Sat - MC Grawhill part 41 pdf

In a 5- 8-inch rectangular photograph, the image of a tree is 3 inches high.. The photograph is then mag-nified until its area is 1,000 square inches?. Cross-multiplying gives x= 15, so

1 If two figures are similar, then their corresponding sides are and their corresponding angles are

2 What are the three sets of conditions of which any one is sufficient to show that two triangles are similar?

a

b

c

3 The hypotenuses of two similar right triangles are 4 centimeters and 6 centimeters long, respectively If the area of the larger triangle is 27, what is the area of the smaller one?

Concept Review 6: Similar Figures

ᐉ1

ᐉ2

A

Note: Figure not drawn to scale.

4 In the figure above, ᐍ1⏐⏐ ᐍ2, AC = 4, BC = 5, and CE = 6 What is the length of DE?

5 In a 5-  8-inch rectangular photograph, the image of a tree is 3 inches high The photograph is then mag-nified until its area is 1,000 square inches What is the height of the tree image in the larger photograph?

1. The ratio of the areas of two squares is 4:1 If

the perimeter of the smaller square is 20, what

is the perimeter of the larger square?

(A) 5

(B) 10

(C) 20

(D) 40

(E) 80

2. A scale drawing of a rectangular patio

mea-sures 5 centimeters by 7 centimeters If the

longer side of the actual patio is 21 feet, what

is the area, in square feet, of the actual patio?

(A) 72

(B) 315

(C) 356

(D) 441

(E) 617

Note: Figure not drawn to scale

3. In the figure above, C and D are the centers of

the two circles with radii of 3 and 2,

respec-tively If the larger shaded region has an area

of 9, what is the area of the smaller shaded

region?

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

3

2

4. In the figure above, 艎1⎥⎥ 艎2 If EF = x, and

EG = y, then which of the following represents the ratio of CD to BC?

(A)

(B) (C)

(D)

(E)

5. A circular cone with a base of radius 5 has been cut as shown in the figure above What is the height of the smaller cone?

(A)

(B) (C)

(D)

(E) 104 5

96 5

96 12

96 13

8 13

8

5 5

1+x y

1−x

y

y

x−1

1+y

x

1− y

x

A

ᐉ1

ᐉ2

SAT Practice 6: Similar Figures

Note: Figure not drawn to scale.

6. In the figure above, what is the perimeter of

the shaded trapezoid?

6 3

10

Note: Figure not drawn to scale

7. In the figure above, BD ––– is parallel to EG –––,

AD = 6, DG = 4, and ΔAEF has an area of 75.

What is the area of ΔABC?

(A) 27 (B) 36 (C) 45 (D) 54 (E) 63

A

G F

E

D C

B

.

1

2

3

4

5

7

8

6

1

0

2

3

4

5

7

8

6

1 0

2 3 4 5

7 8 6

1 0

2 3 4 5

7 8 6

Concept Review 6

1 If two figures are similar, then their

ing sides are proportional and their

correspond-ing angles are equal (or congruent).

2 a two pairs of corresponding angles are equal

b two pairs of corresponding sides are

propor-tional and the included angles are equal

c all three pairs of corresponding sides are

proportional

3 The ratio of the sides is 4:6 or 2:3 The ratio of the

areas is the square of the ratio of sides, which is

4:9 If x is the area of the smaller triangle, then

x/27 = 4/9 Solving for x gives x = 12.

4 If ᐍ1⎥⎥ ᐍ2, then the two triangles must be similar Since corresponding sides are proportional,

AC/AE = BC/DE.

Substituting, this gives 4/10 = 5/DE.

Cross-multiply: 4DE= 50

Divide by 4: DE= 12.5

5 A 5-  8-inch rectangle has an area of 40 square inches The ratio of areas, then, is 40:1,000, or 1:25 This is the square of the ratio of lengths, so the

ratio of lengths must be 1:5 If x is the length of the larger tree image, then 3/x= 1/5 Cross-multiplying

gives x= 15, so the tree is 15 inches high in the larger photograph

Answer Key 6: Similar Figures

ᐉ1

ᐉ2

A

4 5

6

SAT Practice 6

1 D If the ratio of the areas is 4:1, then the ratio

of corresponding lengths is the square root: 2:1

If the perimeter of the smaller square is 20, then

the perimeter of the larger one is twice as big

2 B Find the width of the patio with a proportion:

5/7 = x/21

Cross-multiply: 7x= 105

So the patio is a 15- × 21-foot rectangle, which has

an area of 15 × 21 = 315 square feet

3 A The two regions are similar, because the central

angles are the same The ratio of their

correspond-ing lengths is 3:2, so the ratio of their areas is 9:4

Since the larger area is 9, the smaller area must be 4

4 C If EF has length x and

EG has length y, then FG

must have length y − x,

as shown Since

the two lines

are parallel,

ΔABC is

sim-ilar to ΔAEF

and ΔACD is

similar to

ΔAFG Therefore AC/CF = BC/x and

AC/CF = CD/(y − x) So BC/x = CD/(y − x),

and therefore CD/BC = (y − x)/x = y/x − 1.

5 B The height of the larger cone can be found with the Pythagorean theorem to be

12 (It’s the old 5-12-13 right triangle!) Since the two

trian-gles are similar, x/12 = 8/13

Multiplying by 12 gives

x= 96/13

6 28 The two

trian-gles are similar be-cause their corresponding angles are equal

Since they are right triangles, the missing sides can be found with the Pythagorean theorem Your diagram should look like the one above The perimeter is 3 + 8 + 5 + 12 = 28

7 A Since the lines are parallel, ΔABC is similar to ΔAEF and ΔACD is similar to

ΔAFG Therefore, AD/AG = AC/AF = 6/10 = 3/5.

The ratio of areas between

ΔABC and ΔAEF is the

square of the ratio

of sides, which is (3/5)2= 9/25 Since

ΔAEF has an area of 75, (the area of ΔABC)/75 = 9/25.

So ΔABC has an area of 27.

8

5 5

12

x

A

G F

E

D C

B

6 4

6 3

10 8 12

5

A

ᐉ1

ᐉ2

y

The SAT math section may include a question

or two about volumes Remember two things:

• The volume of a container is nothing more

than the number of “unit cubes” it can hold

• The only volume formulas you will need are

given to you in the “Reference Information”

on every math section

Example:

How many rectangular bricks measuring 2 inches

by 3 inches by 4 inches must be stacked together

(without mortar or any other material) to create

a solid rectangular box that measures 15 inches by

30 inches by 60 inches?

Don’t be too concerned with how the bricks could be

stacked to make the box; there are many possible

arrangements, but the arrangement doesn’t affect the

answer All you need to know is that it can be done If

so, just looking at the volumes is enough: if you use

n bricks, then the box must have a volume that is n

times larger than the volume of one brick Each brick

has a volume of 2 × 3 × 4 = 24 cubic inches The box

has a volume of 15 × 30 × 60 = 27,000 square inches

The number of bricks, then, must be 27,000/24 =

1,125

Lesson 7: Volumes and 3-D Geometry

3-D Distances

If you are trying to find the length of a line seg-ment in three dimensions, look for a right tri-angle that has that segment as its hypotenuse

Example:

The figure at right shows a cube with edges of length 4 If

point C is the midpoint of edge BD, what is the length

of AC –––?

Draw segment CE –– to see that AC ––

is the hypotenuse of right tri-angle ΔAEC.

Leg AE ––has a length of 4, and

leg EC –– is the hypotenuse of right triangle ΔEBC, with legs

of length 2 and 4 Therefore,

so

One possible shortcut for finding lengths in three dimensions is the three-dimensional dis-tance formula:

If you think of point A in the cube above as being the origin (0, 0, 0), then point C can be considered to be (4, 4, 2) The distance from A to C, then, is

d= (x2−x1) +(y −y) +(z −z)

2

2 1 2

2 1 2

AC= ( )202+42 = 20 16+ = 36=6

EC= 22+42 = 4 16+ = 20

A

4

A

4

4 2

E

4 0− 2 4 02 2 02 16 16 4 36 6

1 What is the definition of volume?

2 Write the formula for the volume of a rectangular box

3 Write the 3-D distance formula

4 Graph the points A( −2, 3, 1) and B(2, 1, −2) on an x-y-z graph.

5 What is the distance from point A to point B in the figure above?

6 The two containers with rectangular sides in the figure above have the interior dimensions shown Both con-tainers rest on a flat, horizontal surface Container A is filled completely with water, and then this water is poured, without spilling, into Container B When all of the liquid is poured from Container A into Container

B, what is the depth of the water in Container B?

Concept Review 7: Volumes and 3-D Geometry

12 inches

8 inches

8 inches

Container B Container A

4 inches

6 inches

10 inches

1. The length, width, and height of a rectangular

box, in centimeters, are a, b, and c, where a, b,

and c are all integers The total surface area of

the box, in square centimeters, is s, and the

volume of the box, in cubic centimeters, is v.

Which of the following must be true?

I v is an integer.

II s is an even integer.

III The greatest distance between any two

vertices of the box is

(A) I only

(B) I and II only

(C) I and III only

(D) II and III only

(E) I, II, and III

2. The figure above shows a rectangular box in

which AB = 6, AC = 5, AD = 8, and F is the

midpoint of BE What is the length of the

short-est path from A to F that travels only on the

edges of the box and does not pass through

ei-ther point B or point C?

(A) 27.5

(B) 28.5

(C) 29.5

(D) 30

(E) 30.5

A

B

C

F E

D

a2+b2+c2

3. A pool-filling service charges $2.00 per cubic meter of water for the first 300 cubic meters and $1.50 per cubic meter of water after that

At this rate, how much would it cost to have the service fill a rectangular pool of uniform depth that is 2 meters deep, 20 meters long, and 15 meters wide?

(A) $450 (B) $650 (C) $800 (D) $1,050 (E) $1,200

4. In the figure above, a rectangular box has the

dimensions shown N is a vertex of the box, and M is the midpoint of an edge of the box What is the length of NM –––?

(A) (B) (C) (D) (E) 125 108 98 77 63

M

N

5

6

8

SAT Practice 7: Volumes and 3-D Geometry

5. A cereal company sells oatmeal in two sizes of

cylindrical containers The smaller container

holds 10 ounces of oatmeal If the larger

tainer has twice the radius of the smaller

con-tainer and 1.5 times the height, how many

ounces of oatmeal does the larger container

hold? (The volume of a cylinder is given by the

formula V = πr2h.)

(A) 30

(B) 45

(C) 60

(D) 75

(E) 90

6. The figure above shows a rectangular solid

with a volume of 72 cubic units Base ABCD

has an area of 12 square units What is the area

of rectangle ACEF?

A

D

E F

3

7. The figure above shows a wedge-shaped hold-ing tank that is partially filled with water If the tank is 1/16 full, what is the depth of the water at the deepest part?

(A) 3 (B) 2 (C) 1.5 (D) 1 (E) 0.75

12

16 20

.

1

2

3

4

5

7

8

6

1

0

2

3

4

5

7

8

6

1 0

2 3 4 5

7 8 6

1 0

2 3 4 5

7 8 6

Concept Review 7

1 The volume of a solid is the number of “unit

cubes” that fit inside of it

2 V = lwh

3

4 Your graph should look like this one:

d= (x2−x1) +(y −y) +(z −z)

2

2 1 2

2 1 2

5 Using the 3-D distance formula,

6 Since the water is poured without spilling, the volume of water must remain the same Con-tainer A has a volume of 4 × 6 × 10 = 240 cubic inches Since Container B is larger, the water won’t fill it completely, but will fill it only to a

depth of h inches The volume of the water can

then be calculated as 8 × 8 × h = 64h cubic inches Since the volume must remain the same, 64h=

240, so h= 3.75 inches

d= ( − −( ) ) + −( ) + − −( )

= ( ) + −( ) + −( )

== 16 4 9+ + = 29

Answer Key 7: Volumes and 3-D Geometry

x y

z

A

3

1

−2

B

2

SAT Practice 7

1 E v = abc, so if a, b, and c are integers, v must be

an integer also and statement I is true The total

surface area of the box, s, is 2ab + 2bc + 2ac = 2(ab

+ bc + ac), which is a multiple of 2 and therefore

even So statement II is true Statement III is true

by the 3-D distance formula

3 D The volume of the pool is 2 × 20 × 15 = 600 cubic meters The first 300 cubic meters cost

300 × 2 = $600, and the other 300 cubic meters cost

300 × 1.50 = $450, for a total of $1,050

4 C Draw segment NP ––– as shown It is the hy-potenuse of a right triangle, so you can find its length with the Pythagorean theorem:

NP= 82+52 = 64 25+ = 89

A

B

C

F E

D

8

8

2.5

6

5

M

N

5

6

8

P

3

5

8 89

NM ––––is the hypotenuse of right triangle ΔNPM, so

.

NM= ( )892+32 = 89 9+ = 98

2 C The path shown above is the shortest under

the circumstances The length of the path is

8 + 6 + 5 + 8 + 2.5 = 29.5

5 C If the volume of the smaller container is

V = πr2h, then the volume of the larger container is

π(2r)2(1.5h) = 6πr2h = 6v So the larger container

holds six times as much oatmeal as the smaller

one The smaller container holds 10 ounces of

oat-meal, so the larger one holds 10 × 6 = 60 ounces

6 30 Mark up the diagram as shown If the base has

an area of 12, AB must be 4 If the volume of the

box is 72, then the height must be 72/12 = 6 AC

must be 5, because it’s the hypotenuse of a 3-4-5

triangle So the rectangle has an area of 5 × 6 = 30 7 Aof the tank, the smaller triangle must have an areaIf the volume of the water is 1/16 the volume

1/16 that of the larger triangle The two are simi-lar, so the ratio of the lengths must be 1/4, because 1/16 = (1/4)2

Therefore, the depth of water is 1/4 the depth of the tank: 12/4 = 3

A

D

E F

3

4 5 6