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SAT
GEOMETRY SOLUTIONSI. Two-Sided Figures
1. (D) 72 The degree measure around any point is
360º, just like a circle. Since there are 5 x’s here, 5 x
= 360º. Therefore x = 72º.
2. (A) 65 The angle that is opposite the 50º angle
is also 50º. That angle and the two angles labeled y
will add up to 180º. Therefore,
2 y + 50º = 180º
2 y = 130º
y = 65º
3. (C) 70 Since the lines are parallel, the angle
adjacent to the angle marked 2 x – 30 will equal x.
Since the degree measure of a line is 180º, an
equation can be formed.
x + (2 x – 30) = 180º
3 x – 30 = 180º
3 x = 210º
x = 70º
4. (B) 5 Since lines 1 and 2 are parallel, and x is
95º, the labeled angle must be 95º (See the figure
below.) Therefore, the labeled supplementary angle
must be 85º.
Since lines 3 and 4 are perpendicular, one of theother angles of the triangle must be a right angle.
All of the angles in a triangle must add up to 180º.
Therefore, the angle y can be found by:
y + 90º + 85º = 180º
y + 175º = 180º
y = 5º
5. (B) 80 Since l and m are parallel and w and b
are alternate interior angles, then w = b = 80º. The
angles x, w, and a must add up to 180º, because the
angles all comprise one side of a line. Since a = 20º
and w = 80º, solve for x.
x + 20º + 80º = 180º
x + 100º = 180º
x = 80º
Since x and y are corresponding angles, then x = y
= 80º. Now that the values for x, y and z are known,
find x + y – z.
x + y – z = 80 + 80 – 80 = 80
95º
85º
1
2
3
4
y°
x°
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6. (C) Six Draw 4 lines following the given rules.
Here is one example with the points of intersection
circled.
If drawn correctly according to the given rules there
will always be six points of intersection.
II. The 4 Types of Triangles
1. (D) 105 Remember that when lines intersect,
the angles opposite each other are equal. Therefore,
the angle opposite the 45º angle will also be 45º.Also remember that the sum of the angles in a
triangle is 180º. Therefore,
30º + 45º + xº = 180º
75º + xº = 180º
xº = 105º
2. (B) 105 If BAC = 80º, then ABC must equal
70º because the sum of the angles of any triangle is
180º. Since xº =1
2 BAC and yº =
1
2 ABC , xº = 40
and yº = 35. Also note that xº + yº + z º = 180.
Substitute in values for x and y and then solve for z .
40º + 35º + z º = 180º
z º = 105º
3. (A) 50 Since AB = BC , the triangle is isosceles.
Therefore, BAC = ACB = xº. Since the sum ofthe angles of a triangle equal 180º, make an
equation.
xº + xº + 80º = 180º
2 xº + 80º = 180º
2 xº = 100º
xº = 50º
III. Area of a Triangle
1. (D) 36 The area of a triangle is equal to1
2bh.
Use this formula to solve for height, given the area
of 60 and the base of 10.
60 =1
2(10)h
60 = 5h
12 = h
Since the height is always perpendicular to the base,it creates a right triangle as shown below. Also, note
that the base of the right triangle will be 5, because
the height bisects the base of the isosceles triangle.
Use Pythagorean Theorem to solve for x.
52 + 122 = x2
25 + 144 = x2
169 = x2
13 = x
The perimeter equals x + x + 10 or 13 + 13 + 10 =
36.
2. (A) AOX Since all the triangles have the
same base (OX ), the determining factor is the height.
Since AOX has the least height, it also has the
least area.
IV. The Right Triangle
1. (C) 30 + 12 2 The figure is comprised of two
right triangles. The height of the figure can be found
by using Pythagorean Theorem with the base and
hypotenuse of the left-side triangle. The height ofthe figure is labeled as h.
5
12 x
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52 + h2 = 132
25 + h2 = 169
h2 = 144
h = 12
The right-side triangle has one 90º angle and one 45º
angle. Therefore, the remaining angle must be 45ºand the triangle is a 45º-45º-90º triangle. Since it is a
45º-45º-90º triangle, the remaining two sides of the
triangle are 12 and 12 2 . See the figure below for
all labeled sides.
The perimeter will 5 + 12 + 13 + 12 2 =
30 + 12 2 .
2. (D) 30 3
The area of a triangle is1
2bh. First,
find the height of the triangle. Since the height is
always perpendicular to the base, labeling the height
creates a right triangle. One of the angles is 60º and
the other is 90º, so the remaining angle is 30º.
Therefore, the triangle is a 30º-60º-90º triangle asshown below.
The height of this 30º-60º-90º triangle must be 4 3
because is corresponds to the 60º angle. The area of
the triangle can be found using the formula.
A =1
2bh
A =1
2(15)( 4 3 )
A = (15)( 2 3 )
A = 30 3
V. Similar Triangles
1. (A) 1 Because the two triangles share two
angles (90º and xº), they are similar. Because BC =
CD, the two triangles are not only similar, they are
identical because those sides correspond to oneanother. Since ABC is a right triangle, use the
Pythagorean Theorem to solve for BC .
32 + BC 2 = 52
9 + BC 2 = 25 BC 2 = 16
BC = 4
Since the triangles are identical, then AC = CE = 3.
The length of BE will be equal to the length of BC
minus the length of CE .
BE = BC – CE
BE = 4 – 3 = 1
2. (D) 12.5 Set up a proportion of height raised to
horizontal distance of the corresponding sides. Let x
be the height.
1
8 100
x
100 = 8 x
12.5 = x
3. (A) 5 Since AB = 1 and CD = 3, the two
triangles have a ratio of 1 : 3. The line segments x
and 3 x can then be labeled, since the length of the
larger segment must be 3 times the length of the
smaller. Also, label the hypotenuse of the smaller
triangle as c. See the figure below.
12
12
30º
A B
C D
x
3 x
c
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Since x + 3 x is 4 x and the sum of the line segments
is CB which is 3,
4 x = 3
x =
3
4
Since x =3
4, the length of the hypotenuse of the
smaller triangle can be found with the Pythagorean
Theorem.
c2 =
23
4
+ 12
c2 =9
16+ 1
c2 = 2516
c =5
4
Since the hypotenuse of the larger triangle must be 3
times as long, the length is
5
4× 3 =
15
4
The total length of AD is the sum of the hypotenuses
of both triangles.
5
4 +
15
4 =
20
4= 5
ALTERNATE SOLUTION
A right triangle can be created by drawing the lines
shown below.
This right triangle will have a height equal to CB,
which is 3. The base of the triangle will be equal to
AB + CD, which is 1 + 3 = 4. Using the
Pythagorean Theorem, the length of the hypotenuse
AD can be found.
AD2 = 32 + 42
AD2 = 9 +16
AD2 = 25
AD = 5
VI. Three More Rules
1. (A) 21 and 29 The third, unknown side of this
triangle must be greater than the difference of the
two known sides. The third side must be less than
the sum of the two known sides. Therefore, where x
is the third side,
10 – 5 < x < 10 + 5
5 < x < 15
Since the perimeter is an integer (positive or
negative whole number) x must be a whole number.
Since 5 < x < 15, the largest integer value for x is14. The smallest integer value for x is 6.
If x is as large as possible, the perimeter of the
triangle will be 5 + 10 + 14 = 29. If x is as small as
possible, the perimeter of the triangle will be 5 + 10
+ 6 = 21.
VII. Four-sided Figures
1. (D)
2
2 y
The area of a rectangle equals lw and in
this case, l = x and w = y. Therefore the area is equal
to xy. Since x is half of y, or as an equation x =2
y,
the area is equal to
A B
C D
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2
y(y) =
2
2
y
2. (C) 540º The indicated angles are equal to the
interior angles of the pentagon because they are
corresponding angles. The sum of the angles for any
polygon is equal to (180º)(n – 2) where n is the
number of sides. Since n = 5 in this case, the total
number of degrees equals (180º)(3) or 540º.
3. (B) 8 The perimeter of any rectangle equals 2l+ 2w. In this case, l = 3 x and w = 2 x. Plug these
values into the formula and set it equal to 80.
2l + 2w = 80
2(3 x) + 2(2 x) = 806 x + 4 x = 80
10 x = 80
x = 8
4. (C)18 Divide the shape into 3 rectangularregions as shown below.
Area A is equal to 3 × 3 = 9. Area C is equal to 4 ×2 = 8. Next, find the dimensions of area B. The
height must be equal to 3 – 2 = 1. The base must be
equal to 8 – (4 + 3) = 8 – 7 = 1. Therefore area B is
equal to 1 × 1 = 1. The total area of the figure is
equal to A + B + C = 9 + 8 + 1 = 18.
5. (C) 320 To find the number of squares in the
shaded area, divide the shaded area by the area of
each square.
80 100 8,000
5 5 25
= 320
VIII. Circles
1. (A) 9 Find the area of each circle using the
formula “ Area = r 2”.
r = 6 : (6)2 = 36
r = 2 : (2)2 = 4
Then, divide the larger area by the smaller area.
36
4
= 9
2. (A)9
2
The region that is described is a
semicircle (half of a circle) with a radius equal to the
length of AB, 3. See the figure below.
The area of a semicircle is equal to1
2the area of a
circle, which is “r 2”.
2 2(3) 9
2 2 2
r
3. (D) 112.5º Since the indicated angle is 5 marks
wide, and there are a total of 16 marks, the angle
must be5
16
of the degree measure of a circle. A
circle has a degree measure of 360º. To find the
degree measure of this angle, multiply5
16 by
360º.
5
16× 360º = 112.5º
4.(C)3
The length of an arc is equal to the
circumference multiplied by the degree measure of
the arc divided by 360º. The circumference of acircle equals “2r ”. Solve using the given values.
30
360
× 2(2) =
1
12× 4 =
4
12
=
3
5. (B) 4
The polygon is equilateral, meaning all
the sides are equal. The vertices of the polygon are
A B A
180º
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equally spaced apart. Therefore, the vertices divide
the circle in to equal segments. Here, the circle is
divided into 5 equal segments, and the arc PQR is 2
of those segments. The arc is then2
5 of the entire
circumference.
The arc length can then be solved by multiplying2
5
by the circumference. Remember that circumference
can also be written as “d .”
2
5× d =
2
5× 10 =
20
5
= 4
6. (D) 6r 2 The length and width of the rectangle
can be determined by using the radii of the two
identical circles inscribed inside the rectangle. Seethe diagram below.
Since the length is 3r and the width is 2r , the area
can be found by
3r × 2r = 6r 2
7. (D) 60 First, draw a diagram of the fan.
The length of the fan blade is the radius of a circle.
Find the distance traveled in one revolution by
finding the circumference of the circle.
C = 2r = 2(1.5 ft) = 3 ft
Since the fan spins 1,760 revolutions per minute, the
fan travels
(3 ft)(1,760) = 5,280 ft per minute
Since there are 5,280 feet per mile, the fan travels
miles per minute. There are 60 minutes in an hour,
the fan travels 60 miles per hour.
IX. Solid Figures
1. (D) 128 Start with the formula for volume of a
rectangular solid, V = lwh. Since the volume is 64
and the width is 12
,
64 =1
2lh
Multiply both sides of the equation by 2 to get
128 = lh
Since the area of the shaded face is equal to the
height of the figure times the length of the figure, the
area is 128.
2. (D) 216 Since the surface area of the cube is
given, start with the formula for surface area of a
cube, “Surface Area = 6 s2.” Using this formula,
solve for the length of the side, s.
6 s2 = 216
s2 = 36
s = 6
Now that the length of the side is known, find thevolume of the cube using the formula “Volume = s3.”
s3 = 63 = 216
(Note: The volume and surface area of a cube are
not typically the same number. This occurred here
only because the length of the side is 6. For any
r r
r
r
r
3r
2r
1.5 feet
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other given side length of a cube, the volume and
surface area will be different values.)
3. (A) 1 The maximum height of the cylinder
would be 10 in. Therefore, the 6 in. will form the
circumference of the base. Since the shape of the
base is a circle and the circumference is 6 in, use theformula “C = 2r ” to solve for the radius.
2r = 6
r =6 3
2 1
(Note: This problem did not ask for an exact value
of the radius. Approximate 3 for π.)
4. (D) 2l + 2w + 4h + 20 The ribbon crosses thelength of the box twice, the width twice, and the
height 4 times. The total length of the ribbon is then
twice the length, twice the width, and 4 times the
height of the box plus 20 inches for the bow.
Mathematically, this translates into
2l + 2w + 4h + 20
X. Advanced Volume – Solutions
1. (B) 38 cm3 The volume of a right circular cone is
given by the formula21
3V r h . Using this formula
and the information given by the diagram, compute
the volume. The height of the cone is 4 cm, so h = 4,
and the diameter is 6 cm, so the radius r = 3. Plug
these numbers into the formula to obtain:
V =1
3π (3)2(4) =
1
3π (36) = 12π.
Given the fact that π ≈ 3.14, 12π = 12(3.14) ≈ 38
cm3, or (B).
2. (B) 6 cm The question gives the formula for the
volume of a sphere and then gives the volume of one
such sphere. Given this information, it is possible to
compute the radius, since the question asks for the
diameter.
Set the formula equal to the given volume and solve
accordingly:
4
3πr
3 = 36π
4
3r 3 = 36
r 3 = 27
r = 3
The radius is 3 cm. However, note that the question
asks for the diameter , NOT the radius. To obtain the
diameter, simply double the radius. The diameter is
6 cm, or (B).
3. (C) 6 Since the container is filled to the top, the
first step to finding the weight of the slag is to
calculate the volume of the container. The container
is a half cylinder, so the formula for solving the
volume is V =1
2πr 2h. In the diagram, the height (10
ft.) and the diameter (8 ft.) are given, so r = 4 (half
of the diameter) and h = 10. Plug these numbers into
the formula:
V =
1
2 π (4)2
(10) =
1
2 π (160) = 80π ft3
Now that the volume has been obtained, calculate
the weight of the slag. The slag weighs 50 pounds
per ft3, so the weight of the slag in pounds is 50(80π)
= 4000π pounds.
Note, however, that the question asks for the weight
of the slag in tons. The conversion of 1 ton = 2000
pounds has been provided, so the weight of the slag
is 2π tons, which is about 6 tons, or (C).
XI. Coordinate Geometry
1. (D) (4, 3) The midpoint of any line segment is
equal to the average of the x-coordinates and the y-
coordinates. Find these averages.
( x, y) =4 4 2 8 8 6
, ,2 2 2 2
(4, 3)
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2. (B) l A slope is positive if it moves upward
from left to right. A slope is negative if it moves
downward from left to right. The greatest slope will
be one which is positive and has the sharpest incline
upwards. The least slope will be the one which is
negative and has the sharpest decline downwards.Line r has a negative slope. Line l has a positive
slope. Lines m and n are both positive, but neither
have a sharper incline than l . Line p has a slope of 0.
Therefore, line l has the greatest slope.
3. (D) 3 Using the formula for slope, “m =
2 1
2 1
y y
x x
”, solve for the missing coordinate.
m = 2 1
2 1
y y
x x
2
3 =
7
4 ( 2)
w
2
3 =
7
6
w
From here, cross multiply.
12 = 3(7 – w)
12 = 21 – 3w
3w = 9
w = 3
4. (C) 4 The longer side of the rectangle is equal
to the greater difference of the coordinates. The
difference between the x-coordinates is 5 – 1 = 4.
The difference between the y-coordinates is 4 – 2 =
2. Since 4 is greater than 2, the longer side is 4.
2 1
2 1
12 7 5
8 6 2
y y
x x
XII.Area of Many-SidedPolygons
SAMPLE PROBLEM: (A) 7 The sum of the
angles in an octagon is
(8 – 2)180º = 1080º
Since the figure is equiangular and there are 8
angles, every angle in the figure is equal to
1080
8
= 135º
Since every angle is equal to 135º , every angle can
be divided into a 45º and 90º angle. This allows the
octagon to be divided into five rectangles and four45º-45º-90º right triangles as shown in the figure.
Since each 45º-45º-90º right triangle has a
hypotenuse of 2 , the sides must have a length of 1.
Therefore, the area of each triangle is
A =1
2bh =
1
2(1)(1) =
1
2
Remember that the legs of every right triangle have
a length of 1. This means that the rectangles in the
figure are actually squares with a side length of 1.
Therefore, the area of each square is
A = s2 = (1)2 = 1
Since there are five squares and four triangles, the
entire figure has an area of
5(1) + 41
2
= 5 + 2 = 7
1
11
1
45º
45º
135º
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1. (A) 54 3 The sum of the angles in a hexagon
is
(6 – 2)180º = 720º
Since the figure is equiangular and there are 6
angles, every angle in the figure is equal to
720º
6 = 120º
Therefore, the figure can be divided into six
equilateral triangles as shown in the figure.
The base of each triangle is 6. The height of the
triangle can be determined by dividing the
equilateral triangle into two 30º-60º-90º triangles as
shown below.
The height of the 30º-60º-90º triangle must be 3
times the side opposite the 30º angle. Therefore the
height is 3 3 . Now, the area of each triangle can be determined.
A =1
2bh =
1
2(6) 3 3 = 9 3
Since there are six equilateral triangles, the total area
is
(6) 9 3 = 54 3
XIII. Trigonometry – Solutions
1. (A) 3 Recall that sin( x) =opposite
hypotenuse. Therefore,
sin( P ) =QR
QP . Since sin( P ) = 0.5 and QP = 6, this
means that
6
QR = 0.5
Multiply both sides by 6.
QR = 3
Solving this equation gives QR = 3, so (A) is the
answer.
2. (C) x Recall that cos( x) =adjacent
hypotenuse.
Therefore, in the diagram, cos = x
r . Multiply both
sides by r .
cos =r
r cos = x
The answer is (C).
3. (A) 2 Recall that sin( x) =opposite
hypotenuseand cos( x)
=adjacent
hypotenuse
. The question asks for the value of
sin( ) cos( )
cos( )
A B
B
. Solve for each piece. The sin( A)
=a
cusing the diagram. The cos( B) =
a
caccording to
the figure. Therefore:
120º
60º6
6
6
6
6
6
60º
6
3
30º
60º
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sin( ) cos( )
cos( )
A B
B
=
a a
c ca
c
Simplify by combininga
c
+a
c
.
a a
c ca
c
=
2a
ca
c
Remember that dividing by a fraction is equal to
multiplying by the reciprocal.
2a
ca
c
=2a c
c a
Simplify.
2a c
c a = 2
(A) is the answer.
XIV. Equation of a Circle –
Solutions
1. (B)2
( 5) ( 3) 16 x y The equation is a circle
is given by the formula ( x – x0)2 + ( y – y0)
2 = r 2,
where ( x0, y0) is the center of the circle and r is the
radius. The center of the circle is (5, -3) and the
radius is 4. Therefore, the equation of this circle is ( x
– 5)2 + ( y + 3)2 = (42) = 16, or (B).
2. (D) (8, -3) To find the coordinate of the other end
of the diameter, note that to move from the point A
(2, 1) to (5, -1), we moved 3 units right and 2 unitsdown. To find the other coordinate, move 3 units
right and 2 units down from the center point (5, -1)
to obtain (8, -3), or choice (D). This works because
the two points of the diameter are each the same
distance from the center of the circle and both lie on
the same line with the same slope.
Another way of solving this problem is to note that
the center of the circle is the midpoint of the two
points on the diameter. To find the midpoint,
average the x values and take the average of the y
values. This means that (2+x)/2 = 5 and (1+y)/2 = -
1. You can solve for ( x,y) to find that the other pointis at (8,-3).
3. (A) (0, 10) and (0, -10) The equation of a circle
centered at the origin is x2 + y2 = r 2. Given the
equation x2 + y2 = 100, 100 = r 2 and r = 10. The
points where this circle crosses the y-axis are given
by (0, r ) and (0, -r ). Since r = 10, the two points are
(0, 10) and (0, -10), so (A) is the correct answer.
MATH TRICKS
I. Whole minus Part
SAMPLE PROBLEM 1: (C) 4 –2
The shaded
region is equal to a square minus a semicircle. The
square has side lengths of 2. The semicircle has a
diameter of 2, so it has a radius of 1. The area of the
shaded region can now be found by
Area of square – Area of semi-circle
s2 –2
2
r
(2)2 –2(1)
2
4 –2
SAMPLE PROBLEM 2: (D) mn –2
4
m The
shaded region is equal to a rectangle minus two
semicircles. The width of the rectangle is m and the
length is n. The diameter of the semicircles is m, so
the radius is
2
m. Also, two semicircles are equivalent
to one whole circle. Therefore, the area of the
shaded region can be found by
Area of rectangle – Area of circle
lw – πr 2
mn – π2
2
m
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mn –2
4
m
1.(D) π The shaded area is equal to a semicircle
minus the non-shaded area in the middle. Since a
circle is 360º, the sector with 90º must be90
360
or
1
4 of the circle. Therefore, the area of the shaded
region must be equal to a semi circle minus one
fourth of the circle.
Area of semicircle – Area of1
4 of circle
2 2
2 4
r r
2 2
(2) (2)2 4
4 4
2 4
2π – π = π
2. (A) 4 – π The shaded area is equal to the area of
the square minus a section of the circle. Since all
angles in a square are 90º, the section of the circle
must have an angle of 90º. A circle is 360º, so this
section of the circle must be90
360
or
1
4 of the circle.
Therefore, the shaded region is the area of the square
minus1
4 of the circle. Also, notice that the radius
of the circle is equal to the length of a side of the
square.
Area of the square – Area of1
4 of circle
s2 –2
4
r
(2)2 –
2
(2)4
4 –4
4
= 4 – π
3. (B) 9π – 18 Since the marked angle in the
figure is 90º, this section of the circle must be90
360
or1
4 of the circle. The shaded area is equal to one
fourth of the circle minus the area the right triangle
ABC . Notice that the base and height of the triangle
ABC are both equal to the radius of the circle.
Area of1
4 of circle – Area of triangle
2
4
r –
1
2bh
2(6)
4
–
1
2(6)(6)
36
4
– 18 = 9π – 18
II. Guesstimating
SAMPLE PROBLEM 1: (A) 50(π – 2) The
shaded area is equal to the square ABCD minus the
two white “curved triangles.” First, find the area of
one of the “curved triangles.” One “curved triangle”
equals the area of the square minus the area of1
4 a
circle with radius 10.
Area of square – Area of1
4
a circle
s2 –2
4
r
(10)2 –2(10)
4
100 –100
4
100 – 25π
Again, the area of the shaded region is equal to the
area of the square minus the area of two “curved
triangles.”
Area of the square – Area of two “curved triangles”
s2 – 2(100 – 25π)
102 – 2(100 – 25π)
100 – 200 + 50π 50π – 100
50(π – 2)
SECTION AA: Solutions For SAT Geometry
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SAMPLE PROBLEM 2: (A) 6 3 – 6 In the
triangle ABD, if angle A is 30º and angle B is 90º,
angle D must equal 60º because the sum of the
angles of a triangle equals 180º. Therefore, triangle
ABD is a 30º-60º-90º right triangle. If segment AD is
12, then segment BD is 6 and segment AB is 6 3 .
Since triangle ABC is an isosceles triangle, segment
BC is equal to segment AB which is 6 3 . Notice
that segment DC is equal to segment BC minussegment BD.
DC = BC – BD
DC = 6 3 – 6
1. (C) 120 The angle above the line is equal to 90º
minus 30º, or 60º. Remember that the degree
measure of a straight line is 180º. Therefore,
xº + 60º = 180º xº = 120º
2. (C) 90 Remember that when lines intersect, two
pairs of equal angles are formed. The equal angels
are shown below.
Remember that a triangle has 180º. Therefore,
xº + 45º + 45º = 180º
xº + 90º = 180º
xº = 90º
3. (B) 45 This problem is designed to test the
ability to guesstimate angles. (Note: This type of
problem will not appear on an actual exam. It is only
an exercise to help practice guesstimating.) This
triangle appears to have a 90º angle and two equal
angles, of which x is one of them. It therefore
appears to be a 45º-45º-90º triangle and the best
answer is 45º.
4. (C) 40 Since MN is parallel to OP , angle ACP
is equal to angle xº, which is 100º. Since the triangle
is equilateral, angle ACB is 60º. Notice that angle
ACP is equal to angle ACB plus angle yº.
ACP = ACB + yº
100º = 60º + yº40º = yº
5. (D) 360 Label the interior angles of the triangle
as aº, bº, and cº as shown below.
Since a straight line is 180º, notice that each pair of
angles ( xº and bº, yº and aº, z º and cº) adds up to
180º. Therefore, an equation can be formed.
aº + bº + cº + xº + yº + z º = 180º + 180º + 180º
aº + bº + cº + xº + yº + z º = 540º
Since a triangle is 180º, then aº + bº + cº = 180º.
Now, replace aº + bº + cº with 180º.
180º + xº + yº + z º = 540º
xº + yº + z º = 360º
Guesstimating – PracticeProblems
(Note: Problems 1 through 9 are designed to test the
ability to guesstimate. Problems of this sort will not
appear on an actual exam, but the ability to
guesstimate may help with certain problems on an
exam.)
1. (D) 75 The angle here is less than 90º butgreater than 45º. While 60º is an acceptable guess,
the angle is closer to 75º.
SECTION AA: Solutions For SAT Geometry
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2. (A) 120 The angle here is greater than 90º.
While 95º is an acceptable guess, the angle is closer
to 120º.
3. (E) 90 The angle here appears to be exactly 90º.
4. (A) 15 The angle made up of both x’s looks likeabout 30º together, so each x is likely 15º.
5. (A) 90 The angle here appears to be 90º.
6. (C) 175 The angle here is a little less than half
of a circle. Since a circle is 360º, half of a circle is
180º. Therefore, the angle here is a little less than
180º , making the best guess 175º
7. (B) 120 The angle here appears to be1
3 of a
circle. Since a circle is 360º,1
3 of a circle is 120º.
Therefore, the angle here is likely 120º.
8. (C) 45 The angle here is less than 45º. While
15º is an acceptable guess, the angle is closer to 30º.
9. (D) 720 The triangle appears to be equilateral,
meaning each angle is 60º. Each of the indicated
angles would then be 120º because the degree
measure of a line is 180º, so each angle would be180º – 60º = 120º. The sum would then be 6 × 120º= 720º.
SECTION AA: Solutions For SAT Geometry
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