rectangle is 9 × 8 = 72. Thus, the area of the pentagon is
equal to 72 − 5 − 6 − 5 − 9 = 47 square units.
(7) 18
ID: [B1553]
From the given diagram, we can draw the following diagram:
r
Notice how we can split the regular hexagon into 6 equilateral triangles. In order to find
the area of the hexagon, we can find the area of one of the triangles and then multiply that
by 6. We can assign the following dimensions to the triangle:
√r
3
r
2r
√
3
Now we get that the area of hexagon is
1
2r
6r 2
6· ·r · √ = √ .
2
3
3
2
The area of that Fido can reach is πr . Therefore, the fraction of the yard that Fido can
reach is
√
3
(πr 2 )
=
π.
2
6r
6
√
3
Thus we get a = 3 and b = 6 so ab = 3 · 6 = 18 .
(8) 11
ID: [C34D3]
Consider point A at the center of the diagram. Drawing in lines as shown below divides the
region into 3 parts with equal areas. Because the full circle around point A is divided into 3
angles of equal measure, each of these angles is 120 degrees in measure.
A
Now consider a circle of radius 4 inscribed inside a regular hexagon:
B
M
C
4
A
O
D
F
E
Now, the pieces of area inside the hexagon but outside the circle are identical to the
pieces of area the original region was divided into. There were 3 pieces in the original
diagram, but there are 6 in the hexagon picture. Thus, the area of the original region is the
half the area inside the hexagon but outside the circle.
Because ABO is equilateral, BMO is a 30-60-90 right triangle, so BM = √43 . Thus, the
side length of the equilateral triangle is AB = 2BM = √83 . Now we know the base AB and
√
the height MO so we can find the area of triangle ABO to be 12 · √83 · 4 = √163 = 163 3 . The
entirety of hexagon
ABCDEF
so the area of
√
√ can be divided into 6 such triangles,
16 3
2
ABCDEF is 3 · 6 = 32 3. The area
√ of the circle is π4 = 16π. Thus, the area inside
the
√ heagon but
√outside the circle is 32 3 − 16π. Thus, the area of the original region is
32 3−16π
=
16
3 − 8π.
2
Now we have a = 16, b = 3 and c = −8. Adding, we get 16 + 3 + (−8) = 11 .
(9) 126 degrees
ID: [4A303]
Since AC and BD are line segments that intersect at point O, angle AOD and angle AOB
are supplementary angles and their angle measures must add up to 180 degrees. Since
angle AOD measures 54 degrees, the measure of angle AOB must be 180 − 54 = 126
degrees.
(10) 28 %
ID: [35CC2]
The formula for the volume of a right cylinder is given by πr 2 h, where r is the radius and h
is the height of the cylinder. If the diameter is increased by 25%, then the radius is
increased by the same percentage. Let r ′ and h′ be the new radius and height of the large
cylinder. Then,
2
5
2
2 · (πr h) = π ·
r · (h′ ),
4
2
so 2 · 542 · h =
28% .
32
25 h
= h′ . It follows that
h′
h
=
32
25
=
128
100
= 128%, so h′ has to increase by
(11) 18 square units
ID: [20403]
We first find the length of line segment F G. Since DC has length 6 and DF and GC have
lengths 1 and 2 respectively, F G must have length 3. Next, we notice that DC and AB are
parallel so ∠EF G ∼
= ∠EAB because they are corresponding angles. Similarly,
∼
∠EGF = ∠EBA. Now that we have two pairs of congruent angles, we know that
△F EG ∼ △AEB by Angle-Angle Similarity.
Because the two triangles are similar, we have that the ratio of the altitudes of △F EG
to △AEB equals the ratio of the bases. F G : AB = 3 : 6 = 1 : 2, so the the ratio of the
altitude of △F EG to that of △AEB is also 1 : 2. Thus, the height of the rectangle ABCD
must be half of the altitude of △AEB. Since the height of rectangle ABCD is 3, the
altitude of △AEB must be 6. Now that we know that the base and altitude of △AEB are
both 6, we know that the area of triangle AEB is equal to 12 base × height
= ( 21 )(6)(6) = 18 square units.
(12) 8 chords
ID: [AD253]
The longest chord through P is the diameter, which has length 20.
The shortest chord through P is the chord perpendicular to OQ; let the endpoints of this
chord be A and B as shown.
A
Q
O
P
B
△AOP is right with AO = 10 and OP = 6. Pythagorean theorem on this triangle yields
AP = 8, so AB = 2 · 8 = 16.
Thus, any chord through P has length at most 20 and at least 16. There is one chord
with length 16 (AB), one chord with length 20 (the diameter), and two chords each with
length 17, 18, and 19. Hence there are 8 chords in total.
√
ID: [3D103]
(13) 6 2 cm
If we let x = the side length of the triangle,
then we can find the area of the triangle in
√
terms of x and then set it equal to 16 3 to find x . The base of the triangle has length x .
To find the altitude, we notice that drawing an altitude splits the equilateral triangle into
two 30 − 60 − 90 triangles with the longest
√ of the side
√ side having length x . Since the ratio
x 3
length 2 and the
lengths of a 30 − 60 − 90 triangle
√ is 1 : 2 √3 : 2, the altitude will have √
x 3
x 3
1
area of the triangle will be 2 x 2 = 4 . Setting this equal to 16 3, we have that
√
√
x2 3
3.
=
16
4
Solving for x , we get that x = 8. Since the side length of the triangle is 8 and the square
and triangle have equal perimeters, the square has a side length of 8·3
4 = 6. If we draw the
diagonal of the square, we notice that it splits the square into two 45 − 45 − 90
√ triangles
with legs of length 6. A 45 − 45 − 90 triangle has side length ratios of 1 : 1 : 2, so the
√
diagonal of the square has length 6 2 cm.
(14) 35 inches
ID: [03453]
Since the smaller triangle has hypotenuse 5, we guess that it is a 3-4-5 triangle. Sure
enough, the area of a right triangle with legs of lengths 3 and 4 is (3)(4)/2 = 6, so this
works. The area of the√larger triangle is 150/6 = 25 times the area of the smaller triangle,
so its side lengths are 25 = 5 times as long as the side lengths of the smaller triangle.
Therefore, the sum of the lengths of the legs of the larger triangle is 5(3 + 4) = 35 .
[b]Proof that the only possibility for the smaller triangle is that it is a 3-4-5 triangle[/b]:
Let’s call the legs of the smaller triangle a and b (with b being the longer leg) and the
hypotenuse of the smaller triangle c. Similarly, let’s call the corresponding legs of the larger
triangle A and B and the hypotenuse of the larger triangle C. Since the area of the smaller
triangle is 6 square inches, we can say
1
ab = 6.
2
Additionally, we are told that the hypotenuse of the smaller triangle is 5 inches, so c = 5
and
a2 + b2 = 25.
Because 21 ab = 6, we get ab = 12 or a = 12
b . We can now write the equation in terms of b.
We get
a2 + b2 = 25
2
12
+ b2 = 25
b
122 + b4 = 25b2
Solving for b, we get
b4 − 25b2 + 144 = 0.
b4 − 25b2 + 144 = (b − 4)(b + 4)(b − 3)(b + 3) = 0.
Since we said that b is the longer of the two legs, b = 4 and a = 3. Therefore, the triangle
must be a 3-4-5 right triangle.
(15) 8 sq units
ID: [DC403]
6
4
6
2
3
1
A
2
4
6
B
Consider the three right triangles T1 , T2 , T3 formed by the line AB, the segment
connecting the bottom left corner of the smallest square to the upper right corner of the
largest square, and a side of the smallest, medium, and largest squares, respectively. Since
all three triangles share an angle, it follows that they must be similar. Notice that the base
of T3 is equal to 2 + 4 + 6 = 12, and its height is equal to 6. This, the height-to-base ratio
of each of T1 and T2 is equal to 6/12 = 1/2. Since the base of T1 is 2 and the base of T2
is 2 + 4 = 6, it follows that their heights are, respectively, 2 · (1/2) = 1 and 6 · (1/2) = 3.
The shaded region is a trapezoid with bases 1 and 3 and altitude 4, and area 4(1+3)
= 8.
2
(16) 45
ID: [530D3]
The area of the square is 62 = 36 square centimeters. The area of the four quarter-circles
with radius 3 is equivalent to the area of one circle with radius 3, or π · 32 = 9π. So, the
area of the shaded region is 36 − 9π. Thus, a = 36 and b = 9, so a + b = 45 .
(17) 100 square units
ID: [B2453]
The x and y axis of this graph break it down into four triangles each with the same area.
We find that the x and y intercepts of this graph are (0, 5), (0, −5), (10, 0), and (−10, 0).
This means that the area of each triangle is
1
· 5 · 10 = 25.
2
Therefore, the total area is 4 · 25 = 100 square units.
(18) 28 square units
ID: [41553]
When point A is reflected over the x -axis, we get point B, which is (3, −4). Reflecting
point B over the line y = x , we get that point C is (−4, 3). The distance between A and B
is 8. The distance from point C to the line connecting A and B is 7. Now we can draw the
following diagram:
y
A
x
C
B
We find that the triangle has a height of length 7 and a base of length 8. Therefore, the
area of triangle ABC is equal to
1
1
bh = · 7 · 8 = 28 .
2
2
√
(19) 3 33 (units)
ID: [0B253]
A
9
C
6
x
D
12
2x
B
We use Stewart’s on △ABC:
92 (2x ) + 122 (x ) = 62 (3x ) + (2x )(x )(3x ).
Simplifying yields 198x = 6x 3 ; the nonzero, positive solution is x =
√
BC = 3x = 3 33 .
√
33. It follows that
(20) 56
ID: [C30D3]
Constructing the complete exterior of a hollow cube with edges of 4 units and faces 1 unit
thick is the equivalent of taking the interior 2 × 2 × 2 cube out of a solid 4 × 4 × 4 cube.
So, the number of unit cubes left for the exterior is 4 · 4 · 4 − 2 · 2 · 2 = 64 − 8 = 56 .
√
(21) 2 13 units
ID: [45CC2]
Let P be the foot of the perpendicular from F to the line containing AB.
P A
B
F
X
C
E
D
Since ∠F AB = 120◦ , then ∠P AF = 180 − 120 = 60◦ , and it follows
√ that △P AF is a
30 − 60 − 90 triangle. As AF = 2, it follows that AP = 1 and P F = 3. Also, AB = 2
and so AX = 3AB = 6. Thus, P X = AP + AX = 7. In right triangle F P X, by the
Pythagorean Theorem, it follows that
√
F X 2 = P F 2 + P X 2 = ( 3)2 + (7)2 = 52,
√
√
and F X = 52 = 2 13 .
(22) 102 square units
ID: [4BCC2]
Since E and F are midpoints of the legs of the trapezoid, quadrilateral EF CD is a
trapezoid with half the altitude of the original trapezoid (the altitude of trapezoid EF CD is
12/2 = 6). The length of base CD is still 20, but now we have to find the length of base
EF . Since EF connects the midpoints of the legs of the trapezoid, its length is also the
average of the lengths of AB and CD. Thus, EF has length 8+20
2 = 14. Finally, we can
b1 +b2
find the area of the trapezoid with the formula Area = a
where a is the altitude and
2
b1 and b2 are the lengths of the bases. The area of trapezoid EF CD is
= 6 · 17 = 102 square units.
6 14+20
2
(23) 360 cu cm
ID: [D4CC2]
Suppose that the dimensions of the rectangular prism are given by x , y , and z , such that
x y = 30, y z = 180, and z x = 24. If we multiply all three equations together, we get that
x y · y z · z x = (x y z )2 = 30 · 180 · 24. Using prime factorizations, we find that the
right-hand side is equal to (2 · 3 · 5) × (22 · 32 · 5) × (23 · 3) = 26 · 34 · 52 . Thus,
(x y z )2 = (23 · 32 · 5)2 , so x y z = 360 . This is the formula for the volume of the box.
(24) 3/5
ID: [B1453]
The fly is within 6 feet of the ceiling when its distance from the ceiling is less than or equal
to 6 feet. Since the ceiling is 10 feet high, the probability that the fly is within 6 feet of the
3
6
ceiling is 10
.
=
5
(25) 20
ID: [D0503]
We begin by finding the equation of the line ℓ containing (1, 7) and (13, 16). The slope of
9
3
3
ℓ is 16−7
13−1 = 12 = 4 , so the line has the point-slope form y − 7 = 4 (x − 1). Substituting the
value x = 5, we obtain that y = 7 + 43 (5 − 1) = 10. It follows that the point (5, 10) lies on
the line containing (1, 7) and (13, 16) (for k = 10, we obtain a degenerate triangle). To
minimize the area of the triangle, it follows that k must either be equal to 9 or 11.
Indeed, we claim that both such triangles have the same area. Dropping the
perpendiculars from (5, 9) and (5, 11) to ℓ, we see that the perpendiculars, ℓ, and the line
segment connecting (5, 9) to (5, 11) form two right triangles. By vertical angles, they are
similar, and since they both have a hypotenuse of length 1, they must be congruent. Then,
the height of both triangles must be the same, so both k = 9 and k = 11 yield triangles
with minimal area. The answer is 9 + 11 = 20 .
(26) 6
ID: [BA253]
We draw the pentagon as follows, and draw altitude BG from B to AE. Notice that
AG = GB and ∠AGB = 90◦ .
A
G
E
2
B
2
2
D
C
F
We extend lines BC and ED past points C and D until they intersect
√ at F , creating
2
√
square GBF E. △CF D is a 45-45-90 triangle with CF = F D = 2 = 2. Thus, the side
√
√
length of the square is 2 +√ 2, so AG √
= BG = 2 + 2. It follows that
AE = AG + GE = 2(2 + 2) = 4 + 2 2, and finally a + b = 6 .
(27) (2, 4)
ID: [3BCC2]
When the point (0, 0) is reflected across the line x = 1, it is reflected to point (2, 0) since
the horizontal distance between the original point and the line is 1. If we reflect the new
point (2, 0) across the line y = 2, the vertical distance between the point and the line is 2,
so the coordinates of the resulting point are (2, 4) .
(28) 36 degrees
ID: [B0403]
Since AB = AC, triangle ABC must be an isosceles triangle and the measures of ∠ABC
and ∠ACB must be equal. Continuing, since BD bisects angle ABC, we have that the
measures of ∠ABD and ∠BDC are equal. Finally, since BD = BC, triangle BDC must
also be an isosceles triangle so the measures of ∠BDC = ∠BCD. Now if we consider
triangle BDC, we know that angles BDC and BCD have equal angle measures and angle
DBC has an angle measure that is half that of the other two. Since these three angle
measures must add up to 180◦ , we have that ∠DBC has measure 36◦ and angles BDC
and BCD have measures 72◦ .
Now, since ∠ABC ∼
= ∠ACB and ∠ACB has measure 72◦ , we know that ∠A must have
an angle measure of 180 − 72 − 72 = 36 degrees.
(29) 18 sq cm
ID: [C1CC2]
The area of a triangle is given by the formula 12 bh. Both △ABD and △ADC share the
same height AD. Let [ABD] be the area of △ABD and [ADC] be the area of △ADC. It
1
·BD·h
BD
4
3
3
2
follows that [ABD]
[ADC] = 1 ·DC·h = DC = 3 . Thus, [ADC] = 4 [ABD] = 4 · 24 = 18 .
2
(30) 25pi square inches
ID: [CC3D3]
Let r be the original radius of the pizza. If the diameter increase by 2 inches, then the
radius increases by 1 inch. We can calculate Aor iginal = πr 2 and Af inal = π(r + 1)2 . Using
the information given in the problem, Af inal = Aor iginal · 1.44. Substituting, we get
1.44πr 2 = π(r + 1)2
= π(r 2 + 2r + 1)
1.44r 2 = r 2 + 2r + 1
.44r 2 − 2r − 1 = 0
25(.44r 2 − 2r − 1) = 25(0)
11r 2 − 50r − 25 = 0
(r − 5)(11r + 5) = 0
5
11
r cannot be negative, so r = 5. This means the area of the original circle is π52 = 25π .
r = 5, −
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