SAT Mc Graw Hill 2011

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:

dxx yy zz=−() 21 +−()+−()

2

21

2

21

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 hinches. The volume of the water can
   then be calculated as 8 × 8 ×h= 64 hcubic inches.
   Since the volume must remain the same, 64h=
   240, so h=3.75 inches.

d=−−()()+−()+− −()

= ()+−()+−()

2 2 13 21

423

(^222)
222

==++=16 4 9 29

Answer Key 7: Volumes and 3-D Geometry

x

y

z

A

3

1

− 2

B

2

1

− 2

SAT Practice 7

1. E v=abc,so if a, b,and care integers, vmust be
   an integer also and statement I is true. The total
   surface area of the box, s,is 2ab+ 2 bc+ 2 ac=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=+= += 8225 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 ∆NPM,so

NM= () 89 += += 3 89 9 (^98).
(^22)

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.

398 McGRAW-HILL’S SAT