Geometry Practice Questions — GRE General Test — Page 5

Question 81Question →

In circle $O$ , line segments $AB$ and $CD$ are perpendicular diameters, each of length $12$ . An arc of a second circle, centered at point $A$ with radius $AC$ , is drawn from point $C$ to point $D$ through the interior of circle $O$ . What is the area of the crescent-shaped region bounded by the semicircle $CBD$ of circle $O$ and arc $CD$ of the second circle?

Show answer & explanation

Answer: 36

Answer

36

Circle

O

has radius

r = 6

, giving semicircle

CBD

an area of

\frac{1}{2}\pi(6^2) = 18\pi

. The distance

AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

is the radius of circle

A

. Because

\angle CAD = 90^\circ

, sector

ACD

has area

\frac{90^\circ}{360^\circ}\pi(6\sqrt{2})^2 = 18\pi

. Subtracting the area of triangle

ACD

(

\frac{1}{2} \times 12 \times 6 = 36

) yields a segment area of

18\pi - 36

. Subtracting this segment area from the semicircle area yields

18\pi - (18\pi - 36) = 36

.

Step-by-Step Solution

1

Find the radii of circle

O

and circle

A

.

Radius of circle

O

is

r = 6

. In right triangle

AOC

,

OA = OC = 6

, so radius

AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

.

Perpendicular diameters

AB

and

CD

intersect at center

O

, dividing each diameter into radii of length

6

.

2

Calculate the area of sector

ACD

of circle

A

and triangle

ACD

.

Sector area

= \frac{90^\circ}{360^\circ} \pi (6\sqrt{2})^2 = 18\pi

. Triangle area

= \frac{1}{2} \times 12 \times 6 = 36

.

Angle

\angle CAD = 90^\circ

because

\triangle ACD

is a right isosceles triangle with hypotenuse

CD = 12

.

3

Find the area of the circular segment bounded by chord

CD

and arc

CD

of circle

A

.

Segment Area

= 18\pi - 36

.

The area of a circular segment is equal to the sector area minus the triangle area.

4

Subtract the segment area from the area of semicircle

CBD

of circle

O

.

Region Area

= 18\pi - (18\pi - 36) = 36

.

Semicircle

CBD

has radius

6

and area

\frac{1}{2}\pi(6^2) = 18\pi

. Subtracting the segment area leaves the crescent region.

Key Concept

Area of circular sectors, segments, and compound regions (Lune of Hippocrates)

Question 82Question →

A rectangular park measures $8$ meters in width and $15$ meters in length. A straight walking path extends diagonally across the park from one corner to the opposite corner. What is the length, in meters, of the path?

23

17

289

161

\sqrt{23}

Show answer & explanation

Answer: $17$

Answer

17 meters

The diagonal of a rectangle forms a right triangle where the width (

8

m) and length (

15

m) are legs, and the diagonal path is the hypotenuse. Applying the Pythagorean theorem:

d = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

meters.

Step-by-Step Solution

1

Identify the geometric model and given dimensions

The park is a rectangle with perpendicular sides of length

8

meters and

15

meters. The diagonal forms a right-angled triangle with these two sides as legs.

Adjacent sides of a rectangle meet at right angles (

90^\circ

).

2

Apply the Pythagorean theorem to calculate the diagonal length

d

d^2 = 8^2 + 15^2 = 64 + 225 = 289

In any right triangle, the square of the hypotenuse equals the sum of the squares of the legs (

a^2 + b^2 = c^2

).

3

Solve for

d

by taking the square root

d = \sqrt{289} = 17

The distance

d

must be positive.

Key Concept

Pythagorean Theorem (

a^2 + b^2 = c^2

) and the

8-15-17

Pythagorean Triple

Question 83Question →

In a circle centered at point $O$ , the radius is $12$ units and central angle $\angle AOB$ measures $150^\circ$ . Which of the following statements regarding sector $AOB$ and minor arc $AB$ are correct? Select all such statements.

Select all that apply

The length of minor arc

AB

is

10\pi

units.

The area of sector

AOB

is

60\pi

square units.

The total perimeter of sector

AOB

is

24 + 10\pi

units.

The length of minor arc

AB

is

24\pi

units.

The perimeter of sector

AOB

is equal to

10\pi

units.

Show answer & explanation

Answer: The length of minor arc $AB$ is $10\pi$ units.; The area of sector $AOB$ is $60\pi$ square units.; The total perimeter of sector $AOB$ is $24 + 10\pi$ units.

Answer

The correct statements are those indicating that the minor arc length is

10\pi

units, the sector area is

60\pi

square units, and the perimeter of the sector is

24 + 10\pi

units.

The central angle fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

. Multiplying the full circumference (

24\pi

) by

\frac{5}{12}

gives an arc length of

10\pi

. Multiplying the full circle area (

144\pi

) by

\frac{5}{12}

gives a sector area of

60\pi

. Adding the two radii (

2 \times 12 = 24

) to the arc length (

10\pi

) gives the total sector perimeter of

24 + 10\pi

.

Step-by-Step Solution

1

Determine the central angle fraction of the circle.

The fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

.

Arc length and sector area are proportional to the ratio of the central angle to

360^\circ

.

2

Calculate the circumference of the circle and the arc length of minor arc

AB

.

Circumference =

2\pi(12) = 24\pi

. Arc length =

\frac{5}{12} \times 24\pi = 10\pi

units.

Arc length equals full circumference multiplied by the central angle fraction.

3

Calculate the total area of the circle and the area of sector

AOB

.

Circle Area =

\pi(12^2) = 144\pi

. Sector Area =

\frac{5}{12} \times 144\pi = 60\pi

square units.

Sector area equals total circle area multiplied by the central angle fraction.

4

Calculate the total perimeter of sector

AOB

.

Perimeter =

\text{arc length} + 2r = 10\pi + 2(12) = 24 + 10\pi

units.

The perimeter of a sector consists of the bounding arc plus the two radii.

Key Concept

Circles, Arc Lengths, and Sector Areas

Question 84Question →

A $10$ -foot straight ladder leans against a vertical wall on flat horizontal ground. If the bottom of the ladder is $6$ feet away from the base of the wall, how high up the wall, in feet, does the top of the ladder reach?

Show answer & explanation

Answer: 8

Answer

The top of the ladder reaches

8

feet up the wall.

Applying the Pythagorean theorem

a^2 + b^2 = c^2

with hypotenuse

c = 10

and ground leg

a = 6

gives

6^2 + b^2 = 10^2

, which simplifies to

36 + b^2 = 100

, so

b^2 = 64

and

b = 8

feet.

Step-by-Step Solution

1

Set up the right triangle relationship using the Pythagorean theorem.

a^2 + b^2 = c^2

, where

a = 6

and

c = 10

.

The ladder forms the hypotenuse of a right triangle with the vertical wall and horizontal ground.

2

Substitute the known values into the formula and solve for

b

.

6^2 + b^2 = 10^2 \implies 36 + b^2 = 100 \implies b^2 = 64 \implies b = 8

.

Subtracting

36

from

100

gives

64

, and taking the positive square root yields the vertical height.

Key Concept

Applying the Pythagorean Theorem (

a^2 + b^2 = c^2

) to find an unknown leg length in a right triangle context.

Question 85Question →

A solid sphere of radius $r$ is placed inside a hollow right circular cylinder of base radius $r$ and height $3r$ such that the sphere rests on the bottom base of the cylinder. Water is then poured into the cylinder until the water level just covers the top of the sphere (a height of $2r$ ). What is the ratio of the volume of the empty space remaining in the cylinder above the water level to the volume of the water inside the cylinder?

\frac{1}{3}

\frac{1}{2}

\frac{3}{4}

\frac{3}{2}

2

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

The ratio of the volume of the empty space remaining in the cylinder above the water level to the volume of the water inside the cylinder is

\frac{3}{2}

.

The empty cylinder portion at the top has height

r

, giving a volume of

\pi r^3

. The submerged section up to height

2r

has total volume

2\pi r^3

, of which the sphere accounts for

\frac{4}{3}\pi r^3

. Subtracting the sphere's volume leaves

\frac{2}{3}\pi r^3

of water. The ratio of empty space to water volume is

\frac{\pi r^3}{\frac{2}{3}\pi r^3} = \frac{3}{2}

.

Step-by-Step Solution

1

Calculate the volume of the empty cylindrical space above the water height.

The total height of the cylinder is

3r

and the water level reaches height

2r

, leaving an empty height of

3r - 2r = r

. The volume of empty space is

V_{\text{empty}} = \pi r^2 \cdot r = \pi r^3

.

The empty space forms a smaller cylinder of base radius

r

and height

r

at the top.

2

Calculate the total volume of the cylindrical region filled up to the water height of

2r

.

V_{\text{region}} = \pi r^2 \cdot (2r) = 2\pi r^3

.

This represents the combined volume of the water and the submerged sphere.

3

Calculate the volume of the solid sphere of radius

r

.

V_{\text{sphere}} = \frac{4}{3}\pi r^3

.

Using the standard formula for the volume of a sphere,

V = \frac{4}{3}\pi r^3

.

4

Find the volume of the water by subtracting the sphere's volume from the total filled region.

V_{\text{water}} = 2\pi r^3 - \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3

.

The space taken up by the water is the total volume up to height

2r

minus the displacement volume of the sphere.

5

Compute the required ratio of

V_{\text{empty}}

to

V_{\text{water}}

.

\frac{V_{\text{empty}}}{V_{\text{water}}} = \frac{\pi r^3}{\frac{2}{3}\pi r^3} = \frac{1}{\frac{2}{3}} = \frac{3}{2}

.

Dividing the empty space volume by the water volume simplifies to

\frac{3}{2}

.

Key Concept

Displacement and volume calculations combining 3D geometric figures (cylinders and spheres).

Estimated Time:2m 30s

Question 86Question →

A non-degenerate triangle has integer side lengths $a$ , $b$ , and $c$ such that $a \le b \le c$ . If the perimeter of the triangle is $30$ and its area is an integer, what is the length of the longest side $c$ ?

Show answer & explanation

Answer: 13

Answer

13

Using Heron's formula with semi-perimeter

s = 15

, the area of the triangle is

\sqrt{15(15-a)(15-b)(15-c)}

. By defining

x = 15-a

,

y = 15-b

, and

z = 15-c

, we have

x+y+z = 15

with

x \ge y \ge z \ge 1

. For the area to be an integer,

15xyz

must be a perfect square. The AM-GM inequality bounds

xyz \le 125

, leaving

xyz = 60

as the only valid case (

k=2

). The system

x+y+z=15

and

xyz=60

has a unique positive integer solution

(10, 3, 2)

, giving side lengths

(5, 12, 13)

. Thus, the longest side length

c

is

13

.

Step-by-Step Solution

1

Find the semi-perimeter of the triangle.

s = \frac{a+b+c}{2} = \frac{30}{2} = 15

Heron's formula requires the semi-perimeter

s

to compute the area of a triangle.

2

Set up Heron's formula for the area.

\text{Area} = \sqrt{15(15-a)(15-b)(15-c)}

Heron's formula relates the side lengths of a triangle directly to its area.

3

Define variables for the differences from the semi-perimeter.

Let

x = 15-a

,

y = 15-b

, and

z = 15-c

. Then

x+y+z = 15

and

x \ge y \ge z \ge 1

.

Because

a \le b \le c

, their complements with respect to

15

satisfy

x \ge y \ge z

.

4

Determine the condition for the area to be an integer.

\text{Area} = \sqrt{15xyz}

must be an integer, so

15xyz = m^2

for some integer

m

, implying

xyz = 15k^2

for an integer

k \ge 1

.

For

\sqrt{15xyz}

to be an integer,

15xyz

must be a perfect square containing

15

as a factor.

5

Bound

xyz

using the AM-GM inequality.

xyz \le \left(\frac{x+y+z}{3}\right)^3 = 5^3 = 125 \implies 15k^2 \le 125 \implies k=1 \text{ or } k=2

The AM-GM inequality provides a maximum upper bound for the product of three positive numbers with a fixed sum.

6

Evaluate the possible values of

k

.

If

k=1

,

xyz = 15

, which has no integer solutions for

x+y+z=15

. If

k=2

,

xyz = 60

, which yields

(x, y, z) = (10, 3, 2)

.

Solving the system

x+y+z=15

and

xyz=60

under

x \ge y \ge z \ge 1

gives

z=2

,

y=3

,

x=10

uniquely.

7

Calculate the side lengths and identify the longest side

c

.

a = 5

,

b = 12

,

c = 13

, so the longest side length is

13

.

Since

c = 15 - z = 15 - 2 = 13

, the longest side of the right triangle

(5, 12, 13)

is

13

.

Key Concept

Heron's formula, integer partitioning, and the AM-GM inequality for triangle area optimization

Question 87Question →

The function $f(x)$ is defined by $f(x) = |x - 3| - 2$ . The graph of $y = f(x)$ in the $xy$ -plane is reflected across the $y$ -axis to produce the graph of $y = g(x)$ . The graph of $y = g(x)$ is then translated $1$ unit to the left and $4$ units downward to produce the graph of $y = h(x)$ . Which of the following statements about the graph of $y = h(x)$ must be true? Select all that apply.

Select all that apply

The

y

-intercept of the graph of

y = h(x)

is

(0, -2)

.

The graph of

y = h(x)

is symmetric with respect to the line

x = -4

.

The area of the triangular region bounded by the graph of

y = h(x)

and the

x

-axis is

36

.

The minimum value of

h(x)

is

-2

.

The distance between the two

x

-intercepts of the graph of

y = h(x)

is

6

.

Show answer & explanation

Answer: The $y$ -intercept of the graph of $y = h(x)$ is $(0, -2)$ .; The graph of $y = h(x)$ is symmetric with respect to the line $x = -4$ .; The area of the triangular region bounded by the graph of $y = h(x)$ and the $x$ -axis is $36$ .

Answer

The statements establishing that the

y

-intercept is

(0, -2)

, that the line of symmetry is

x = -4

, and that the bounded region with the

x

-axis has an area of

36

are all correct.

Reflecting

f(x) = |x - 3| - 2

across the

y

-axis yields

g(x) = |x + 3| - 2

. Translating

g(x)

left by

1

unit and down by

4

units yields

h(x) = |x + 4| - 6

. From

h(x) = |x + 4| - 6

, evaluating

h(0) = -2

verifies the

y

-intercept of

(0, -2)

. The vertex at

(-4, -6)

defines the axis of symmetry at

x = -4

. Setting

h(x) = 0

gives

x

-intercepts at

-10

and

2

, producing a triangle bounded by the

x

-axis with base

12

and height

6

, which has an area of

\frac{1}{2} \times 12 \times 6 = 36

.

Step-by-Step Solution

1

Apply reflection across the

y

-axis to determine

g(x)

.

g(x) = f(-x) = |-x - 3| - 2 = |x + 3| - 2

Reflecting

y = f(x)

across the

y

-axis replaces every occurrence of

x

with

-x

.

2

Apply horizontal and vertical translations to determine

h(x)

.

h(x) = g(x + 1) - 4 = |(x + 1) + 3| - 2 - 4 = |x + 4| - 6

Translating a graph

1

unit left adds

1

to the input variable, and translating

4

units down subtracts

4

from the expression.

3

Find the

y

-intercept of

y = h(x)

.

h(0) = |0 + 4| - 6 = 4 - 6 = -2

, so the intercept point is

(0, -2)

.

The

y

-intercept is found by setting

x = 0

.

4

Determine the vertex, axis of symmetry, and minimum value.

Vertex is at

(-4, -6)

, line of symmetry is

x = -4

, and minimum value is

-6

.

For

y = a|x - h| + k

, the vertex is at

(h, k)

and the axis of symmetry is

x = h

.

5

Find the

x

-intercepts and compute the bounded triangular area.

|x + 4| - 6 = 0 \implies |x + 4| = 6 \implies x = 2

or

x = -10

. Base

= 2 - (-10) = 12

, height

= |-6| = 6

. Area

= \frac{1}{2} \times 12 \times 6 = 36

.

The region bounded by the V-shaped graph below the

x

-axis forms a triangle with base on the

x

-axis and height equal to the magnitude of the vertex's

y

-coordinate.

Key Concept

Coordinate transformations of algebraic functions including

y

-axis reflection, horizontal and vertical translations, and analyzing key geometric features of the resulting graph.

Question 88Question →

A regular polygon has interior angles that each measure $140^\circ$ . What is the total number of diagonals that can be drawn in this polygon?

9

14

27

36

54

Show answer & explanation

Answer: 27

Answer

The total number of diagonals that can be drawn in the polygon is 27.

To find the total number of diagonals in a regular polygon, first determine its number of sides

n

. Using the interior angle formula

\frac{(n-2) \times 180^\circ}{n} = 140^\circ

, we find

n = 9

. Next, applying the diagonal formula

D = \frac{n(n-3)}{2}

for

n = 9

gives

D = \frac{9 \times 6}{2} = 27

. Thus, the value 27 is correct.

Step-by-Step Solution

1

Find the number of sides

n

of the regular polygon using the interior angle measure.

The polygon has

n = 9

sides.

Each interior angle of a regular

n

-gon is given by

\frac{(n - 2) \times 180^\circ}{n} = 140^\circ

. Solving for

n

:

180n - 360 = 140n \implies 40n = 360 \implies n = 9

.

2

Calculate the total number of diagonals using the formula

D = \frac{n(n - 3)}{2}

.

The total number of diagonals is 27.

Substituting

n = 9

into the formula yields

D = \frac{9(9 - 3)}{2} = \frac{9 \times 6}{2} = 27

.

Key Concept

Interior Angle Measure and Diagonal Formula for Regular Polygons

Estimated Time:1m 0s

Question 89Question →

The side lengths of triangle $ABC$ , in units, are given by $AB = x$ , $BC = x + 2$ , and $CA = x + 4$ , where $x$ is a positive integer. If the perimeter of triangle $ABC$ is strictly less than $36$ , how many possible integer values of $x$ exist such that triangle $ABC$ is an acute triangle?

2

3

4

5

7

Show answer & explanation

Answer: 3

Answer

There are 3 possible integer values of

x

.

To form an acute triangle with side lengths

x

,

x+2

, and

x+4

, three conditions must be met: the perimeter bound (

3x+6 < 36 \implies x < 10

), the non-degeneracy condition (

x + x + 2 > x + 4 \implies x > 2

), and the acute angle condition (

(x+4)^2 < x^2 + (x+2)^2 \implies x^2 - 4x - 12 > 0 \implies x > 6

). Combining these bounds restricts

x

to integer values in the range

6 < x < 10

, giving exactly three valid integer values:

7, 8,

and

9

.

Step-by-Step Solution

1

Apply the perimeter constraint to find an upper bound for

x

.

The perimeter is

P = x + (x + 2) + (x + 4) = 3x + 6

. Given

3x + 6 < 36

, subtracting

6

yields

3x < 30

, so

x < 10

.

The total length around the triangle must be strictly less than

36

.

2

Apply the triangle inequality theorem to find a lower bound for non-degenerate triangles.

AB + BC > CA \implies x + (x + 2) > x + 4 \implies 2x + 2 > x + 4 \implies x > 2

.

The sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.

3

Formulate and solve the acute triangle condition.

For an acute triangle with longest side

CA = x + 4

, we require

CA^2 < AB^2 + BC^2

.

(x + 4)^2 < x^2 + (x + 2)^2

x^2 + 8x + 16 < x^2 + x^2 + 4x + 4

x^2 + 8x + 16 < 2x^2 + 4x + 4

0 < x^2 - 4x - 12

(x - 6)(x + 2) > 0

Since

x

is a positive integer,

x + 2 > 0

, which implies

x - 6 > 0

, so

x > 6

.

In any acute triangle, the square of the longest side is strictly less than the sum of the squares of the remaining two sides.

4

Combine the bounds to determine the number of valid integer values for

x

.

Combining

x > 6

and

x < 10

gives integer values

x \in \{7, 8, 9\}

. There are

3

such values.

Only integer values in the open interval

(6, 10)

satisfy all conditions simultaneously.

Key Concept

Acute Triangle Criteria and Triangle Inequality

Question 90Question →

In acute triangle $ABC$ , point $D$ lies on side $BC$ such that segment $AD$ is perpendicular to $BC$ . The length of side $BC$ is $15$ , and the ratio of the area of triangle $ABD$ to the area of triangle $ADC$ is $2 : 3$ . If the lengths of sides $AB$ and $AC$ are both integers, what is the perimeter of triangle $ABC$ ?

30

45

54

60

75

Show answer & explanation

Answer: $60$

Answer

The perimeter of triangle

ABC

is

60

.

The correct answer is derived by recognizing that the altitude splits the base into segments of lengths 6 and 9 based on the area ratio 2:3. Applying the Pythagorean theorem to both right triangles yields

AC^2 - AB^2 = 45

. Factoring 45 into positive integer pairs shows that the only valid side lengths yielding a real, positive height are

AB = 22

and

AC = 23

, giving a total perimeter of

15 + 22 + 23 = 60

.

Step-by-Step Solution

1

Determine the lengths of base segments

BD

and

DC

.

BD = 6

and

DC = 9

.

Triangles

ABD

and

ADC

share the common height

AD = h

. The ratio of their areas is equal to the ratio of their bases:

\frac{\text{Area}(ABD)}{\text{Area}(ADC)} = \frac{BD}{DC} = \frac{2}{3}

. Since

BD + DC = 15

, we have

BD = 6

and

DC = 9

.

2

Express the square of height

h^2

using the Pythagorean theorem in right triangles

ABD

and

ADC

.

h^2 = AB^2 - 36 = AC^2 - 81

.

In right triangle

ABD

,

AB^2 = BD^2 + h^2 = 36 + h^2

. In right triangle

ADC

,

AC^2 = DC^2 + h^2 = 81 + h^2

.

3

Set up a difference of squares equation for side lengths

AB

and

AC

.

(AC - AB)(AC + AB) = 45

.

Equating the two expressions for

h^2

gives

AC^2 - 81 = AB^2 - 36 \implies AC^2 - AB^2 = 45

.

4

Find positive integer solutions for

AB

and

AC

.

AB = 22

and

AC = 23

.

Since

AB

and

AC

are positive integers and

AC > AB

, we analyze factor pairs

(AC - AB, AC + AB)

of

45

with same parity (both odd):
- Pair

(1, 45)

:

AC - AB = 1

and

AC + AB = 45 \implies AC = 23, AB = 22

. Here

h^2 = 22^2 - 36 = 448 > 0

, giving a valid non-degenerate triangle.
- Pair

(3, 15)

:

AC - AB = 3

and

AC + AB = 15 \implies AC = 9, AB = 6

. Here

h^2 = 6^2 - 36 = 0

, which means

h = 0

(degenerate line segment, invalid).
- Pair

(5, 9)

:

AC - AB = 5

and

AC + AB = 9 \implies AC = 7, AB = 2

. Here

h^2 = 2^2 - 36 = -32 < 0

(impossible).

5

Calculate the total perimeter of triangle

ABC

.

Perimeter

= 15 + 22 + 23 = 60

.

Summing all three side lengths gives

BC + AB + AC = 15 + 22 + 23 = 60

.

Key Concept

Triangles: Properties, Perimeter, and Area

Question 91Question →

In quadrilateral $ABCD$ , diagonal $AC$ divides the quadrilateral into two right triangles, $\triangle ABC$ and $\triangle ADC$ . The measure of angle $ABC$ is $90^\circ$ , the measure of angle $ADC$ is $90^\circ$ , and the measure of angle $BAC$ is $30^\circ$ . If $AB = 12$ and $AD = CD$ , what is the area of quadrilateral $ABCD$ ?

48 + 24\sqrt{3}

36 + 18\sqrt{3}

192 + 24\sqrt{3}

96 + 48\sqrt{3}

24 + 48\sqrt{3}

Show answer & explanation

Answer: $48 + 24\sqrt{3}$

Answer

48 + 24\sqrt{3}

The total area of quadrilateral ABCD is the sum of the areas of right triangles ABC and ADC. Using 30-60-90 triangle properties on ABC gives

BC = 4\sqrt{3}

and hypotenuse

AC = 8\sqrt{3}

, resulting in

\text{Area}(\triangle ABC) = 24\sqrt{3}

. Triangle ADC is a 45-45-90 triangle with hypotenuse

8\sqrt{3}

, yielding leg lengths

4\sqrt{6}

and

\text{Area}(\triangle ADC) = 48

. Adding both areas gives

48 + 24\sqrt{3}

.

Step-by-Step Solution

1

Analyze right triangle ABC to find side lengths BC and hypotenuse AC.

BC = 4\sqrt{3}

and

AC = 8\sqrt{3}

In 30-60-90 triangle ABC,

\cos(30^\circ) = \frac{AB}{AC} \implies \frac{\sqrt{3}}{2} = \frac{12}{AC} \implies AC = 8\sqrt{3}

. Also,

\tan(30^\circ) = \frac{BC}{AB} \implies \frac{1}{\sqrt{3}} = \frac{BC}{12} \implies BC = 4\sqrt{3}

.

2

Calculate the area of right triangle ABC.

\text{Area}(\triangle ABC) = 24\sqrt{3}

\text{Area}(\triangle ABC) = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 12 \times 4\sqrt{3} = 24\sqrt{3}

.

3

Analyze isosceles right triangle ADC to find leg length

s = AD = CD

.

s = 4\sqrt{6}

In 45-45-90 triangle ADC, the hypotenuse

AC = s\sqrt{2} = 8\sqrt{3} \implies s = \frac{8\sqrt{3}}{\sqrt{2}} = 4\sqrt{6}

.

4

Calculate the area of right triangle ADC.

\text{Area}(\triangle ADC) = 48

\text{Area}(\triangle ADC) = \frac{1}{2} \times s^2 = \frac{1}{2} \times (4\sqrt{6})^2 = \frac{1}{2} \times 96 = 48

.

5

Sum the areas of the two triangles to get total area of quadrilateral ABCD.

\text{Total Area} = 48 + 24\sqrt{3}

\text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 24\sqrt{3} + 48 = 48 + 24\sqrt{3}

.

Key Concept

Special Right Triangle Ratios (30-60-90 and 45-45-90)

Alternative Method

Use trigonometric ratios directly:

\text{Area}(\triangle ABC) = \frac{1}{2} (12)^2 \tan(30^\circ) = 72 \times \frac{1}{\sqrt{3}} = 24\sqrt{3}

, and

\text{Area}(\triangle ADC) = \frac{1}{4} AC^2 = \frac{1}{4} (8\sqrt{3})^2 = 48

.

Estimated Time:2m 30s

Question 92Question →

In isosceles triangle $ABC$ , side $AB$ is equal in length to side $AC$ . The perimeter of triangle $ABC$ is $36$ , and the length of the altitude from vertex $A$ to base $BC$ is $12$ . What is the area of triangle $ABC$ ?

Show answer & explanation

Answer: 60

Answer

The area of triangle

ABC

is

60

.

Let

x

be the length of the two equal sides

AB

and

AC

, and

b

be the length of base

BC

. The perimeter is

2x + b = 36

, yielding

x = 18 - \frac{b}{2}

. The altitude from

A

to

BC

has length

12

and bisects

BC

into two segments of length

\frac{b}{2}

. Applying the Pythagorean theorem to one of the right triangles gives

x^2 = 12^2 + \left(\frac{b}{2}\right)^2

. Substituting

x = 18 - \frac{b}{2}

gives

\left(18 - \frac{b}{2}\right)^2 = 144 + \frac{b^2}{4} \implies 324 - 18b + \frac{b^2}{4} = 144 + \frac{b^2}{4} \implies 18b = 180 \implies b = 10

. The area is

\frac{1}{2} \times 10 \times 12 = 60

.

Step-by-Step Solution

1

Set up an equation for the side lengths using the perimeter.

Let

b

be the length of base

BC

, and

x

be the length of sides

AB

and

AC

. Since the perimeter is

36

,

2x + b = 36

, which gives

x = 18 - \frac{b}{2}

.

An isosceles triangle has two sides of equal length, and perimeter is the sum of all three side lengths.

2

Apply the Pythagorean theorem to the right triangle formed by the altitude.

The altitude of length

12

drops perpendicularly to base

BC

, bisecting it into two equal segments of length

\frac{b}{2}

. Thus,

x^2 = 12^2 + \left(\frac{b}{2}\right)^2 = 144 + \frac{b^2}{4}

.

In an isosceles triangle, the altitude to the base bisects the base and creates two congruent right-angled triangles.

3

Solve for the base length

b

.

Substitute

x = 18 - \frac{b}{2}

into the equation:

\left(18 - \frac{b}{2}\right)^2 = 144 + \frac{b^2}{4} \implies 324 - 18b + \frac{b^2}{4} = 144 + \frac{b^2}{4} \implies 18b = 180 \implies b = 10

.

Expanding the squared binomial allows the

\frac{b^2}{4}

terms to cancel out, resulting in a linear equation for

b

.

4

Calculate the area of the triangle.

\text{Area} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 10 \times 12 = 60.

The area of a triangle is evaluated using half the product of its base and corresponding altitude.

Key Concept

Isosceles triangle properties, altitude-to-base bisector property, Pythagorean theorem, and triangle area calculation.

Question 93Question →

In the coordinate plane, segment $AB$ lies along the positive $x$ -axis with point $A$ at the origin $(0,0)$ and point $B$ at $(6\sqrt{3}, 0)$ . Point $C$ is located in the first quadrant such that $\triangle ABC$ is a right triangle with $\angle ACB = 90^\circ$ and $\angle CAB = 30^\circ$ . Point $D$ is also located in the first quadrant such that $\triangle ABD$ is an isosceles right triangle with hypotenuse $AB$ and $\angle ADB = 90^\circ$ .

Which of the following statements regarding this figure are true? Select all that apply.

Select all that apply

The length of segment

AC

is

9

.

The length of segment

AD

is

3\sqrt{6}

.

The area of triangle

ABD

is

27

.

The length of segment

BC

is

9

.

The length of segment

AD

is

6\sqrt{6}

.

Show answer & explanation

Answer: The length of segment $AC$ is $9$ .; The length of segment $AD$ is $3\sqrt{6}$ .; The area of triangle $ABD$ is $27$ .

Answer

The correct statements are: the length of segment

AC

is

9

, the length of segment

AD

is

3\sqrt{6}

, and the area of triangle

ABD

is

27

.

The statement that

AC = 9

is correct because in

\triangle ABC

,

AC = AB \cos(30^\circ) = 6\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 9

. The statement that

AD = 3\sqrt{6}

is correct because in isosceles right

\triangle ABD

,

AD = \frac{AB}{\sqrt{2}} = \frac{6\sqrt{3}}{\sqrt{2}} = 3\sqrt{6}

. The statement that the area of

\triangle ABD

is

27

is correct because

\frac{1}{2}(3\sqrt{6})(3\sqrt{6}) = 27

.

Step-by-Step Solution

1

Determine the side lengths of 30-60-90 triangle ABC

Hypotenuse

AB = 6\sqrt{3}

. Leg

BC

opposite

30^\circ

is

\frac{1}{2}(6\sqrt{3}) = 3\sqrt{3}

. Leg

AC

opposite

60^\circ

is

3\sqrt{3} \cdot \sqrt{3} = 9

.

In a

30^\circ-60^\circ-90^\circ

triangle, side ratios are

1 : \sqrt{3} : 2

relative to angles

30^\circ : 60^\circ : 90^\circ

.

2

Determine the side lengths of 45-45-90 triangle ABD

Hypotenuse

AB = 6\sqrt{3}

. Legs

AD = BD = \frac{6\sqrt{3}}{\sqrt{2}} = 3\sqrt{6}

.

In a

45^\circ-45^\circ-90^\circ

isosceles right triangle, side ratios are

1 : 1 : \sqrt{2}

, so leg length equals hypotenuse divided by

\sqrt{2}

.

3

Calculate the area of right triangle ABD

Area

(\triangle ABD) = \frac{1}{2} \cdot AD \cdot BD = \frac{1}{2} (3\sqrt{6})(3\sqrt{6}) = \frac{1}{2} (54) = 27

.

The area of a right triangle is half the product of its perpendicular legs.

4

Evaluate each given option against computed values

Segment

AC = 9

is true. Segment

AD = 3\sqrt{6}

is true. Area of

\triangle ABD = 27

is true. Segment

BC = 9

is false (

BC = 3\sqrt{3}

). Segment

AD = 6\sqrt{6}

is false (

AD = 3\sqrt{6}

).

Direct comparison with calculated geometric dimensions.

Key Concept

Side ratios of 30-60-90 (

1:\sqrt{3}:2

) and 45-45-90 (

1:1:\sqrt{2}

) special right triangles

Question 94Question →

A rectangle has a length of $8$ units and a width of $6$ units. Which of the following statements about this rectangle must be true? Select all that apply.

Select all that apply

The area of the rectangle is

48

square units.

The length of each diagonal of the rectangle is

10

units.

The perimeter of the rectangle is

28

units.

The length of each diagonal of the rectangle is

14

units.

The area of the rectangle is

28

square units.

Show answer & explanation

Answer: The area of the rectangle is $48$ square units.; The length of each diagonal of the rectangle is $10$ units.; The perimeter of the rectangle is $28$ units.

Answer

The correct statements are that the area of the rectangle is 48 square units, the length of each diagonal is 10 units, and the perimeter of the rectangle is 28 units.

The statement regarding the area being 48 square units is correct because

8 \times 6 = 48

. The statement regarding the diagonal being 10 units is correct because

\sqrt{8^2 + 6^2} = 10

. The statement regarding the perimeter being 28 units is correct because

2 \times (8 + 6) = 28

.

Step-by-Step Solution

1

Calculate the area of the rectangle

\text{Area} = 8 \times 6 = 48

square units

The area formula for a rectangle is length multiplied by width.

2

Calculate the length of the diagonal using the Pythagorean theorem

\text{Diagonal} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10

units

The sides and diagonal of a rectangle form a right triangle where the diagonal is the hypotenuse.

3

Calculate the perimeter of the rectangle

\text{Perimeter} = 2 \times (8 + 6) = 28

units

The perimeter formula for a rectangle is twice the sum of its length and width.

Key Concept

Basic geometric properties of rectangles including area, perimeter, and diagonal calculation via the Pythagorean theorem.

Question 95Question →

In a circle centered at point $O$ , the radius is $18$ units. Points $P$ and $Q$ lie on the circle such that the area of sector $POQ$ is $54\pi$ square units. What is the ratio of the length of minor arc $PQ$ to the total perimeter of sector $POQ$ ?

\frac{\pi}{\pi + 6}

\frac{1}{6}

\frac{\pi}{\pi + 3}

\frac{6\pi}{\pi + 6}

\frac{9\pi}{\pi + 6}

Show answer & explanation

Answer: $\frac{\pi}{\pi + 6}$

Answer

The ratio of the length of minor arc

PQ

to the total perimeter of sector

POQ

is

\frac{\pi}{\pi + 6}

.

The full circle area is

324\pi

, making the sector

54\pi / 324\pi = 1/6

of the circle. The arc length is

1/6 \times 36\pi = 6\pi

. The sector perimeter is the arc length plus two radii (

6\pi + 36

). Taking the ratio of arc length to sector perimeter yields

6\pi / (6\pi + 36) = \pi / (\pi + 6)

.

Step-by-Step Solution

1

Calculate the total area of the circle and determine the fractional size of sector

POQ

.

Total area =

\pi(18)^2 = 324\pi

. Sector fraction =

\frac{54\pi}{324\pi} = \frac{1}{6}

.

Determining the fraction of the circle represented by the sector is required to find the arc length.

2

Calculate the length of minor arc

PQ

and the total perimeter of sector

POQ

.

Minor arc

PQ = \frac{1}{6} \times 2\pi(18) = 6\pi

. Sector perimeter =

6\pi + 2(18) = 6\pi + 36

.

The sector perimeter consists of the curved arc length plus the two straight radii

OP

and

OQ

.

3

Form and simplify the ratio of arc length to sector perimeter.

\frac{6\pi}{6\pi + 36} = \frac{6\pi}{6(\pi + 6)} = \frac{\pi}{\pi + 6}

.

Factoring out 6 from the numerator and denominator simplifies the expression to its lowest form.

Key Concept

Arc Length and Sector Perimeter Calculations

Question 96Question →

A hollow metallic spherical shell has an inner radius of $3$ centimeters and an outer radius of $r$ centimeters, where $r > 3$ . If the spherical shell is melted down and completely recast into a solid right circular cylinder with base radius $r$ centimeters and height $7$ centimeters, what is the value of $r$ ?

4.5

5.25

6

7.5

9

Show answer & explanation

Answer: $6$

Answer

6

The volume of metal in the hollow spherical shell is

\frac{4}{3}\pi(r^3 - 3^3) = \frac{4}{3}\pi(r^3 - 27)

. The volume of the recast cylinder is

\pi r^2 h = 7\pi r^2

. Equating these volumes gives

\frac{4}{3}(r^3 - 27) = 7r^2

, which simplifies to

4r^3 - 21r^2 - 108 = 0

. Factoring this cubic equation gives

(r - 6)(4r^2 + 3r + 18) = 0

. Since the quadratic term has no real roots, the only real solution is

r = 6

.

Step-by-Step Solution

1

Set up the formula for the volume of metal in the hollow spherical shell.

V_{\text{shell}} = \frac{4}{3}\pi \left(r^3 - 3^3\right) = \frac{4}{3}\pi \left(r^3 - 27\right)

The metal occupies only the region between the inner sphere of radius 3 cm and outer sphere of radius r cm.

2

Set up the formula for the volume of the recast solid right circular cylinder.

V_{\text{cylinder}} = \pi r^2 h = 7\pi r^2

The cylinder has base radius r cm and height 7 cm.

3

Equate the two volumes since no metal is lost during melting and recasting.

\frac{4}{3}\pi \left(r^3 - 27\right) = 7\pi r^2

Conservation of volume during recasting.

4

Simplify the equation and solve for r.

4(r^3 - 27) = 21r^2 \implies 4r^3 - 21r^2 - 108 = 0

Divide both sides by

\pi

and multiply by 3 to clear the fraction.

5

Factor the cubic polynomial

4r^3 - 21r^2 - 108 = 0

.

(r - 6)(4r^2 + 3r + 18) = 0 \implies r = 6

Testing

r = 6

gives

4(216) - 21(36) - 108 = 864 - 756 - 108 = 0

. The quadratic factor

4r^2 + 3r + 18

has a negative discriminant and produces no real roots.

Key Concept

Volume formulas for hollow spheres and right circular cylinders

Question 97Question →

A triangle has side lengths of $8$ , $11$ , and $x$ , where $x$ is an integer. If the perimeter of the triangle is a positive integer multiple of $5$ , which of the following could be the value of $x$ ? Select all such values.

Select all that apply

1

6

11

16

21

Show answer & explanation

Answer: $6$ ; $11$ ; $16$

Answer

The possible values for

x

are

6

,

11

, and

16

.

According to the Triangle Inequality Theorem, the third side

x

must be strictly greater than

11 - 8 = 3

and strictly less than

11 + 8 = 19

. The perimeter of the triangle is

8 + 11 + x = 19 + x

. For

19 + x

to be a positive multiple of

5

,

19 + x

can be

25

,

30

, or

35

within the allowed range for

x

, giving

x = 6

,

x = 11

, and

x = 16

.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to determine the valid range for the unknown side

x

.

The length of

x

must satisfy

11 - 8 < x < 11 + 8

, which simplifies to

3 < x < 19

.

The sum of any two side lengths of a non-degenerate triangle must be strictly greater than the third side length.

2

Set up an expression for the perimeter

P

of the triangle.

P = 8 + 11 + x = 19 + x

.

Perimeter is the total sum of all three side lengths.

3

Determine which values of

x

within the range

3 < x < 19

make

P = 19 + x

a multiple of

5

.

If

x = 6

,

P = 25

(multiple of

5

). If

x = 11

,

P = 30

(multiple of

5

). If

x = 16

,

P = 35

(multiple of

5

).

Adding

19

to

6

,

11

, and

16

gives multiples of

5

within the strict inequality bounds.

4

Test boundary values outside the inequality bounds.

x = 1

gives

P = 20

, but

1 \le 3

(invalid).

x = 21

gives

P = 40

, but

21 \ge 19

(invalid).

Values outside

3 < x < 19

cannot form a valid triangle.

Key Concept

Triangle Inequality Theorem and Perimeter Constraints

Question 98Question →

In the $xy$ -plane, line $L$ is given by the equation $3x + 2y = 18$ . Line $N$ is perpendicular to line $L$ and passes through the point $(-1, 4)$ . If line $L$ and line $N$ intersect at the point $(p, q)$ , what is the value of $p + q$ ?

Show answer & explanation

Answer: 8

Answer

8

Converting the given equation

3x + 2y = 18

into slope-intercept form yields

y = -\frac{3}{2}x + 9

, so line

L

has slope

-\frac{3}{2}

. A perpendicular line must have a slope equal to the negative reciprocal, which is

\frac{2}{3}

. Using point

(-1, 4)

in the point-slope form gives

y - 4 = \frac{2}{3}(x + 1)

, simplifying to

2x - 3y = -14

. Solving the system of equations formed by line

L

(

3x + 2y = 18

) and line

N

(

2x - 3y = -14

) via elimination yields

x = 2

and

y = 6

. Therefore, the intersection point is

(2, 6)

, and

p + q = 2 + 6 = 8

.

Step-by-Step Solution

1

Determine the slope of line

L

Slope of line

L

is

m_L = -\frac{3}{2}

Convert

3x + 2y = 18

to

y = -\frac{3}{2}x + 9

to identify the slope coefficient of

x

.

2

Calculate the perpendicular slope for line

N

Slope of line

N

is

m_N = \frac{2}{3}

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Derive the equation for line

N

Equation of line

N

is

2x - 3y = -14

Apply point-slope form

y - 4 = \frac{2}{3}(x + 1)

and rearrange into standard linear form.

4

Solve the system of equations to find the intersection point

(p, q)

p = 2

and

q = 6

, giving point

(2, 6)

Eliminate variable

y

by adding

3 \times (3x + 2y = 18)

and

2 \times (2x - 3y = -14)

to get

13x = 26

.

5

Sum the coordinates

p

and

q

p + q = 8

Evaluate

2 + 6 = 8

as required by the stem.

Key Concept

Perpendicular line slope relationships and linear system intersection

Estimated Time:2m 0s

Question 99Question →

In triangle $ABC$ , the ratio of the side lengths $AB : BC : AC$ is $3 : 4 : 5$ , and the total area of triangle $ABC$ is $24$ square units. A line segment $DE$ is drawn parallel to side $BC$ , where point $D$ lies on side $AB$ and point $E$ lies on side $AC$ . If the perimeter of triangle $ADE$ is exactly half the perimeter of triangle $ABC$ , what is the area of trapezoid $DBCE$ in square units?

6

12

16

18

20

Show answer & explanation

Answer: 18

Answer

The area of trapezoid

DBCE

is

18

square units.

Because line segment

DE

is parallel to side

BC

, triangle

ADE

is similar to triangle

ABC

. Given that the perimeter of triangle

ADE

is half the perimeter of triangle

ABC

, the ratio of their side lengths (the linear scale factor) is

1/2

. The area ratio of similar triangles is the square of the linear scale factor, which is

(1/2)^2 = 1/4

. Thus, the area of triangle

ADE

is

1/4 \times 24 = 6

square units. Subtracting this from the total area gives the area of trapezoid

DBCE

:

24 - 6 = 18

square units.

Step-by-Step Solution

1

Determine the linear scale factor between triangle

ADE

and triangle

ABC

.

Linear scale factor

k = \frac{\text{Perimeter}(ADE)}{\text{Perimeter}(ABC)} = \frac{1}{2}

.

Since

DE \parallel BC

, triangle

ADE

is similar to triangle

ABC

, so the ratio of their perimeters equals the ratio of corresponding side lengths.

2

Calculate the area of triangle

ADE

using the area scale factor

k^2

.

\text{Area}(ADE) = k^2 \times \text{Area}(ABC) = \left(\frac{1}{2}\right)^2 \times 24 = \frac{1}{4} \times 24 = 6

square units.

The ratio of the areas of two similar figures is the square of their linear scale factor.

3

Subtract the area of triangle

ADE

from the area of triangle

ABC

to find the area of trapezoid

DBCE

.

\text{Area}(DBCE) = \text{Area}(ABC) - \text{Area}(ADE) = 24 - 6 = 18

square units.

Trapezoid

DBCE

is formed by removing triangle

ADE

from triangle

ABC

.

Key Concept

Properties of Similar Triangles and Area Scaling

Estimated Time:1m 30s

Question 100Question →

Trapezoid $PQRS$ has parallel sides $PQ$ and $RS$ with lengths of $7$ centimeters and $13$ centimeters, respectively. If the perpendicular height between these parallel sides is $4$ centimeters, what is the area of trapezoid $PQRS$ , in square centimeters?

24

28

40

52

80

Show answer & explanation

Answer: $40$

Answer

The area of trapezoid

PQRS

is

40

square centimeters.

The area of a trapezoid is found by averaging the lengths of the two parallel bases and multiplying by the perpendicular height:

\text{Area} = \frac{7 + 13}{2} \times 4 = 10 \times 4 = 40

square centimeters.

Step-by-Step Solution

1

Identify the formula for the area of a trapezoid

\text{Area} = \frac{b_1 + b_2}{2} \times h

, where

b_1

and

b_2

are the lengths of the parallel bases and

h

is the height.

The area of any trapezoid is equal to the average of its parallel bases multiplied by its perpendicular height.

2

Substitute the given dimensions into the formula

\text{Area} = \frac{7 + 13}{2} \times 4

The given bases are

b_1 = 7\text{ cm}

and

b_2 = 13\text{ cm}

, and the height is

h = 4\text{ cm}

.

3

Calculate the average base length and multiply by the height

\frac{20}{2} \times 4 = 10 \times 4 = 40\text{ cm}^2

Simplifying

\frac{7 + 13}{2}

gives

10

, and

10 \times 4 = 40

.

Key Concept

Trapezoid Area Formula

Estimated Time:45s

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