Cracking the SAT subject test in math 2, 2nd edition

Cracking the SAT Subject Test in Math 2, 2nd Edition CHAPTER 7 PLANE AND SOLID GEOMETRY DRILL EXPLANATIONS Drill 1 Triangles in rectangular solids 17 D Use the formula for the long diagonal of a cube[.]

DRILL EXPLANATIONS

Drill 1: Triangles in rectangular solids

17 D Use the formula for the long diagonal of a cube Given that the

cube’s edge is the cube root of 27, or 3, the formula will be 32 +

32 + 32 = d2 If you simplify this, you’ll see that the cube’s long diagonal must be 5.2

22 E Be careful here; you won’t be using the Super Pythagorean

Theorem The sides of this triangle are the diagonals of three of

the solid’s faces BD is the hypotenuse of a 3:4:5 triangle, and

Theorem with a = 3 and b = 12: The sum of the three sides is 5 + 12.37 + 12.65 = 30.02

25 A This is a long-diagonal question, with a twist Each edge of the

cube is 1, but you’re actually finding the long diagonal of a quarter of the cube Think of it as finding the long diagonal of a rectangular solid with dimensions Plug those three numbers into the Super Pythagorean Theorem You could also

solve this by finding the length of CN and then using the Pythagorean Theorem on triangle MCN.

41 B Plug In! Make s = 2 Each side of ΔDEG is a diagonal of a square

with side length 2; therefore, DE, EG, and DG are all equal to 2 This makes ΔDEG an equilateral triangle Use the equation

, where s is the side of triangle, to find the area:

is the target Plug s = 2 into each

answer choice, and you find that only (B) equals 2

Drill 2: Volume

1 D PITA! Quickly move through the answer choices (starting in this

case with the smallest, easiest numbers), calculating the volumes and surface areas of each The only answer choice that makes these quantities equal is (D)

2 B If the cube’s surface area is 6x, then x is the area of one face.

Pick an easy number, and Plug In! Suppose x is 4 That means

that the length of any edge of the cube is 2, and that the cube’s

volume is 8 Just plug x = 4 into all of the answer choices, and

find the one that gives you 8 (B) does the trick Choice (E) is

the formula for the volume of a cube with edge x.

10 E This one’s a pain The only way to do it is to try out the various

possibilities The edges of the solid must be three factors which multiply to 30, such as 2, 3, and 5 That solid would have a surface area of 62 The solid could also have dimensions of 1, 5, and 6, which would give it a surface area of 82 Keep experimenting, and you find that the solid with the greatest surface area has the dimensions 1, 1, and 30, giving it a surface area of 122

15 C So you don’t know the formula for the area of a pentagon? You

= Bh.

26 C Plug In! Suppose the sphere’s original radius is r = 2, which

would give it a surface area of 50.3 If that radius is then

increased by b = 1, the new radius is 3 The sphere would then have a surface area of 113.1 That’s an increase of 62.8 Plug r =

2 and b = 1 into the answer choices; the one that gives you 62.8

is correct That’s (C)

35 B Just Plug In any values for b and h that obey the proportion b =

2h Then plug those values into the formula for the volume of a

pyramid

42 B The sphere has a volume of 4.19 When submerged, it will push

up a layer of water having equal volume The volume of this layer of water is the product of the area of the circular surface (50.27) and the height to which it’s lifted—it’s like calculating

the volume of a very flat cylinder You get the equation 50.27h = 4.19 Solve, and you find that h = 0.083.

Drill 3: Inscribed Solids

17 B The long diagonal of the rectangular solid is the diameter of the

sphere Just find the length of the long diagonal and divide it in half

19 C Calculate the volume of each shape separately The cube’s

volume is 8; the cylinder, with a radius of 1 and a height of 2, has a volume of 6.28 The difference between them is 1.72

25 B The cube must have the dimensions 1 by 1 by 1 That means that

the cone’s base has a radius of 0.5 and that the cone’s height is

1 Plug these numbers into the formula for the volume of a cone, and you should get 0.26

37 D If the volume of Cube A is 1,000, then the edge length is

That makes the diameter of Sphere S also 10 Cube B’s diagonal will be equal to 10; to find the edge of Cube B, use the Super Pythagorean Theorem: a2 + b2 + c2 = d2 Because

all the edges are equal, this becomes 3a2 = d2, or 3a2 = 102, so a

= 5.774 To find the surface area, use the formula SA = 6s2 : 6(5.7742) = 200, which is (D)

Drill 4: Rotation Solids

19 C This rotation will generate a cylinder with a radius of 2 and a

height of 5 Its volume is 62.8

29 D This rotation will generate a cone lying on its side, with a height

of 5 and a radius of 3 Its volume is 47.12

40 D Here’s an odd one The best way to think about this one is as

two triangles, base to base, being rotated The rotation will generate two cones placed base to base, one right-side-up and one upside-down Each cone has a radius of 3 and a height of 3 The volume of each cone is 28.27 The volume of the two together is 56.55

Drill 5: Changing Dimensions

3 E Just Plug In a value for the radius of sphere A, say 2 So the

radius of sphere B is 6 Use the volume formula: A has volume =

, and B has volume = If you

make a ratio, the cancels, and you have

5 D Plug In for the dimensions of the rectangular solid so that the

volume is 24 So pick l = 3, w = 4, and h = 2 The volume of the

solid which the question asks for will then be

10 B Plug In for the length of the edge of the cube—try 2 So the

surface area formula gives you 6(2)2 = 24, and the volume of the cube is 8 Increasing this by a factor of 2.25 gives you a new surface area of 54 Setting the surface area formula equal to this gives you a length of 3 for the new, increased edge So the new volume is 27 27 ÷ 8 gives you 3.375 for the increase between the old and new volumes, which approximates to (B)

33 C Plug In for the dimensions of the original cylinder Make the

radius 3 and the height 4 This makes the original volume = π(32)(4) = 36π = x Doubling the radius and halving the height makes r = 6 and h = 2 Find the new area: = π(62)(2) = 72π

Plug x = 36π into each answer choice; the only choice which

equals 72π is (C)

46 C The volume of a prism is Bh, where B is the area of the base and

h is the height perpendicular to that base Plug In for the hexagonal base and the height; make B = 3 and h = 2, so the

initial volume is 6 Doubling the dimensions of each rectangular face will double the length of each side of the hexagonal bases

If you double the dimensions of a 2-dimensional figure, you increase its area by a factor of four, so the new base area is 12 The lengths of the rectangular faces are also the height of the prism, so once the length is doubled, the new height is 4 Using the new base and height, you find that the new volume is (12) (4)= 48, which is 8 times bigger than the original volume of 6,

so choose (C)

3 A First, you need the height of the cone to find the volume To

find the height, use the radius and the given distance from the base to the apex to form a right triangle with a leg of 4 and a

hypotenuse of 5 Using the Pythagorean Theorem, a2 + 42 = 52; the height is 3 (Note this is a Pythagorean Triplet: 32 + 42 = 52 Recognizing these can save you time!) Next, use the formula for

volume of a cone: πr2h, so π(42)(3) = 50.265, which is (A)

14 D You could use the distance formula to find the sides of the

original rectangle, solve for area, triple the coordinates, find the distance, solve for area, and compare Or, remember your dimension change rules: change in area is the square of the change in length Therefore, if the lengths triple, the area increases by a factor of 9, which is (D)

19 E Draw it:

(8.196, 12) is 6 units above the rightmost vertex This creates a line segment parallel to and the same length as the two leftmost

points; this would create a parallelogram Eliminate (C) and (D) (8.196, 0) is similar, just 6 units below the rightmost vertex This point would also create a parallelogram; eliminate (A) Finally, (−2.196, 6) is 5.196 units to the left of the line created by (3, 3) and (3, 9), whereas (8.196, 6) is 5.196 units to the right Because this new point would create another equilateral triangle with (3, 3) and (3, 9), the resulting shape would be a parallelogram Choose (E)

25 C The best way to approach this problem is to draw it Draw two

hexagons and connect each vertex to its corresponding vertex in the other hexagon:

Then count the lines

Each hexagon is 6 edges, for a total of 12 The rectangular faces each share two edges with the hexagon bases, so you only need

to count the edges between each rectangular face There are 6 distinct edges between rectangular shapes, each between corresponding vertices of the hexagon bases This brings the total to 18, which is (C)

26 C First, the exposed surface area of the cylinder is the product of

the circumference and the height: 60 × 40 = 2,400 Eliminate (A) and (B), because (B) would only cover the exposed part of

the cylinder (and we haven’t figured out the cone yet), and (A) is even smaller than that To find the lateral surface area of the

cone, use the formula πrl, where r is the radius and l is the slant height Use 2πr with the circumference to find the radius: 2πr =

60, so r = 9.549 Make a right triangle with the radius and

height to find the slant height using the Pythagorean Theorem:

202 + 9.5492 = c2, so c = 22.164 The lateral surface area of the

cone is π(9.549)(22.163)=664.884 Add this to the cylinder to have a total volume of 3064.884; divide by 100 to find you need 30.649 gallons, (C)

32 A Plug In! Make x = 3 Using the volume of a sphere formula, you

find the volume of both shapes is π(3)3 = 36π Next, use the

volume of a cone formula, using y for the radius and x = 3 for the height: π(y)2(3) = 36π Solve for y and you find y = 6 Plug

in y = 6 into each answer and look for the choice which equals

3 Only (A) works; choose (A)

38 D To find the area of the equilateral triangle, all you need is a side

and the formula The distance between the first two points can be determined by using the distance formula:

If the side is 10, then the area is , which is (D)

41 D If the diameter is equal to the height and the radius is r, then 2r

= h The surface area is of a cylinder is 2(πr)2 + 2πrh This is equal to the volume, so 2(πr)2 + 2πrh = πr2h Plug In 2r for h, then solve for r:

To find the longest distance within the cylinder, make a right triangle with one leg the height and the other the diameter Because height equals diameter, they are both equal to 6 The hypotenuse of an isosceles right triangle is times the length

of the legs, so the longest distance is 6

If you were stuck on this question, you could Ballpark away a few answers If the diameter equals the height, and you want the longest distance within the cylinder, you know you’re looking for the hypotenuse of an isosceles right triangle From there, you can eliminate (A), (C), and (E), because the hypotenuse of the isosceles right triangle will be times the leg length At this point you can Plug In the Answers and see which answer fits the criteria of the question

45 A If the shape is rotated around AB, then the height of the

resulting cylinder will be the length of AB Only (A) has a length

of 7 for AB Alternatively, the segments perpendicular to AB will

be the resulting cylinder’s radius; if the cylinder has a diameter

of 9, you need a radius of 4.5 Once again, only (A) fits

48 B In polar coordinates, the first number is the distance from the

origin Because one point is the origin and the other two known

points are (2, ) and (4, 0), you know the two sides of the parallelogram are 2 and 4 gives you one angle in the

parallelogram because one side of the parallelogram is the x-axis (the θ value is 0) and the other side is at radians from the x-axis Use the formula for finding the area of a parallelogram using trigonometry: absinθ Therefore, the area is (2)(4) sin =

4, which is (B) If you chose (A), your calculator is in degrees, not radians!

50 A Plug In for s Make s = 2 To find the height of the pyramid,

create a right triangle One leg will be the height of the pyramid, the other leg will extend from a vertex of the base of the pyramid to the center of the base, and the hypotenuse will be the edge of the pyramid from the vertex of the base to the top

You already know the hypotenuse is s = 2 The base is a square;

the leg on the base is half the diagonal of the square If the side

of the square is 2, then the diagonal is 2 and the leg is

You now have the hypotenuse and one leg of the triangle; use the Pythagorean Theorem to find the remaining leg, which is the height of the pyramid: