Tài liệu Master the Gre 2010 - Part 30 pdf

You can apply the trapezoid formula to determine the area of this piece: A 5 8 312 2 5 48 The total area of the two shaded strips is 12 square feet, so the total area of the floor is 60

7 If a square’s sides are each increased by 50%, by what percent does the

square’s area increase?

(A) 75%

(B) 100%

(C) 125%

(D) 150%

(E) 200%

The correct answer is (C) Letting s 5 the length of each side before the

increase, area 5 s2 If 3

2s 5 the length of each side after the increase, the new area 5S3

2sD2

5 9

4s

2

The increase from s2to 9

4s

2

is5

4, or 125%.

GRE questions involving non-square rectangles also come in many possible flavors

For example, a question might ask you to determine area based on perimeter, or vice

versa

8 The length of a rectangle with area 12 is three times the rectangle’s width.

What is the perimeter of the rectangle?

inches Enter a number in the box

The correct answer is (16) The ratio of length to width is 3:1 The ratio 6:2 is

equivalent, and 6 3 2 5 12 (the area) Thus, the perimeter 5 (2)(6) 1 (2)(2) 5 16

A question might involve the properties of a square or rectangle as well as those of

another geometric figure, such as a right, isosceles, or equilateral triangle

9. A rectangular block of wood is to be cut into sections to form

either two triangular enclosures or three rectangular enclosures All wood sections must be equal in length on all sides

The total area of the two

proposed triangular enclosures

The total area of the three proposed rectangular enclosures

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

The correct answer is (A) Since the sections must all be equal in length, the

triangles must each be equilateral and the rectangles must each be square The

easiest way to make the comparison is to assign to the length of the original block

a number that is the least common multiple of 6 and 12 (the total number of sides

among the triangles and squares, respectively) That multiple, of course, is 12

Each triangle side = 2 Applying the area formula for equilateral triangles, the

area of each triangle is 2 3

4 3

2

= , and the total area for both triangles is

2 3, or about 3.4 (using 1.7 as an approximate value for 3) Given a block

length of 12, and 12 sides for three squares, the length of each side of each square

= 12

12=1 Accordingly, the area of each square is 1, and the total area of all three squares is 3

Or, a question might require you to determine a combined perimeter or area of adjoining rectangles

10.

In the figure above, all intersecting line segments are perpendicular

What is the area of the shaded region, in square units?

(A) 84 (B) 118 (C) 128 (D) 139 (E) 238 The correct answer is (C) The figure provides the perimeters you need to

calculate the area One way to find the area of the shaded region is to consider it

as what remains when a rectangular shape is cut out of a larger rectangle The area of the entire figure without the “cut-out” is 14 3 17 5 238 The “cut-out” rectangle has a length of 11, and its width is equal to 17 2 4 2 3 5 10 Thus, the area of the cut-out is 11 3 10 5 110 Accordingly, the area of the shaded region is

238 2 110 5 128

Another way to solve the problem is to partition the shaded region into three smaller rectangles, as shown in the next figure, and sum up the area of each

A GRE question about a non-rectangular parallelogram might focus on angle mea-sures These questions are easy to answer In any parallelogram, opposite angles are congruent, and adjacent angles are supplementary (Their measures total 180°.) So if

one of a parallelogram’s angles measures 65°, then the opposite angle must also

measure 65°, while the two other angles each measure 115°

11.

z y

WXZY is a parallelogram

r = 180 2 s

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

The correct answer is (D) In the figure, r° and s° are opposite angles, and

therefore r = s Given that r = 180 2 s, r and s must both equal 90 Since the

diagonals of WXZY form right angles, WXZY must be a rhombus (a parallelogram

in which all four sides are equal in length) However, p = q if and only if WXZY is

a square

A more difficult question about a non-rectangular parallelogram might focus on area

To determine the parallelogram’s altitude, you might need to apply the Pythagorean

theorem (or one of the side or angle triplets)

12.

In the figure above, AB i CD and AD i BC If BC is 4 units long and CD is

2 units long, what is the area of quadrilateral ABCD?

(A) 4

(B) 4=2

(C) 6

(D) 8

(E) 6=2

The correct answer is (B) Since ABCD is a parallelogram, its area 5 base (4)

3 altitude To determine altitude (a), draw a vertical line segment connecting

point A to BC, which creates a 45°-45°-90° triangle The ratio of the triangle’s

ALERT!

A GRE geometry figure will not necessarily be drawn to scale—unless the figure is accompanied by a note stating that it is.

TIP

A non-rectangular parallelogram in which all four sides are congruent (called a

rhombus) has the following in

common with a square:

Perimeter 5 4s;

Area 5 one half the product

of the diagonals.

hypotenuse to each leg is=2:1 The hypotenuse AB 5 2 Thus, the altitude (a) of

ABCD is 2

=2, or=2 Accordingly, the area of ABCD 5 4 3=2, or 4=2

Trapezoids

A trapezoid is a special type of quadrilateral The next figure shows a trapezoid Referring to this figure, all trapezoids share these four properties:

Only one pair of opposite sides are parallel (BC i AD)

The sum of the measures of all four angles is 360°

Perimeter 5 AB 1 BC 1 CD 1 AD Area 5BC 1 AD

2 3 altitude (that is, one half the sum of the two parallel sides multiplied by the altitude)

On the GRE, a trapezoid problem might require you to determine the altitude, the area, or both

To cover the floor of an entry hall, a 1' 3 12' strip of carpet is cut into two

pieces, shown as the shaded strips in the figure above, and each piece is

connected to a third carpet piece as shown If the 1' strips run parallel to

each other, what is the total area of the carpeted floor, in square feet?

(A) 46

(B) 48

(C) 52.5

(D) 56

(E) 60

The correct answer is (E) The altitude of the trapezoidal piece is 8 The sum

of the two parallel sides of this piece is 12' (the length of the 1' 3 12' strip before

it was cut) You can apply the trapezoid formula to determine the area of this

piece:

A 5 8 312

2 5 48

The total area of the two shaded strips is 12 square feet, so the total area of

the floor is 60 square feet

A GRE trapezoid problem might require you to find the trapezoid’s altitude by the

Pythagorean theorem

14. B C

3 5

4

120°

In the figure above, BC i AD What is the area of quadrilateral ABCD?

(A) 5=2

(B) 9=3 2

(C) 27=3

4

(D) 27

2

(E) 16 The correct answer is (C) The figure shows a trapezoid To find its area, first

determine its altitude by creating a right triangle:

3 5

120 60

3 3 2 4

√

°

This right triangle conforms to the 30°-60°-90° Pythagorean angle triplet Thus, the ratio of the three sides is 1:=3:2 The hypotenuse is given as 3,

so the trapezoid’s altitude is 3=3

2 Now you can calculate the area of the trapezoid:

S1

2D~4 1 5!S3=3

2 D5S9

2DS3=3

2 D527=3

4

For the GRE, you’ll need to know the following basic terminology involving circles:

Circumference: The distance around the circle (its “perimeter”).

Radius: The distance from a circle’s center to any point on the circle’s

cir-cumference

Diameter: The greatest distance from one point to another on the circle’s

circumference (twice the length of the radius)

Chord: A line segment connecting two points on the circle’s circumference

(a circle’s longest possible chord is its diameter, passing through the circle’s

center)

Finding a Circle’s Circumference or Area

In solving most GRE circle problems, you apply one or both of the following two

formulas (r 5 radius, d 5 diameter):

Circumference 5 2pr, or pd

Area 5 pr2

Note that the value of p is approximately 3.14, or22

7 For the GRE, you won’t need to work with a value for p any more precise In fact, in most circle problems, the solution

is expressed in terms of p rather than numerically

With the two formulas, all you need is one value—area, circumference, diameter, or

radius—and you can determine all the others For example:

Given a circle with a diameter of 6:

Radius 5 3

Circumference 5 (2)(3)p 5 6p

Area 5 p (3)2 5 9p

15 If a circle’s circumference is 10p centimeters long, what is the area of the

circle, in square centimeters?

(A) 12.5

(B) 5p

(C) 22.5

(D) 25p

(E) 10p

The correct answer is (D) First, determine the circle’s radius Applying the

circumference formula C 5 2pr, solve for r:

TIP

If answering a GRE question requires you to use a number value for p, rest assured that a rough approximation—e.g., a little greater than 3—will suffice.

10p 5 2pr

5 5 r Then, apply the area formula, with 5 as the value of r:

A 5 p~5!25 25p

Arcs and Degree Measures of a Circle

An arc is a segment of a circle’s circumference A minor arc is the shortest arc

connecting two points on a circle’s circumference For example, in the next figure, minor arc AB is the one formed by the 60° angle from the circle’s center (O)

A

B

A circle, by definition, contains a total of 360° The length of an arc relative to the circle’s circumference is directly proportionate to the arc’s degree measure as a fraction of the circle’s total degree measure of 360° For example, in the preceding figure, minor arc AB accounts for 60

360, or

1

6, of the circle’s circumference.

16.

Circle O, as shown in the figure above, has diameters of DB and AC and has

a circumference of 9 What is the length of minor arc BC?

7 2

(D)13

TIP

An arc of a circle can be

defined either as a length (a

portion of the circle’s

circumference) or as a degree

measure of the central angle

that determines the arc.

The correct answer is (C) Since AO and OB are both radii, DAOB is an

isosceles triangle, and therefore m∠BAO 5 70° It follows that m∠AOB 5

40° ∠BOC is supplementary to ∠AOB, therefore m∠BOC 5 140°

(Remember: angles from a circle’s center are proportionate to the arcs they

create.) Since m∠BOC accounts for140

360, or

7

18 of the circle’s circumference,

we have the length of minor arc BC 5S7

18D~9! 57

2.

Circles and Inscribed Polygons

A polygon is inscribed in a circle if each vertex of the polygon lies on the circle’s

circumference

A test question might require you to visualize the possible shapes or proportions of a

type of triangle or quadrilateral inscribed in a circle

17. An equilateral triangle is inscribed in a circle such that

each vertex of the triangle lies along the circle’s circumference

The length of any side of

the triangle

The circle’s diameter

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

The correct answer is (B) If any of the triangle’s sides were the length of the

circle’s diameter, that side would pass through the circle’s center In this case,

however, it would be impossible to construct an inscribed equilateral triangle:

Each of the other two sides would need to extend beyond the circle’s

circum-ference

The next figure shows an inscribed square The square is partitioned into four

congruent triangles, each with one vertex at the circle’s center (O)

O

Look at any one of the four congruent triangles—for example, DABO Notice that

DABO is a right triangle with the 90° angle at the circle’s center The length of each of the triangle’s two legs (AO and BO) equals the circle’s radius (r) Accordingly, DABO is

a right isosceles triangle, m∠OAB 5 m∠OBA 5 45°, and AB 5 r=2 (The ratio of the triangle’s sides is 1:1:=2.) Since AB is also the side of the square, the area of a

square inscribed in a circle is ~r=2!2, or 2r2 (The area of DABO isr

2

2, or one fourth the area of the square.)

You can also determine relationships between the inscribed square and the circle:

• The ratio of the inscribed square’s area to the circle’s area is 2:p

• The difference between the two areas—the total shaded area—is pr2

2 2r2

• The area of each crescent-shaped shaded area is1

4~pr

2

2 2r2!

The next figure shows a circle with an inscribed regular hexagon (In a regular polygon, all sides are congruent.) The hexagon is partitioned into six congruent triangles, each with one vertex at the circle’s center (O)

B

Look at any one of the six congruent triangles—for example, DABO Since all six

triangles are congruent, m∠AOB 5 60° (one sixth of 360°) You can see that the length

of AO and BO each equals the circle’s radius (r) Accordingly, m∠OAB 5 m∠OBA 5

60°, DABO is an equilateral triangle, and length of AB 5 r.

Applying the area formula for equilateral triangles: Area of DABO 5r

2=3

4 The area

of the hexagon is 6 times the area of DABO, or 3r

2=3

2 You can also determine

relationships between the inscribed hexagon and the circle For example, the

dif-ference between the two areas—the total shaded area—is pr223r

2=3

2 .