67 terms

# Circle

#### Terms in this set (...)

Together with triangles, circles comprise most of the GMAT Geometry problems.

A circle is the set of all points on a plane at the same distance from a single point ("the center").

The boundary line of a circle is called the circumference.
There are 360° around the center point of a circle.

If you see a circular region on the GMAT you may assume it is a circle. Also, if a point in a diagram seems inside, outside, or on the circumference of the circle, you may assume it is that way.

However, do not assume that a point inside a circle is its center, unless the problem specifically states so.

We will talk about the different properties of circles and the way they are tested on the GMAT in the future.
Q1...Note: Figure NOT drawn to scale.

If point A is found outside of a circle centered at P, and point B is found inside that circle, and the radius of the circle is 2, which of the following must be true?

PA / PB > 2

PA / PB > 1

PA − PB = 2

PB / PA = 2

PA2 + PB2 = 4
B

Correct.

PA is longer than the radius, while PB is shorter than the radius, therefore, PA must be longer than PB, thus PA/PB has to be greater than 1.
A radius is the distance from the center of the circle to its boundary line. All radii of the same circle are equal to each other.
A chord is a line between any two points on the outer line of the circle.
A diameter is a chord that passes through the center. It is also the longest chord. The diameter measures two radii.
Q2...Three identical circles are arranged such that their diameters overlap the diameter of a larger circle, as shown above. What is the ratio of the radius of the larger circle to the diameter of a smaller circle?

4/3

1.5

3

6

9
B

Very good.

The awkward phrasing of the question leaves much room for confusion. Plug in to simplify the problem. Use good numbers: plug in 1 for the diameter of a small circle. The diameter of the larger circle is then 3. Now, what did the question ask? Compare the radius of the larger circle (half of 3=1.5) to the diameter of the smaller circle (1), and get a ratio of 1.5 to 1. Since all we need is a ratio, there's no need to plug in again - the ratio will always be 1.5 to 1, or answer choice B.
Q3...A circle passes through the center of a larger circle, as shown above. The diameter of the smaller circle is how many times the radius of the larger circle?

1/4

1/2

1

2

4
C

Correct.

Note that the diameter of the small circle is simply the radius of the large circle. Since the two are equal, the diameter of the small circle is 1 times the size of the radius of the large circle.

If you're not sure, plug in 1 for the diameter of the smaller circle and calculate the radius of the larger circle.
Q4...If the circles shown above are centered at the three vertices of a triangle whose sides are 3, 4 and 5, what is the radius of the largest circle?

2

2.5

3

3.5

4
C

Correct.

For convenience's sake, call the large, medium and small circles L, M and S, respectively. Hence, S's radius is x, M's radius is 3-x and L's radius is 4-x.

From the question stem, you know that

The hypotenuse = L's radius + M's Radius = 5. Form an equation with x:
(4-x) + (3-x) = 5
--> 7 - 2x = 5
--> 2x = 2
--> x = 1

L's radius is 4-x = 4-1 = 3.
Q5...If the diagonal of a square of side √6 is the side of an equilateral triangle, what is the area of this triangle?

(3/8)·√3

(3/4)·√3

(3/2)·√3

2√6

3·√3
E

Correct.

Remember that the diagonal of a square divides the square into two 45:45:90 right isosceles triangles. Use the recycled 45:45:90 ratio of the sides of a right isosceles triangle: s:s:s√2 to find the length of the diagonal i.e. the side of the equilateral triangle. Then, save time finding the area of the equilateral triangle by using the unique area formula for an equilateral triangle of side a:

Area: a^2·√3/4

Given that the side of the square is √6, the diagonal of this square is √6√2 = √12.

The area of an equilateral triangle with side s = (s2·√3) / 4

area of triangle = ((√12)2·√3) / 4

area of triangle = (12·√3) / 4

area of triangle = 3√3

Hence, this is the correct answer.
Q6...In the figure below, what is the value of x?

20°

35°

70°

110°

145°
E

Correct.

Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember that in a triangle angles opposite equal sides are also equal.

AEB = 180° - 90° - 20° = 70°. Based on this AED = 180° - 70° = 110°. Since AED is an isosceles triangle, EDA = EAD = [(180° - 110°) / 2] = 35°. Hence, x = 180° - 35° = 145°.
Q7...If in the figure above, l1 and l2 are parallel, what is x?

40°

35°

25°

20°

10°
D

Correct.

Lines given in a parallel lines diagram can be extended to make it easier to solve the question.

Extend l2 backwards and extend the line intersecting both parallel lines. The resulting triangle is shown in the figure. The supplementary angle of 40 is 140. Given that l1 and l2 are parallel lines, the alternate angle of x (on line l1) is also x (on left side of l2). Since the sum of all angles in a triangle is 180, 2x = 180 - 140 = 40. Hence, x = 40/2 = 20°.
Q8...If ABCDE is a regular pentagon, what is the value of x in the figure above?

36

42

45

48

72
A

Correct.

A regular pentagon's internal angle=108º. The supplementary angle is therefore 180-108=72º. The triangle is an isosceles triangle with two 72º angles. x, the third angle is 180-(2·72)=180-144=36º.
If the length of the longest chord of a certain circle is 10, what is the radius of that certain circle?

2.5

5

10

15

20
B

Correct.

The longest chord in a circle is the circle's diameter. If the diameter=2r is 10, the radius (r) is 5.
The circumference is a special word for the perimeter of a circle.
The circumference of a circle is given in the formula Circumference=2πr=π·d.
In any circle the relationship between the circumference and the diameter is π. Most of you probably remember that π=3.14. Some of you geeks out there may remember π up to the nth decimal. In GMATs it is sufficient to remember that π is a little more than 3. In other words, the circumference is somewhat more than three times the diameter.
Q9...If the diameter of semicircle C is π, then the perimeter of C is which of the following?

π2/2

(π2/2)+(π/2)

(π2/2)+π

(π/2)+π2
C

Correct.

Remember, a perimeter is always a closed shape (think of a call encircling a castle in the shape of the semicircle above). The perimeter of the semicircle is composed of the arc AB and the diameter.

Arc AB is half the circumference of a circle with diameter n, thus its length is 1/2·πd = 1/2·π·π = π2/2 .The perimeter of C is thus π2/2 + π.
Q10...In the figure above, ABCD is a square, and P and Q are the centers of identical semicircles BF and AE, respectively. If DE=EA=AF=FB=1, then what is the perimeter of the shaded region?

4+π/2

6

8

6+π

6+2π
D

Correct.

Since AE=EF=1, the side of the square is 1+1=2.

Thus, AF + ED + CD + BC = 1+1+2+2 = 6

semicircles BF and AE are together a full circle with a radius of 1/2 (one half of the diameter of 1), and thus a circumference of 2πr = 2π·(1/2) = π

The required perimeter is 6 + π.
Q11..If the perimeter of semicircle C is 4+4π, what is the radius of C?

4+4/π

4

(4+4π)/(2+π)

2+2/(√π)

2
E

Close enough - you took 1 minutes and 29 seconds to answer this question.

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

B

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

A

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

D

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

C

Correct.

--> 2r + πr = 4+4π

--> r(2+π) = 4+4π

--> r = (4+4π)/(2+π)
If the circumference of a certain circle is equal in length to the perimeter of a certain square, what is the length of the side of the square in terms of d, the diameter of the circle?

π/2×d

√2×d

π/2√2×d

π/4×d

1/√2×d
D

Correct.

According to the question, π⋅d=4s, where s is the side of the square and d is the diameter of the circle. Since the question asks for the length of the side of the square, divide both sides of the equation by 4 to get s=π⋅d/4.
The area of a circle is the two-dimensional region within the boundary lines of the circle.
The area of a circle is given in the formula Area=πr2, where π is the relationship between the circumference and the diameter in any circle.
In GMAT land it is sufficient to remember that π is a little more than 3.
If three times the circumference of circle P is double the circumference of circle Q, then the diameter of circle Q is how many times the radius of circle P?

6

3

1.5

4/3

2/3
B

Very good.

The phrasing of the question is confusing, and there is much potential for careless algebra mistakes. Make the algebra go away - plug in!

Use good numbers: 2 for the diameter of circle P, and 3 for the diameter of circle Q. That way, the circumference of P is 2π, and the circumference of Q is 3π. These numbers satisfy the question's requirements - 3×2π = 2×3π= 6π.

Now, what did the question ask again? ah, compare the diameter of circle Q (3) to the radius of circle P (radius=half of diameter=2/2=1), and find the relationship of 3 to 1. Since all we need is the ratio, there's no need to plug in additional numbers - the ratio must always be 3 to 1, or answer choice B.
Tina and Rebecca had the same number of candies. Tina gave Rebecca 24 candies so that now Rebecca has five times as many candies as Tina has. Rebecca has now how many candies?

36

48

54

55

60
E

You grossly underestimated the time this question took you. You actually solved it in 6 minutes and 43 seconds.

Correct.

Numbers in the answer choices and a specific question ("How many candies...") call for Plugging In The Answers. You may feel like jotting down an equation or more. This is just your algebraic urge, which is another stop sign for Reverse PI problems.

Start with answer choice C - assume the amount in the answer choice is the number of candies Rebecca has now, and then follow the story in the problem. If everything fits - stop. Pick it. Otherwise - POE and move on, until you find an answer that works.

Start with answer choice C. If Rebecca has 54 candies, then she had 54-24=30 candies to begin with. Since the question states that Tina and Rebecca had the same number of candies originally, this means that Tina also had 30 candies. Therefore, after giving 24 candies to Rebecca, Tina should now have 30-24=6 left and Rebecca 54, but that means that Rebecca has 9 times more than Tina, not 5 (as the question stipulates). So you should eliminate C. Which direction should you go now?

The answer is 'not sure', so test either way. Check answer choice B. If Rebecca has 48 candies now, she and Tina had 24 each to begin with, so now Rebecca has all the candies and Tina has none, which doesn't fit the question. POE B and A as well, because A would result in even less candies for Tina. Plug in D or E to check which is correct.

If Rebecca has 60 candies, it means that she and Tina had 36 candies. Now Tina has 36-24=12, exactly a fifth of the amount Rebecca has.

Note: knowing that the number of Rebecca's candies is a multiple of 5, you can also automatically eliminate all the answer choices which are not divisible by 5. Noticing that makes your life a whole lot easier, as it leaves you with D or E.
Q12...If the radii of the two concentric circles shown above are 8 and 4, what fraction of the area of the larger circle is shaded?

1/4

1/3

1/2

2/3

3/4
E

Correct.

Let's call the large circle L and the small circle S. The area of the shaded region can be calculated by subtracting the unshaded region (area of S) from the area of the large circle (area of L).

The radius of S is 4, therefore S's area = πr2 = π×42 = 16π
The radius of L is 8, therefore L's area = πr2 = π×82 = 64π

16π/64π = 1/4 of the large circle is NOT shaded, therefore 3/4 IS shaded.
Q13... A circle passes through the center of a larger circle, as shown above. The circumference of the larger circle is how many times the circumference of the smaller circle?

√2

2

3

4

9
B

Correct

The large circle's radius (r) is the small circle's diameter. Therefore the large circle's circumference is 2πr, while that of the small circle is πr, and the circumference of the larger circle is twice that of the smaller circle. If you're not sure, plug in a good number such as R=2 for the larger circle and work out the problem with numbers.
Q14...Four isosceles right triangles were combined as shown above. What is the ratio of the hypotenuse of the largest triangle to the leg of the smallest triangle?

2√2

4

3√2

2+2√2

4√2
B

Remember the ratio of the sides of a 45, 45, 90 triangle is 1, 1, √2.

Correct.

Plug in 1 for the side of the smallest triangle so the hypotenuse is √2. Since the leg of the second triangle is the hypotenuse of the first, the legs of the second triangle are √2, √2 and the hypotenuse is 2. Likewise, the legs and hypotenuse of the third triangle are 2, 2 and 2√2. Lastly, the legs and the hypotenuse of the largest triangle are 2√2, 2√2 and 4. Hence the hypotenuse of the largest triangle is 4 times the leg of the smallest triangle.
A central angle in a circle originates from the center and its rays are radii.
...
If the area of a certain square is equal to the area of a certain circle, what is the length of the side of the square in terms of d, the diameter of the circle?

πd

√π×d

π/2 × d

√π/√2 × d

√π/2 × d
E

Correct.

According to the question, π⋅r2=s2, where s is the side of the square and r is the radius of the circle. Since the question asks for the length of the side of the square, take the square root out of both sides of the equation s=√π⋅r. Substitute d/2 for r to get s=√π⋅d/2.
If half the area of a certain circle, in square inches, is equal to its circumference in inches, what is the diameter of that certain circle?

2

4

8

C

Correct.

According to the question, (π⋅r2)/2=2πr, where r is the radius of the circle. Since the question asks for the diameter of the circle, solve for r: divide both sides of the equation by πr get r/2=2, or r=4. Therefore, the diameter is 2r=8.
An inscribed or peripheral angle originates from the circumference. Its rays are chords.
GMAT Geometry problems refer to several areas inside a circle;

A sector is a portion of the circle's area defined by radii (Think "pizza slice").

A quadrant is a sector with a right central angle.

A segment is a portion of the circle's area defined by a chord.

A semicircle is a segment defined by a diameter
A tangent is a line outside the circle that touches the outer boundary at a single point.

A radius drawn to the point of tangency is always perpendicular to the tangent.
...
Q15...If the radius of the circle shown above is 6, And A lies at the point of tangency with tangent AB, what is the length of line segment AB?

12

6√3

3√3

6/√3

3/√3
B

Correct.

Draw a radius from the center to the point of tangency A.

AOD is a recycled 30:60:90 triangle - angle BAO is a right angle because the tangent is perpendicular to the radius at the tangency point. AO (the radius of the circle, which is also the little leg in the triangle) = 6, thus AB, the big leg, equals 6√3 (according to the 30:60:90 side ratio of a:a√3:2a).
A circumscribed circle or a circumcircle passes through all the vertices of a polygon, such that the entire polygon is contained within the circle.
All triangles and regular polygons have circumcircles.
Q16...Vertex A of equilateral triangle ABC is the center of the circle above. If the side of the triangle is 2, what is the radius of the circle?

1/(2√3)

1/2

(√3)/2

1

√3
E

Correct.

A radius drawn to the point of tangency is always perpendicular to the tangent, thus line AD (in the figure below) is the height of triangle ABC. The height of an equilateral triangle divides it into two recycled 30:60:90 triangles (as shown in the figure). The recycled side ratio of this triangle is a:a√3:2a. Since the height of an equilateral triangle is also the median, BD = half the side of triangle ABC = 1. Therefore, AD (the large leg, which is also the radius of the circle) is 1·√3.
What is the side of an equilateral triangle whose area is √3?

1

√2

√3

2

3
D

Correct.

Save time by using the unique area formula for an equilateral triangle of side a: .

(a2√3) / 4 = √3 /:√3

a2 / 4 =1 /×4

a2 = 4

Since the side of a triangle can't be negative, a = 2.
An arc is a portion of a circle's circumference.
When an arc is defined by three points, (e.g. arc ACB,) it refers to the part of the circumference that passes through the three points. Alternatively, the arc can be defined by only two points and an indication whether the arc is the major or the minor part of the whole circle.
Arcs can be measured using length units or angles. In the following, we are going to concentrate on arcs measured in angles.

To sum up:

"major arc AB" is the greater of the two parts of the circle that are defined by A and B
"minor arc AB" is the lesser of the two parts of the circle that are defined by A and B
"arc ABC" is the part of the circle on which points A,B,C lie in this order
The measure of the whole circumference is 360°
An arc may be measured in angles, as a fraction of a circle.
The circumference of a full circle is actually an arc of 360 degrees. An arc defined by two rays is a fractional part of a circle. For example, the measure of an arc defined by a quarter of a circle is simply a quarter of 360 = 90 degrees.
Points A, B, C, D lie in this order on the circumference of a circle. Minor arc AC is 160°, and minor arc BD is 140°. If B bisects minor arc AC, then what is the measure of minor arc AD?

80°

120°

140°

160°

220°
A

Incorrect.

This is a question about the measure of an arc in angles.

Draw the circle and go on to indicate ABCD and the rest of the information on it.

Point B bisects minor arc AC, so minor arc AB and minor arc BC are 80° each.

Minor arc BD of 140° is the total of minor arcs BC (=80°) and CD, therefore, minor arc CD=60°.

Calculate arc AD, then find minor arc AD. Recall that the minor of an arc larger than 180° is the arc that completes it to 360°.

C

Correct.

This is a question about the measure of an arc in angles.

Draw the circle and go on to indicate ABCD and the rest of the information on it.

Point B bisects minor arc AC, so minor arc AB and minor arc BC are 80° each.

Minor arc BD of 140° is the total of minor arcs BC (=80°) and CD, therefore, minor arc CD=60°.

Finally, arc AD is the total of minor arcs AB, BC and CD, i.e., 80+80+60=220°. Obviously, this is the measure of major arc AD. The minor arc completes the major arc to 360°, so minor AD is 360−220=140°.
Q17...A,B and C are points on a circle as shown above, and AC is a diameter. If the measure of minor arc AB is double the measure of minor arc BC, then what is the measure of minor arc AB?

160°

150°

120°

60°

45°
C

Correct.

This is a question about the measure of an arc. Recall that the whole circumference is an arc of 360°.

Arc AC, defined by a diameter, is 180°. It is divided into two arcs of measures x° and 2x°. Therefore

x + 2x = 180 --> x = 60

Arc AB is 2x° = 120°.
An Inscribed circle is the largest possible circle that can be drawn within a polygon, so that each side of the polygon is tangent to the circle.
Concentric circles share the same center.
A central angle in a circle defines an arc -
An arc can be measured as a fraction of a circle, or as a function of the central angle which defines it. The measure of an arc defined by a central angle xº is simply xº.
Q18...If the center of a circle lies on a smaller circle, as shown above, what portion of the area of the larger circle is shaded?

5/8

2/3

11/16

3/4

7/9
D

Correct.

Let's call the large circle L and the small circle S. The area of the shaded region can be calculated by subtracting the unshaded region (area of S) from the area of the large circle (area of L). Note that the diameter of S is the radius of L. Plug in numbers that reflect this, so if the radius of L = 2, then the radius of S = 1.

The area of S = πr2 = π
The area of L = πr2 = 4π

Hence, π/4π = 1/4 of L is NOT shaded, so 3/4 of it IS shaded.
Q19...The circle above is circumscribed around square ABCD. Points E and F lie on minor arcs AB and CD respectively. The measure of minor arc EF is between

90° and 270°

90° and 180°

60° and 120°

0° and 90°

0° and 180°
B

Correct.

This is a question about the measure of an arc - its minimum and maximum, under the limitations imposed on E and F by the other points on the circle. Recall that minor arc is always the smaller arc - the one less than 180.

Points E and F cannot get any closer than points A and D. Thus When E and F are closest and minor arc EF is least, it looks so

When E and F are furthest apart, then EF is a diameter and arc EF is 180°. Thus, When E and F are furthest and minor arc EF is greatest, it looks so

Use this to find the minimum and maximum measure of arc EF.

Recall that the whole circumference is an arc of 360°. Square ABCD divides the circle into four 360/4= 90° arcs. Therefore, minor arc EF is at least 90° (one arc defined by the square) but no more than 180° (two arcs defined by the square). Therefore, EF's measure lies between 90° and 180°.
Some GMAT Geometry problems require you to find weird values, such as the area of a sector ("pizza slice"), or the length of an arc. It is hard to accomplish, since these are portions of a circle.
The most important factor in an arc or a sector is the central angle that defines it. The larger the angle, the larger the size of the arc or sector, and vice versa.

The proportion of the central angle from the entire circle is what matters in determining the measure of an arc.

Remember that

x°/360° = Arc Length / Circumference

The proportion of the central angle from the entire circle is what matters in determining the area of a sector.

Remember that

Sector Area / Circle Area = x°/360°

Let's review the relationship between an arc, a central angle, and a sector. Remember that

The size of the sector or arc is proportional to the measure of the central angle that defines it.

Arc Length / Circumference = x°/360° = Sector Area / Circle Area
Q20...In the figure above, square ABCD is inscribed in the circle. If the length of arc ABCD is 90 cm, what is the approximate radius of the circle, in cm?

10

15

20

30

40
A

Incorrect.

Draw the figure in your notebook - the square divides the circumference into four equal arcs of 90º, 3 of which are covered by the given arc ABCD as shown in the figure below:

Recall that the circumference of a circle is given in the formula 2πr. This allows you to bridge the gap between the length of the arc given in the question and the length of the radius required by it.

B

Incorrect.

C

Correct.

If the length of arc ABCD is 90 cm, then each of the equal 90º arcs equals a third of 90 = 30 cm. The last arc is also equal to 30 cm, bringing the full circumference to 90+30 = 120 cm.

Therefore, 2πr = 120, and r = 120/2π.

Ballpark π ≈ 3 to get r ≈ 120/2·3 = 20.
Q21...Noisy the dog walks from his doghouse at A, to his food bowl at B. Usually he runs the 24 meter distance in a straight line, but on Tuesday there was a little puddle in his way, and he decided to go around it by running along a circular path centered at P, as shown in the figure above. If the radius of the circular bypass is 4, what is the distance Noisy ran when going from A to B?

16+6π

24−4√2+6π

16+8π

24−4√2+8π

24+6π
B

Correct.

Form the general expression to find out what you need to know:

Noisy ran 24 meters, less the portion of the line he skipped, plus the portion of the circumference of the circle he ran over.

Of the straight line AB, noisy did not run the part which is also the hypotenuse of the red 45:45:90 triangle. The legs of the triangle, which are also radii in the circle, are 4. Thus, The missing part from AB is 4√2.

Now, the circular part is 3/4 of the circumference of the circle (the missing part is 1/4, since it is formed by a 90 degree central angle, as shown above). The length of the circular part is thus (3/4)·2πr = (3/4)·2π·4 = 3·2π = 6π.

Noisy thus ran 24 - 4√2 + 6π.
Q22...If the side of the regular pentagon shown above is 5, what is the length of arc AD, centered at E?

1.5π

(5/3)π

(10/3)π
D

Correct.

Arc AD is part of an imaginary circle centered at E. Find Angle E, which is an internal angle of a regular pentagon, and also a central angle of that imaginary circle:

Since in a regular polygon of n sides , the sum of the angles is 180°·(n-2), each equal angle in a regular polygon is of measure 180°·(n-2)/n

Therefore, the sum of the angles in the regular pentagon is 180(5-2) = 180(3) = 540. Divide this by 5 angles to get 540/5 = 108° for each angle.

Therefore, the ratio between the arc's length and the circumference of the imagined circle equals the ratio of 108/360. Because the side of the pentagon is also the radius of the imagined circle, the circumference of the circle = 2Πr=2·Π·5=10Π.
So the arc's length is (108/360)·10Π=(108/36)·Π=3Π.
An inscribed angle in a circle defines an arc.

The measure of an arc defined by an inscribed angle xº is 2xº.
All the inscribed angles that define the same arc of 2x° are equal to each other, and equal to x°.

Recall that the central angle defining an arc is equal to the measure of the arc. It follows that when an inscribed angle and a central angle define the same arc, the central angle is double the inscribed angle.
Q23... In the regular pentagon above, what is x?

24°

36°

42°

45°

72°
B

Great work!

Draw the pentagon inside a circle. The pentagon divides the circle into five arcs, each of which measures 360°/5 or 72°. Each inscribed angle equals half the arc segment it creates. Therefore, ∠EBD, or x, is half of arc segment DE, or 36°.
Q24...The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?

180°

135°

120°

100°

90°
A

Incorrect.

The issue of the question is the relation of inscribed angles to the arcs they define. An arc that is defined by an inscribed angle is double that angle, hence, the inscribed angle is half the arc it defines.

The square divides the whole circumference into 4 equal arcs of 90° each. Inscribed angles w° and y° define two arcs whose total measure is 90°, as do angles x° and z°. Use the above-mentioned relation to find the value of w°+x°+y°+z°.

E

The issue of the question is the relation of inscribed angles to the arcs they define. An arc that is defined by an inscribed angle is double that angle, hence, the inscribed angle is half the arc it defines.

The square divides the whole circumference into 4 equal arcs of 90° each. Inscribed angles w° and y° define two arcs whose total measure is 90°. Therefore w°+y°=45°. Arcs x° and z° also define two arcs whose total measure is 90°. Therefore x°+z°=45°. It follows that w°+x°+y°+z°=45+45=90°.
Q25...In the regular octagon above, what is the value of central angle x?

67.5°

120°

135°

145°

150°
C

Very good!

Draw the octagon inside a circle. The octagon divides the circle into eight arcs, each of which measures 360°/8 or 45°. Each central angle is equal to the arc segment defined by its rays. Therefore, ∠BMG is equal to arc segments BA+AH+HG, or 135°.
Q26...If O is the center of the circle of radius 6 shown above, what is the area of the shaded region?

1.2π

3.6π

36π
B

Correct.

The central angle that 'belongs' to the dark sector is vertical to the 36 degree angle, and thus the areas of the two small sectors in the figure are identical.

If the radius of the circle is 6, then the area of the circle = πr2 = 36π.

--> 36/360 = sector area/36π

--> sector area = (36·36π)/360 = (1/10)·36π = 3.6π
Q27...Sector AOB is a quarter of the area of a circle centered at O, and DE is a bisector of angle AOB. If the length of arc AEB is 0.8, what is the length of arc BCD?

1.6

1.2

1.1

1.0

0.9
B

Correct.

When the problem asks for arc length, focus on the central angles.

Angle AOB = 90 and lies on arc AEB.
Angle EOB, which is half of AOB, is 45.
Angle DOB, which lies on the wanted arc BCD, supplements EOB, and therefore equals 180-45=135.

At this point you can use the arc length of AB to find the circle's circumference and find the portion of the circumference covered by a central angle of 135 out of 360, but finding the circumference is an unnecessary step. 135=90+45, or 1.5·90. The ratio between arcs is the same as the ratio between their corresponding central angles; therefore arc BCD is 1.5 times arc AEB = 1.5·0.8=1.2
Q28...Arc AB is centered at C and the area of the shaded region above is 10π. If the length of line segment CA is 5, what is the value of x?

30

36

45

135

144
B

Correct.

The size of the sector or arc is proportional to the measure of the central angle that defines it. Here you're given the area of a sector and the radius of a circle. This data can help calculate angle ACB, from which we can find x. Line AC is the radius=5, thus the area of the circle is πr²=25π. The shaded sector is 10π, thus it is 10π/25π=2/5 from the circle's area. Therefore, central angle ACB is also 2/5 of 360:
ACB = (2/5)*360= 144.
x complements ACB and is therefore 180-144=36 degrees.
Within a given circle, arcs of equal length are arcs of equal degrees.

The opposite is true as well: Within a given circle, arcs of equal degrees are arcs of equal length.
In particular:

Equal central angles define equal sections on the circumference of the circle.

And equal inscribed angles define equal sections on the circumference of the circle, and vice versa.
Q29...A circle is circumscribed around square ABDC, as shown above. What is the value of x?

30

36

40

45

60
D

Correct.

COPY the figure, INDICATE the measure of every arc next to it.

The square's vertices divide the circle's circumference (360°) into 4 equal arcs of 90° each. The angle x is an inscribed angle that defines an arc of 90°. Recall that the arc is twice the inscribed angle that defines it, therefore the angle x is half the 90° arc, i.e. x=45°.
A diameter is a central angle of 180°, therefore an inscribed angle on the same arc length measures 90°.
In fact, All inscribed angles lying on a diameter equal 90°.
Q30...If a circle of area 2π is circumscribed around an isosceles right triangle, what is the area of the triangle?

√2

2

2√2

π

π·√2
B

Correct.

To find the area of a right triangle, you need to find the two legs of the right angle. In a right isosceles triangle, all you need is one side of the triangle - the other two can be found using the 45:45:90 ratio.

The right angle of the triangle is also an inscribed angle in the circle. A right inscribed angle always lies on the diameter. Therefore, the diameter of the circle is also the hypotenuse of the triangle. Find the diameter from the given area of the circle.

The area of the circle is 2π:

--> πr2 = 2π

--> r2 = 2

--> r = √2

--> 2r = 2√2 = diameter

The side ratio of a 45:45:90 triangle is x:x:x√2, thus the legs of the triangle are 2√2/√2 = 2.

The area of a right isosceles triangle = 1/2·base·height = leg2/2 = 22/2 = 2.
Q31...If P is the center of the circle shown above, and BAC=30º, and the area of triangle ABC is 6, what is the area of the circle?

(√3)π

(2√3)π

(4√3)π
C

Incorrect.

You're asked to find the area of the circle.
ΔABC is a recycled 30:60:90 triangle - ABC is an inscribed angle lying on the diameter and therefore it is a right angle, and BAC = 30º.
The area of ΔABC = ½×BASE×HEIGHT = ½×AB×BC. The side ratio in this triangle is x:x√3:2x.

E

Correct.

If AC, the diameter, which is also the triangle's hypotenuse is 2r, then the little leg, BC, is r and AB is r√3.
The triangle's area = ½×BASE×HEIGHT = ½×AB×BC = ½×r√3×r = ½×√3×r² = 6.

Isolate r²: multiply by 2 and divide √3:

--> r²=12/√3=4·3/√3 ->4√3

The area of the circle = πr²=(4√3)π
Q32...If D is the center of circle D shown above, what is the area of circle D?

2.5π

6.25π

20

12π

12.5π
B

∠CAB is an inscribed angle lying on the diameter, therefore it equals 90°.

ΔCAB is a simple recycled 3:4:5 right triangle. Since its legs are 3 and 4, the hypotenuse, which is also the diameter of the circle, is 5. Therfore, the radius is 5/2=2.5. The area of D is πr²=(2.5)²π=6.25π.
Q33...If in the figure above, AB is the diameter of a circle of radius 6, what is the perimeter of the shaded region?

π+6+6√3

π+12+6√3

2π+9+6√3

2π+12+6√3

2π+12+12√3
D

You underestimated the time this question took you. You actually solved it in 1 minutes and 54 seconds.

Correct.

The perimeter of the shaded region is composed of the length of arc BC, and the length of the sides AC and AB.

To find AC, connect B and C to create the ABC recycled 30:60:90 triangle. ABC is a recycled 30:60:90 triangle because it has a 30 degree angle from the original figure, and a 90 degree angle, ∠ACB, is an inscribed angle on the diameter.

Use the 30:60:90 ratio: the hypotenuse, which is also the diameter, is 2r=12. Thus the little leg, CB, is 12/2 = 6 and the big leg, AC is 6√3.

Now we need the minor arc BC. Recall that The measure of an arc is twice the measure of an inscribed angle that defines it. Angle CAB is a 30 degrees inscribed angle, so the measure of minor arc CB is 2·30=60°. 60 is 1/6 of 360, so small arc BC is 1/6 of the circle's circumference. Plug in the radius=6 into the circumference formula to get

1/6×(2πr) = 1/6×(2×π×6) = 2π.

Thus, the total perimeter of the shaded region is (minor arc CB)+AB+AC = 2π+12+6√3.
Q34...If AC=BC=3, and ∠C is a right angle, what is the circumference of the circle above?

3√2

3√2 / 2 × π

3√2π

4×1/2π
D

Correct.

Note: Since AC=BC and ∠ACB = 90°, ΔABC is a recycled 45:45:90 right isosceles triangle. Use the recycled ratio to ease calculations.

The side ratio is x:x:x√2, therefore AB=3√2. Since inscribed angle ∠ACB = 90°, the chord defined by its rays must be the the diameter. AB=3√2 is the diameter d of the circle, the circle's circumference is therefore πd=3√2π
Q35...Point P is the center of the circle in the figure above. If ∠PDE=66º and PA=AE=CD=15, what is the length of arc ABC?

π

C

Correct.

In order to find the length of an arc in a circle, you need two pieces of data: the circle's radius and a central or inscribed angle lying on the said arc. Here you have central angle DPE. Triangle PDE is an isosceles triangle, since PA and PC are also equal radii.

Since triangle PDE is an isosceles, angle PED = angle PDE = 66º. Thus, angle DPE = 180 - 66 - 66 = 48º.

The radius of the circle, PA, is 15, so the circumference is 2πr = 30π

Now plug in the circumference and the central angle to find the arc length:

48/360 = arc/30π

--> arc = (30π·48)/360 = (π·48)/12 = 4π
Q36...Answer the following Non-GMAT question -

If O is the center of the circle above, and minor arc AB is 115°, then what is the measure of minor arc BC?

115°

95°

75°

65°

45°
D

Correct.

This is a question about the measure of an arc. Recall that the whole circumference is an arc of 360°.

Line segment AC is a diameter of the circle (It passes through the circle's center.) Therefore arc AB is 180°. The 180° is the total of two arcs, minor arc AB=115° and minor arc BC. It follows that minor arc CB=180−115=65°.
The following is the general method for dealing with any GMAT geometry question involving angles within a circle. Use the following work order.

1. Copy the figure to your noteboard.

2. Indicate the measure of every arc on the circle.

3. Then solve for x.
When indicating the measures of the arcs, use the following rules -

An arc measures, in angles, as much as the central angle that defines it;

An arc measures, in angles, twice as much as the inscribed angle that defines it.

Let's exercise that a bit.
Q37...If O is the center of the circle shown above, what is the measure of x?

30°

40°

50°

80°

100°
B

Well done.

After you copy the figure to your noteboard, indicate the measure of every arc on the circle. This is what your noteboard should look like:

The central 100° angle defines a 100° arc.

The inscribed angle x defines a 2x arc.

However, the total of the 2x arc and the 100° arc must be 180°. (The diameter divides the whole circle to two semicircles of 180°.)

Now, solve for x: 100°+2x=180° --> x = 40º
To sum up our approach to questions involving arcs and angles,
COPY the figure. INDICATE the measures of all known arcs. Then SOLVE.
An arc measures, in angles, as much as the central angle that defines it.
An arc measures, in angles, twice as much as the inscribed angle that defines it.
Q38...Circle P shown above is centered at P. If the length of arc ABC is 40, what is the area of circle P?

10000 / π²

10000 / π

10000

10000 π

10000 π²
B

Correct.

Find the circumference of the circle, to find the radius and the area:

--> 72/360 = 40/circumference

--> circumference = 40·360/72

Reduce 360/72 to 5/1

--> circumference = 40·5 = 200

Since the circumference is given by the formula 2πr, it is now possible to find r, and from there, the area of the circle.

--> 200 = 2πr

--> r = 200/(2π) = 100/π
--> area of circle = πr2 = π(100/π)2 = 10000/π
Q39... If AE=ED and AB=BC=CD, what is the value of x in the figure above?

105

130

135

140

145
B

Correct.

COPY the figure, INDICATE the measure of every arc next to it:

Angle E defines a 150° arc ABCD. Therefore arc DEA is 360-150=210°.

Each of these angles is further divided: Arc ABCD is divided in three equal parts, 50° each.

Once all these arc measures are indicated on the drawing, the way forward is clear: x defines an arc of (50+105+105)°=260°. Recall that the arc is twice the inscribed angle that defines it, therefore the angle x is half of 260° i.e., x=130°
Q40...If BD=DC, what is the value of x in the figure above?

75

95

97.5

100

105
E

Correct.

The issue of this question is arcs and angles in a circle.

Connect B and C to form an isosceles triangle BCD. The base angles of this triangle are (180°-30°)/2=75° each. Now, indicate the measure of each arc next to it, and solve for x. Recall that an arc is double the inscribed angle that defines it.

Inscribed angle x defines a (60°+150°)=210° arc, and therefore its measure is 210°/2=105°.
Q41...Each vertex of the regular hexagon shown above is the center of a circle. If the side of the hexagon is 2, what is the total length of the arcs?

12π

16π
C

You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 32 seconds.

Correct.

Focus on one arc - the other arcs are identical. The hexagon's internal angle, 120 degrees, is a central angle that lies on this arc. When you find the length of an arc, you'll want to count the number of arcs and multiply, but wait - it is easy to get confused and miss an arc in the figure. Instead use the fact that each arc is centered at a single vertex.

120 is a third of 360, therefore the arc is a third of the circumference of a circle with a radius of 2 (the side of the hexagon)=2πr/3=(4/3)π. A hexagon has 6 vertices, so we have 6 identical arcs. Thus their total length is 6*(4/3)π=8π.
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