Geometry Practice Questions — GRE General Test — Page 6

Question 101Question →

In a circle centered at point $O$ , the radius is $6$ units and central angle $\angle POQ$ measures $120^\circ$ . Which of the following statements regarding sector $POQ$ and minor arc $PQ$ must be true? Select all such statements.

Select all that apply

The length of minor arc

PQ

is

4\pi

units.

The area of sector

POQ

is

12\pi

square units.

The area of sector

POQ

is

36\pi

square units.

The length of minor arc

PQ

is

12\pi

units.

The perimeter of sector

POQ

is

4\pi + 6

units.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ units.; The area of sector $POQ$ is $12\pi$ square units.

Answer

The correct statements are that the length of minor arc

PQ

is

4\pi

units and the area of sector

POQ

is

12\pi

square units.

The correct statements accurately apply the arc length and sector area formulas by multiplying the full circumference (

12\pi

) and full area (

36\pi

) by the central angle fraction

\frac{120^\circ}{360^\circ} = \frac{1}{3}

, yielding an arc length of

4\pi

units and a sector area of

12\pi

square units.

Step-by-Step Solution

1

Calculate the central angle fraction of the circle.

The fraction of the circle represented by sector

POQ

is

\frac{120^\circ}{360^\circ} = \frac{1}{3}

.

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

360^\circ

.

2

Calculate the arc length of minor arc

PQ

.

\text{Arc length} = \frac{1}{3} \times 2\pi(6) = 4\pi \text{ units}.

Arc length equals the central angle fraction times the total circumference

2\pi r

.

3

Calculate the area of sector

POQ

.

\text{Sector area} = \frac{1}{3} \times \pi(6^2) = 12\pi \text{ square units}.

Sector area equals the central angle fraction times the total circle area

\pi r^2

.

4

Evaluate the given statements against the calculated values.

Statements asserting an arc length of

4\pi

units and a sector area of

12\pi

square units are true. Other statements miscalculate by omitting the fraction or adding incorrect boundary components.

Comparing calculated values confirms the valid choices.

Key Concept

Arc Length and Sector Area Formulas

Question 102Question →

In the $xy$ -plane, triangle $PQR$ has vertices at $P(0, 0)$ , $Q(8, 0)$ , and $R(2, 6)$ . Point $S$ lies on segment $PQ$ such that segment $RS$ divides triangle $PQR$ into two regions of equal area. Point $T$ lies on segment $QR$ such that segment $ST$ is parallel to segment $PR$ . What is the area of triangle $QST$ ?

Show answer & explanation

Answer: 6

Answer

6

The area of triangle

PQR

is calculated using base

PQ = 8

and height

h = 6

, yielding

\frac{1}{2} \times 8 \times 6 = 24

. Since segment

RS

divides triangle

PQR

into two regions of equal area that share the height from vertex

R

, point

S

must be the midpoint of

PQ

, making

QS = 4

. Because segment

ST

is parallel to segment

PR

, triangle

QST

is similar to triangle

QPR

with a side length ratio of

\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}

. The ratio of the areas of similar triangles is the square of the side ratio,

\left(\frac{1}{2}\right)^2 = \frac{1}{4}

. Multiplying the total area

24

by

\frac{1}{4}

gives the area of triangle

QST

as

6

.

Step-by-Step Solution

1

Calculate the area of the main triangle

PQR

.

\text{Area}(PQR) = \frac{1}{2} \times 8 \times 6 = 24

.

The base

PQ

lies along the x-axis with length

8 - 0 = 8

, and the perpendicular height from vertex

R(2,6)

to the base is

6

.

2

Determine the length of segment

QS

.

QS = 4

.

Line segment

RS

splits

\triangle PQR

into two smaller triangles,

\triangle PSR

and

\triangle QSR

, which share the same altitude from vertex

R

. For their areas to be equal, their base lengths

PS

and

SQ

must be equal. Therefore,

S

is the midpoint of

PQ

, giving

QS = \frac{8}{2} = 4

.

3

Establish the similarity relationship and scale factor between

\triangle QST

and

\triangle QPR

.

\triangle QST \sim \triangle QPR

with scale factor

k = \frac{1}{2}

.

Because segment

ST

is parallel to segment

PR

, corresponding angles are equal (

\angle QST = \angle QPR

and

\angle QTS = \angle QRP

). Thus,

\triangle QST

is similar to

\triangle QPR

. The ratio of corresponding side lengths is

\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}

.

4

Compute the area of triangle

QST

.

\text{Area}(QST) = 6

.

The ratio of the areas of similar triangles is equal to the square of their linear scale factor:

\text{Area}(QST) = \left(\frac{1}{2}\right)^2 \times \text{Area}(PQR) = \frac{1}{4} \times 24 = 6

.

Key Concept

Area of triangles, midpoint area partitioning, and area ratio scaling in similar triangles

Estimated Time:2m 0s

Question 103Question →

In $\triangle ABC$ , the measure of $\angle A$ is $45^\circ$ , the measure of $\angle C$ is $30^\circ$ , and the length of side $AC$ is $6 + 2\sqrt{3}$ . What is the length of side $AB$ ?

2\sqrt{6}

6\sqrt{2}

4\sqrt{3}

2\sqrt{3}

3\sqrt{2} + \sqrt{6}

Show answer & explanation

Answer: $2\sqrt{6}$

Answer

The length of side

AB

is

2\sqrt{6}

.

The correct answer is

2\sqrt{6}

. By drawing altitude

BD

from vertex

B

to side

AC

,

\triangle ABC

is decomposed into right triangle

BDA

(a

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

triangle) and right triangle

BDC

(a

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

triangle). Setting altitude

BD = x

gives

AD = x

and

CD = x\sqrt{3}

. Combining these gives

AC = x + x\sqrt{3} = x(1 + \sqrt{3})

. Equating this to

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

gives

x = 2\sqrt{3}

. The hypotenuse

AB

of the

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

triangle is

x\sqrt{2} = (2\sqrt{3})\sqrt{2} = 2\sqrt{6}

.

Step-by-Step Solution

1

Draw altitude

BD

perpendicular to side

AC

with point

D

lying on segment

AC

.

\triangle ABC

is partitioned into two adjacent right triangles:

\triangle BDA

and

\triangle BDC

.

Constructing an interior altitude allows the application of special right triangle ratio rules.

2

Analyze

\triangle BDC

(

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

right triangle).

If

BD = x

, then

CD = x\sqrt{3}

and

BC = 2x

.

In a

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

triangle, sides opposite the angles are in the ratio

1 : \sqrt{3} : 2

.

3

Analyze

\triangle BDA

(

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

right triangle).

AD = BD = x

, and hypotenuse

AB = x\sqrt{2}

.

In a

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

isosceles right triangle, legs are equal and the hypotenuse is

\text{leg} \times \sqrt{2}

.

4

Set up an equation for total side length

AC = AD + CD

.

x + x\sqrt{3} = 6 + 2\sqrt{3} \implies x(1 + \sqrt{3}) = 2\sqrt{3}(1 + \sqrt{3}) \implies x = 2\sqrt{3}

.

Segment addition postulate combines

AD

and

CD

to match given total length

AC

.

5

Calculate requested side length

AB

.

AB = x\sqrt{2} = (2\sqrt{3})(\sqrt{2}) = 2\sqrt{6}

.

Substitute

x = 2\sqrt{3}

into the expression for hypotenuse

AB

.

Key Concept

Partitioning non-right triangles into

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

and

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

special right triangles by constructing an altitude.

Question 104Question →

In the $xy$ -plane, line $k$ passes through the origin $(0,0)$ and the point $(4, 3)$ . Line $m$ is perpendicular to line $k$ and intersects line $k$ at point $(4, 3)$ . If line $m$ intersects the $x$ -axis at point $Q$ and the $y$ -axis at point $R$ , what is the distance between point $Q$ and point $R$ ?

\frac{35}{12}

\frac{25}{12}

10

\frac{125}{12}

\frac{175}{12}

Show answer & explanation

Answer: $\frac{125}{12}$

Answer

\frac{125}{12}

The correct answer is derived by finding the slope of line

k

as

\frac{3}{4}

, taking its negative reciprocal

-\frac{4}{3}

as the slope of perpendicular line

m

, using the point

(4,3)

to form line

m

's equation

y = -\frac{4}{3}x + \frac{25}{3}

, identifying intercepts

Q\left(\frac{25}{4}, 0\right)

and

R\left(0, \frac{25}{3}\right)

, and applying the distance formula to obtain

\sqrt{\left(\frac{25}{4}\right)^2 + \left(\frac{25}{3}\right)^2} = \frac{125}{12}

.

Step-by-Step Solution

1

Determine the slope of line

k

The slope of line

k

is

m_k = \frac{3 - 0}{4 - 0} = \frac{3}{4}

.

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Find the equation of line

m

Line

m

has slope

m_m = -\frac{4}{3}

and equation

y - 3 = -\frac{4}{3}(x - 4) \implies y = -\frac{4}{3}x + \frac{25}{3}

.

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

m_m = -\frac{1}{m_k}

). Point-slope form is used with point

(4, 3)

.

3

Calculate the intercepts

Q

and

R

x

-intercept

Q = \left(\frac{25}{4}, 0\right)

and

y

-intercept

R = \left(0, \frac{25}{3}\right)

.

Setting

y = 0

in

y = -\frac{4}{3}x + \frac{25}{3}

gives

x = \frac{25}{4}

. Setting

x = 0

gives

y = \frac{25}{3}

.

4

Compute the distance between

Q

and

R

Distance

d = \sqrt{\left(\frac{25}{4}\right)^2 + \left(\frac{25}{3}\right)^2} = 25 \sqrt{\frac{1}{16} + \frac{1}{9}} = 25 \sqrt{\frac{25}{144}} = 25 \times \frac{5}{12} = \frac{125}{12}

.

Applying the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

for points

Q

and

R

.

Key Concept

Perpendicular slopes and distance formula in coordinate geometry

Estimated Time:2m 0s

Question 105Question →

An altitude of an acute triangle divides its base into two adjacent segments of lengths $5$ and $9$ . If the area of the triangle is $84$ square units, what is the perimeter of the triangle?

28

36

42

48

52

Show answer & explanation

Answer: $42$

Answer

The perimeter of the triangle is

42

.

The base of the triangle is the sum of the two adjacent segments,

5 + 9 = 14

. Setting up the area formula gives

84 = \frac{1}{2} \times 14 \times h

, which yields an altitude height of

h = 12

. The altitude creates two right triangles: one with legs

5

and

12

(hypotenuse

\sqrt{5^2 + 12^2} = 13

) and another with legs

9

and

12

(hypotenuse

\sqrt{9^2 + 12^2} = 15

). Adding all three side lengths (

14 + 13 + 15

) yields

42

.

Step-by-Step Solution

1

Find the total base length and calculate the height (altitude) of the triangle.

Base

= 5 + 9 = 14

. Height

h = \frac{2 \times \text{Area}}{\text{base}} = \frac{2 \times 84}{14} = 12

.

The area of a triangle is given by

A = \frac{1}{2} b h

.

2

Calculate the lengths of the two non-base sides using the Pythagorean theorem on the two right triangles formed by the altitude.

Left side

= \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13

. Right side

= \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = 15

.

An altitude drawn to the base creates two right triangles sharing the altitude as a common leg.

3

Sum all three boundary sides to find the perimeter.

Perimeter

= 14 + 13 + 15 = 42

.

Perimeter is the total length around the outside of the triangle.

Key Concept

Area and perimeter of triangles split by an altitude using the Pythagorean theorem.

Estimated Time:1m 30s

Question 106Question →

Three adjacent faces of a right rectangular prism have surface areas of $24$ , $36$ , and $54$ square centimeters. What is the volume of the prism, in cubic centimeters?

Show answer & explanation

Answer: 216

Answer

The volume of the rectangular prism is

216

cubic centimeters.

Let the edge lengths of the right rectangular prism be

l

,

w

, and

h

. The areas of three adjacent faces meeting at a common vertex are given by

lw = 24

,

wh = 36

, and

lh = 54

. Multiplying these three equations together yields

(lw)(wh)(lh) = (lwh)^2 = 24 \times 36 \times 54 = 46,656

. Since the volume

V

of a right rectangular prism is equal to

lwh

, taking the positive square root gives

V = \sqrt{46,656} = 216

cubic centimeters.

Step-by-Step Solution

1

Define variables for the three dimensions of the rectangular prism.

Let

l

,

w

, and

h

represent the length, width, and height of the prism.

The area of each rectangular face is the product of two adjacent dimensions.

2

Express the given face areas in terms of the dimensions.

lw = 24

,

wh = 36

, and

lh = 54

.

Adjacent faces share pairwise dimensions.

3

Multiply all three face area equations together.

(lw)(wh)(lh) = l^2 w^2 h^2 = (lwh)^2 = 24 \times 36 \times 54 = 46,656

.

Multiplying the three face areas yields the square of the prism's volume (

V = lwh

).

4

Calculate the square root of the product to determine the volume.

V = lwh = \sqrt{46,656} = 216

.

Taking the square root gives the volume directly without needing to solve for individual edge lengths.

Key Concept

For a right rectangular prism with adjacent face areas

A_1

,

A_2

, and

A_3

, the volume is given by

V = \sqrt{A_1 \times A_2 \times A_3}

.

Estimated Time:1m 15s

Question 107Question →

In right triangle $ABC$ , the measure of angle $C$ is $90^\circ$ . Point $D$ lies on segment $AC$ such that $AD = 11$ and $BD = 13$ . If $DC = 5$ , what is the perimeter of triangle $ABD$ ?

Show answer & explanation

Answer: 44

Answer

The perimeter of triangle

ABD

is

44

.

To find the perimeter of triangle

ABD

, we need the lengths of its three sides:

AD

,

BD

, and

AB

. We are given

AD = 11

and

BD = 13

. To find

AB

, we first analyze right triangle

BCD

with right angle at

C

, hypotenuse

BD = 13

, and leg

DC = 5

. Using the Pythagorean theorem, leg

BC = \sqrt{13^2 - 5^2} = 12

. Next, the full leg

AC = AD + DC = 11 + 5 = 16

. In right triangle

ABC

, legs are

AC = 16

and

BC = 12

, making hypotenuse

AB = \sqrt{16^2 + 12^2} = 20

. Finally, the perimeter of triangle

ABD

is

11 + 13 + 20 = 44

.

Step-by-Step Solution

1

Find the length of side

BC

using right triangle

BCD

BC = 12

In right triangle

BCD

, angle

C = 90^\circ

, hypotenuse

BD = 13

, and leg

DC = 5

. By the Pythagorean theorem,

BC = \sqrt{13^2 - 5^2} = \sqrt{144} = 12

.

2

Determine the total length of side

AC

AC = 16

Since point

D

lies on segment

AC

,

AC = AD + DC = 11 + 5 = 16

.

3

Calculate the hypotenuse

AB

of the main right triangle

ABC

AB = 20

In right triangle

ABC

, legs are

AC = 16

and

BC = 12

. By the Pythagorean theorem,

AB = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

.

4

Calculate the perimeter of triangle

ABD

Perimeter

= 44

The sides of triangle

ABD

are

AD = 11

,

BD = 13

, and

AB = 20

. Adding these gives

11 + 13 + 20 = 44

.

Key Concept

Pythagorean Theorem and Multi-Step Triangle Properties

Question 108Question →

An isosceles triangle has two sides of length $10$ units each and a third side of integer length $x$ units. If the area of the triangle is strictly greater than $24$ square units and less than or equal to $48$ square units, which of the following could be the value of $x$ ? Select all such values.

Select all that apply

4

6

14

16

20

Show answer & explanation

Answer: 6; 16

Answer

The values 6 and 16 are the valid side lengths.

The values 6 and 16 produce valid areas of approximately 28.62 and exactly 48 square units respectively, both of which satisfy the given condition that the area must be strictly greater than 24 and less than or equal to 48.

Step-by-Step Solution

1

Express the height and area of the isosceles triangle in terms of

x

.

Height

h = \sqrt{10^2 - (x/2)^2} = \sqrt{100 - \frac{x^2}{4}}

, so Area

A = \frac{1}{2} x \sqrt{100 - \frac{x^2}{4}} = \frac{1}{4} x \sqrt{400 - x^2}

.

The altitude to the base of an isosceles triangle bisects the base into two equal segments of length

\frac{x}{2}

.

2

Set up the inequality for the area constraints

24 < A \le 48

.

24 < \frac{1}{4} x \sqrt{400 - x^2} \le 48 \implies 96 < x \sqrt{400 - x^2} \le 192

.

Multiplying all parts by

4

isolates the radical expression.

3

Square all terms to analyze the function

f(x^2) = x^2(400 - x^2)

.

9216 < x^2(400 - x^2) \le 36864

.

Squaring positive quantities preserves the inequality direction.

4

Evaluate the area function for each given option choice.

For

x=4

:

A \approx 19.6

(too small). For

x=6

:

A \approx 28.6

(valid). For

x=14

:

A \approx 50.0

(too large). For

x=16

:

A = 48

(valid). For

x=20

: degenerate triangle with

A = 0

(invalid).

Direct evaluation identifies which integer choices satisfy

24 < A \le 48

.

Key Concept

Properties of isosceles triangles, Pythagorean theorem for altitude, area bounds, and triangle inequality.

Question 109Question →

In the right rectangular solid $ABCDEFGH$ , the base $ABCD$ is a rectangle with edge lengths $AB = 6$ and $BC = 6\sqrt{3}$ . The vertical edge $CG = 12$ . Point $M$ is the midpoint of edge $CG$ . What is the perimeter of triangle $BDM$ ?

24 + 6\sqrt{2}

12 + 12\sqrt{2}

24 + 6\sqrt{3}

18 + 6\sqrt{2} + 6\sqrt{3}

12 + 6\sqrt{2} + 6\sqrt{5}

Show answer & explanation

Answer: $24 + 6\sqrt{2}$

Answer

The perimeter of triangle

BDM

is

24 + 6\sqrt{2}

.

The correct answer is obtained by recognizing three right triangles within the 3D figure:

\triangle BCD

has legs

6

and

6\sqrt{3}

giving

BD = 12

;

\triangle BCM

has legs

6\sqrt{3}

and

6

giving

BM = 12

; and

\triangle DCM

has legs

6

and

6

giving

DM = 6\sqrt{2}

. Summing the three sides yields a perimeter of

24 + 6\sqrt{2}

.

Step-by-Step Solution

1

Calculate the length of base diagonal

BD

using the right triangle

\triangle BCD

.

BD = 12

In right triangle

\triangle BCD

, legs are

CD = AB = 6

and

BC = 6\sqrt{3}

. Using the

30^\circ\text{--}60^\circ\text{--}90^\circ

ratio

(1 : \sqrt{3} : 2)

, hypotenuse

BD = 2 \times 6 = 12

(or via Pythagorean theorem:

BD = \sqrt{6^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12

).

2

Calculate the length of segment

BM

using the right triangle

\triangle BCM

.

BM = 12

Since

M

is the midpoint of vertical edge

CG = 12

,

CM = 6

. Vertical edge

CG

is perpendicular to base

ABCD

, so

\triangle BCM

is a right triangle at

C

. The legs are

CM = 6

and

BC = 6\sqrt{3}

. Applying the

30^\circ\text{--}60^\circ\text{--}90^\circ

ratio

(1 : \sqrt{3} : 2)

, hypotenuse

BM = 2 \times 6 = 12

.

3

Calculate the length of segment

DM

using the right triangle

\triangle DCM

.

DM = 6\sqrt{2}

In right triangle

\triangle DCM

, legs are

CD = 6

and

CM = 6

. Since the legs are equal,

\triangle DCM

is a

45^\circ\text{--}45^\circ\text{--}90^\circ

isosceles right triangle with side ratio

1 : 1 : \sqrt{2}

. Thus, hypotenuse

DM = 6\sqrt{2}

.

4

Sum the three side lengths to find the perimeter of

\triangle BDM

.

Perimeter =

12 + 12 + 6\sqrt{2} = 24 + 6\sqrt{2}

The perimeter of

\triangle BDM

is

BD + BM + DM

.

Key Concept

Applying special right triangle ratios (

30^\circ\text{--}60^\circ\text{--}90^\circ

and

45^\circ\text{--}45^\circ\text{--}90^\circ

) to 3D rectangular solids.

Question 110Question →

A solid cube has a total surface area of $S$ and a volume of $V$ . A solid right circular cylinder has a height equal to its base diameter. If the total surface area of the cylinder is also equal to $S$ , what is the volume of the cylinder in terms of $V$ ?

\frac{2}{\sqrt{\pi}} V

\frac{\sqrt{\pi}}{2} V

\frac{1}{\sqrt{\pi}} V

\sqrt{\frac{27}{2\pi}} V

\frac{2}{\pi} V

Show answer & explanation

Answer: $\frac{2}{\sqrt{\pi}} V$

Answer

\frac{2}{\sqrt{\pi}} V

The total surface area of a cube with edge

a

is

6a^2

and its volume is

V = a^3

. For a cylinder with base radius

r

and height

h = 2r

, the total surface area is

2\pi r^2 + 2\pi r(2r) = 6\pi r^2

. Setting

6\pi r^2 = 6a^2

yields

r = \frac{a}{\sqrt{\pi}}

and

h = \frac{2a}{\sqrt{\pi}}

. Substituting these into the volume formula

V_{\text{cyl}} = \pi r^2 h

produces

\pi \left(\frac{a^2}{\pi}\right) \left(\frac{2a}{\sqrt{\pi}}\right) = \frac{2a^3}{\sqrt{\pi}} = \frac{2}{\sqrt{\pi}} V

.

Step-by-Step Solution

1

Express the surface area and volume of the cube in terms of its side length

a

.

Surface area

S = 6a^2

and volume

V = a^3

.

A cube with edge length

a

has

6

identical square faces of area

a^2

and volume

a^3

.

2

Set up the total surface area formula for the cylinder with radius

r

and height

h = 2r

, and equate it to

S

.

Total surface area

S_{\text{cyl}} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r (2r) = 6\pi r^2 = 6a^2

.

The cylinder's height is equal to its base diameter (

2r

). Equating surface areas gives

6\pi r^2 = 6a^2

.

3

Solve for radius

r

in terms of edge length

a

.

r^2 = \frac{a^2}{\pi} \implies r = \frac{a}{\sqrt{\pi}}

.

Dividing both sides by

6\pi

and taking the square root isolates

r

.

4

Calculate the volume of the cylinder in terms of

V

.

V_{\text{cyl}} = \pi r^2 h = \pi \left(\frac{a^2}{\pi}\right) \left(\frac{2a}{\sqrt{\pi}}\right) = \frac{2a^3}{\sqrt{\pi}} = \frac{2}{\sqrt{\pi}} V

.

Substituting

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

and

h = \frac{2a}{\sqrt{\pi}}

into the cylinder volume formula

\pi r^2 h

yields the answer in terms of

V = a^3

.

Key Concept

Volume and surface area relationship between geometric solids

Estimated Time:2m 0s

Question 111Question →

In triangle $ABC$ , the length of side $AB$ is $16$ units and the length of side $AC$ is $13$ units. If the area of triangle $ABC$ is $96$ square units and the altitude from vertex $C$ to side $AB$ intersects the line segment $AB$ at point $D$ , what is the length of segment $AD$ ?

5

11

12

7

1

Show answer & explanation

Answer: $5$

Answer

The length of segment

AD

is

5

units.

The correct answer is

5

. First, find altitude

CD

using the triangle area equation:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 96 = \frac{1}{2} \times 16 \times CD \implies CD = 12

. Next, because

CD \perp AB

, triangle

ADC

is a right triangle with hypotenuse

AC = 13

and leg

CD = 12

. Applying the Pythagorean theorem yields

AD = \sqrt{13^2 - 12^2} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Calculate the length of altitude

CD

using the triangle area formula.

CD = 12

units.

The area of triangle

ABC

is given by

\text{Area} = \frac{1}{2} \times AB \times CD

. Substituting the given values:

96 = \frac{1}{2} \times 16 \times CD \implies 96 = 8 \times CD \implies CD = 12

.

2

Apply the Pythagorean theorem in right triangle

ADC

to find

AD

.

AD = 5

units.

Altitude

CD

is perpendicular to

AB

, forming right triangle

ADC

with hypotenuse

AC = 13

and leg

CD = 12

. By the Pythagorean theorem,

AD^2 + CD^2 = AC^2 \implies AD^2 + 12^2 = 13^2 \implies AD^2 + 144 = 169 \implies AD^2 = 25 \implies AD = 5

.

Key Concept

Triangles: Properties, Perimeter, and Area

Estimated Time:1m 30s

Question 112Question →

The sum of the measures of all interior angles of a convex polygon, excluding one interior angle $\theta$ , is equal to $2190^\circ$ . If the degree measure of $\theta$ is an integer, what is the perimeter of a regular polygon with $n$ sides, where $n$ is the number of sides of the original polygon and each side length is $\frac{\theta}{10}$ units?

195

210

225

240

255

Show answer & explanation

Answer: 225

Answer

The perimeter of the regular polygon is 225.

The sum of the interior angles of a convex polygon with

n

sides is

(n-2) \times 180^\circ

. Adding the excluded interior angle

\theta

to

2190^\circ

yields the total sum

S

. Because

0^\circ < \theta < 180^\circ

,

S

must fall strictly between

2190^\circ

and

2370^\circ

. The only multiple of

180^\circ

in this interval is

2340^\circ

. Setting

(n-2)\times 180^\circ = 2340^\circ

yields

n = 15

. Solving for

\theta

gives

\theta = 2340^\circ - 2190^\circ = 150^\circ

. The side length is

\frac{150}{10} = 15

, making the perimeter

15 \times 15 = 225

.

Step-by-Step Solution

1

Set up the inequality for the sum of interior angles of a convex polygon.

The total sum of interior angles for a convex

n

-gon is

S = (n-2) \times 180^\circ

. Given

S - \theta = 2190^\circ

, we have

S = 2190^\circ + \theta

.

The sum of interior angles of any convex polygon with

n

sides is

(n-2) \times 180^\circ

.

2

Determine the value of

S

using the bounds for an interior angle of a convex polygon.

Since

0^\circ < \theta < 180^\circ

, it follows that

2190^\circ < S < 2190^\circ + 180^\circ = 2370^\circ

. The only multiple of

180^\circ

in this range is

2340^\circ

.

S

must be an integer multiple of

180^\circ

and

\theta

must be strictly between

0^\circ

and

180^\circ

.

3

Calculate the number of sides

n

and the missing angle

\theta

.

(n-2) \times 180^\circ = 2340^\circ \implies n - 2 = 13 \implies n = 15

. Then

\theta = 2340^\circ - 2190^\circ = 150^\circ

.

Solving the linear equations gives exact values for the number of sides and the excluded angle.

4

Compute the perimeter of the regular regular

n

-gon.

Side length

= \frac{\theta}{10} = \frac{150}{10} = 15

. Perimeter

= n \times \text{side length} = 15 \times 15 = 225

.

The perimeter of a regular polygon is the product of its number of sides and its individual side length.

Key Concept

Sum of Interior Angles of Convex Polygons

Estimated Time:2m 0s

Question 113Question →

A triangle has side lengths of $x$ , $x + 4$ , and $14$ units, where $x$ is an integer. Which of the following could be the perimeter of the triangle? Select all that apply.

Select all that apply

26

28

30

34

40

Show answer & explanation

Answer: $30$ ; $34$ ; $40$

Answer

The possible perimeters of the triangle are 30, 34, and 40.

The Triangle Inequality Theorem dictates that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the remaining side. For side lengths

x

,

x + 4

, and

14

, setting

x + (x + 4) > 14

yields

2x > 10

, so

x > 5

. Since

x

is specified as an integer,

x

can be any integer greater than or equal to 6. Substituting valid values of

x

into the perimeter formula

P = 2x + 18

gives possible perimeters of 30 (when

x = 6

), 34 (when

x = 8

), and 40 (when

x = 11

). Thus, the values 30, 34, and 40 are all valid perimeters.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem

x > 5

According to the Triangle Inequality Theorem, the sum of any two side lengths must be strictly greater than the third side length. Therefore,

x + (x + 4) > 14

, which simplifies to

2x + 4 > 14 \implies 2x > 10 \implies x > 5

. The other two inequality conditions (

(x + 4) + 14 > x

and

x + 14 > x + 4

) are satisfied for all positive values of

x

.

2

Express the perimeter in terms of

x

P = 2x + 18

The perimeter

P

is the sum of the three side lengths:

P = x + (x + 4) + 14 = 2x + 18

.

3

Evaluate the given choices against the valid bounds of

x

Valid integer values of

x \ge 6

correspond to perimeters

P \ge 30

that are even numbers.

Since

x

is an integer and

x > 5

, the minimum integer value for

x

is 6, which yields a minimum perimeter of

P = 2(6) + 18 = 30

. Testing each option:
- For 26:

2x + 18 = 26 \implies x = 4

(Invalid,

x \le 5

)
- For 28:

2x + 18 = 28 \implies x = 5

(Invalid,

x \le 5

)
- For 30:

2x + 18 = 30 \implies x = 6

(Valid)
- For 34:

2x + 18 = 34 \implies x = 8

(Valid)
- For 40:

2x + 18 = 40 \implies x = 11

(Valid)

Key Concept

Triangle Inequality Theorem and Perimeter Calculation

Question 114Question →

In the $xy$ -plane, line $L_1$ is defined by the equation $2x - 3y = 6$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the point $P(4, -1)$ . Point $Q(a, b)$ lies on line $L_2$ such that the distance between point $P$ and point $Q$ is $\sqrt{13}$ . Which of the following statements regarding point $Q$ or line $L_2$ could be true? Select all such statements.

Select all that apply

Point

Q

lies in Quadrant I.

The sum of the coordinates of point

Q

,

a + b

, is equal to

4

.

The distance from point

Q

to the origin is

2\sqrt{13}

.

The

y

-intercept of line

L_2

is

(0, -6)

.

Point

Q

could have coordinates

(4, 2)

.

Show answer & explanation

Answer: Point $Q$ lies in Quadrant I.; The sum of the coordinates of point $Q$ , $a + b$ , is equal to $4$ .; The distance from point $Q$ to the origin is $2\sqrt{13}$ .

Answer

The correct statements are that point Q can lie in Quadrant I, the sum of the coordinates of point Q can equal 4, and the distance from point Q to the origin can be 2√13.

Solving the system formed by line

L_2

(

3x + 2y = 10

) and the distance constraint

(a-4)^2 + (b+1)^2 = 13

yields two possible points:

(2, 2)

and

(6, -4)

. The point

(2, 2)

lies in Quadrant I and has a coordinate sum of

2 + 2 = 4

. The point

(6, -4)

has a distance to the origin of

\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}

.

Step-by-Step Solution

1

Determine the slope and equation of line

L_2

.

Line

L_1

has slope

m_1 = \frac{2}{3}

. Therefore, perpendicular line

L_2

has slope

m_2 = -\frac{3}{2}

. Using point

P(4, -1)

, the equation of

L_2

is

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

, which simplifies to

y = -\frac{3}{2}x + 5

or

3x + 2y = 10

.

Perpendicular lines in the coordinate plane have negative reciprocal slopes.

2

Express the distance constraint between

P(4, -1)

and

Q(a, b)

algebraically.

Since

Q(a, b)

lies on

L_2

,

b = -\frac{3}{2}a + 5

. The distance squared is

(a - 4)^2 + (b + 1)^2 = 13

. Substituting

b + 1 = -\frac{3}{2}(a - 4)

yields

(a - 4)^2 + \left(-\frac{3}{2}(a - 4)\right)^2 = 13

, which simplifies to

\frac{13}{4}(a - 4)^2 = 13

, so

(a - 4)^2 = 4

.

Applying the distance formula and substituting the line equation reduces the problem to a quadratic equation in one variable.

3

Solve for the possible coordinates of point

Q

.

Taking square roots gives

a - 4 = 2

or

a - 4 = -2

, resulting in

a = 6

or

a = 2

. Correspondingly,

b = -4

or

b = 2

. Thus,

Q

can be

(6, -4)

or

(2, 2)

.

Quadratic equations of the form

(x-h)^2 = k

have two real solutions.

4

Evaluate each statement against the possible coordinates

Q(6, -4)

and

Q(2, 2)

.

For

Q(2, 2)

: it lies in Quadrant I (valid), its coordinate sum is

2 + 2 = 4

(valid). For

Q(6, -4)

: its distance to the origin is

\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}

(valid). The

y

-intercept of

L_2

is

(0, 5)

, and

(4, 2)

is not on

L_2

.

Direct verification confirms which properties hold for the two solved points.

Key Concept

Perpendicular Slopes and Distance Formula in Coordinate Geometry

Question 115Question →

A right circular cylinder has a base radius of $r$ and a height of $h$ . If the base radius is doubled and the height is halved to create a new right circular cylinder, which of the following statements must be true? Select all that apply.

Select all that apply

The volume of the new cylinder is twice the volume of the original cylinder.

The lateral surface area of the new cylinder is equal to the lateral surface area of the original cylinder.

The total surface area of the new cylinder is twice the total surface area of the original cylinder.

The volume of the new cylinder is four times the volume of the original cylinder.

The area of the circular base of the new cylinder is four times the area of the circular base of the original cylinder.

Show answer & explanation

Answer: The volume of the new cylinder is twice the volume of the original cylinder.; The lateral surface area of the new cylinder is equal to the lateral surface area of the original cylinder.; The area of the circular base of the new cylinder is four times the area of the circular base of the original cylinder.

Answer

The correct statements are: the volume of the new cylinder is twice the volume of the original cylinder, the lateral surface area of the new cylinder is equal to the lateral surface area of the original cylinder, and the area of the circular base of the new cylinder is four times the area of the circular base of the original cylinder.

The volume formula

V = \pi r^2 h

scales by

2^2 \times \frac{1}{2} = 2

, so volume doubles. The lateral surface area formula

L = 2\pi r h

scales by

2 \times \frac{1}{2} = 1

, remaining unchanged. The base area formula

A = \pi r^2

scales by

2^2 = 4

, quadrupling the area.

Step-by-Step Solution

1

Write down the original formulas for a cylinder with radius

r

and height

h

.

Base Area

A_1 = \pi r^2

, Lateral Surface Area

L_1 = 2\pi r h

, Total Surface Area

T_1 = 2\pi r h + 2\pi r^2

, Volume

V_1 = \pi r^2 h

.

Establish baseline geometric properties.

2

Substitute the new dimensions

r' = 2r

and

h' = \frac{h}{2}

into each formula.

New Base Area

A_2 = \pi (2r)^2 = 4\pi r^2 = 4A_1

.
New Lateral Surface Area

L_2 = 2\pi (2r) \left(\frac{h}{2}\right) = 2\pi r h = L_1

.
New Volume

V_2 = \pi (2r)^2 \left(\frac{h}{2}\right) = \pi (4r^2) \left(\frac{h}{2}\right) = 2\pi r^2 h = 2V_1

.

Calculate the updated values for each geometric property.

3

Evaluate each statement against the calculated values.

Volume doubles (

V_2 = 2V_1

), lateral surface area remains equal (

L_2 = L_1

), and base area quadruples (

A_2 = 4A_1

).

Determine which options state correct mathematical relationships.

Key Concept

Scaling effects on 3D geometry formulas for right circular cylinders

Question 116Question →

In the $xy$ -plane, line $L_1$ has a positive slope $m$ and a $y$ -intercept of $(0, 4)$ . The perpendicular distance from the origin $(0, 0)$ to line $L_1$ is $2\sqrt{2}$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the points $(2, 1)$ and $(6, k)$ . What is the value of $k$ ?

-7

-5

-3

3

5

Show answer & explanation

Answer: $-3$

Answer

-3

The equation of line

L_1

is

mx - y + 4 = 0

. Using the distance formula from the origin to a line,

\frac{4}{\sqrt{m^2 + 1}} = 2\sqrt{2}

, which simplifies to

m^2 + 1 = 2

, yielding

m = 1

. The slope of line

L_2

is the negative reciprocal of

1

, which is

-1

. Using the slope equation

\frac{k - 1}{6 - 2} = -1

, we find

k - 1 = -4

, giving

k = -3

.

Step-by-Step Solution

1

Express line

L_1

in standard form and set up the distance formula from the origin.

Line

L_1

has slope

m

and

y

-intercept

(0,4)

, so its equation is

y = mx + 4

, or

mx - y + 4 = 0

. The perpendicular distance

d

from

(0,0)

to

Ax + By + C = 0

is

d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{4}{\sqrt{m^2 + 1}}

.

The shortest distance from a point to a line is given by the perpendicular distance formula.

2

Solve for the positive slope

m

of line

L_1

.

Setting

\frac{4}{\sqrt{m^2 + 1}} = 2\sqrt{2}

yields

\sqrt{m^2 + 1} = \sqrt{2}

. Squaring both sides gives

m^2 + 1 = 2

, so

m^2 = 1

. Since

m > 0

,

m = 1

.

The problem specifies that line

L_1

has a positive slope.

3

Determine the slope of line

L_2

.

Since line

L_2

is perpendicular to line

L_1

, its slope

m_2

is the negative reciprocal of

m

:

m_2 = -\frac{1}{1} = -1

.

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

4

Use the slope formula with points

(2, 1)

and

(6, k)

to solve for

k

.

m_2 = \frac{k - 1}{6 - 2} = \frac{k - 1}{4} = -1 \implies k - 1 = -4 \implies k = -3

.

Equating the slope calculated from two points to the known perpendicular slope solves for the missing coordinate.

Key Concept

Perpendicular Distance and Perpendicular Lines in Coordinate Geometry

Question 117Question →

In the $xy$ -plane, the parabola defined by $y = 2x^2 - 8x + 5$ is reflected across the line $y = x$ . The resulting curve is then translated $3$ units to the right and $4$ units downward to produce curve $C$ . If the point $(a, -1)$ lies on curve $C$ , what is the value of $a$ ?

Show answer & explanation

Answer: 2

Answer

The value of

a

is

2

.

Completing the square yields

y = 2(x - 2)^2 - 3

. Reflecting across

y = x

interchanges

x

and

y

, producing

x = 2(y - 2)^2 - 3

. Shifting this curve

3

units to the right replaces

x

with

x - 3

, and shifting

4

units down replaces

y

with

y + 4

. The equation simplifies to

x = 2(y + 2)^2

. Setting

y = -1

gives

a = 2(-1 + 2)^2 = 2

.

Step-by-Step Solution

1

Rewrite the original parabola equation in vertex form.

y = 2(x - 2)^2 - 3

Expressing the parabola in vertex form makes tracking coordinate transformations straightforward.

2

Apply the reflection across the line

y = x

.

x = 2(y - 2)^2 - 3

Reflecting across

y = x

interchanges the

x

and

y

coordinates of every point on the graph.

3

Apply the translations to find the equation of curve

C

.

x = 2(y + 2)^2

Translating right by

3

replaces

x

with

x - 3

, and translating down by

4

replaces

y

with

y + 4

.

4

Substitute

y = -1

into the equation for curve

C

to calculate

a

.

a = 2

Since the point

(a, -1)

lies on curve

C

, substituting

x = a

and

y = -1

yields

a = 2(1)^2 = 2

.

Key Concept

Reflection across

y = x

and rigid horizontal/vertical translations of conic sections

Question 118Question →

In the $xy$ -plane, point $M(3, 3)$ is the midpoint of line segment $AB$ , where point $A$ has coordinates $(1, 2)$ . Line $L$ is perpendicular to line segment $AB$ and passes through point $B$ . What is the $x$ -intercept of line $L$ ?

Show answer & explanation

Answer: 7

Answer

7

Using the midpoint formula with

A(1, 2)

and

M(3, 3)

gives endpoint

B(5, 4)

. The slope of segment

AB

is

\frac{4-2}{5-1} = \frac{1}{2}

, so line

L

, being perpendicular to

AB

, has slope

-2

. The equation of line

L

passing through

(5, 4)

is

y - 4 = -2(x - 5)

, which simplifies to

y = -2x + 14

. Setting

y = 0

gives

0 = -2x + 14

, so the

x

-intercept is

7

.

Step-by-Step Solution

1

Calculate the coordinates of endpoint B using the midpoint formula.

Point B has coordinates

(5, 4)

.

Midpoint

M(x_m, y_m) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)

. Solving

\frac{1 + x_B}{2} = 3

yields

x_B = 5

, and solving

\frac{2 + y_B}{2} = 3

yields

y_B = 4

.

2

Determine the slope of segment AB.

The slope

m_{AB} = \frac{1}{2}

.

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

between

(1, 2)

and

(5, 4)

gives

\frac{4 - 2}{5 - 1} = \frac{2}{4} = \frac{1}{2}

.

3

Determine the slope of line L.

The slope of line L is

-2

.

Perpendicular lines have negative reciprocal slopes. The negative reciprocal of

\frac{1}{2}

is

-2

.

4

Find the equation of line L and solve for its x-intercept.

The x-intercept is 7.

Using point-slope form with

B(5, 4)

and

m = -2

:

y - 4 = -2(x - 5) \implies y = -2x + 14

. Setting

y = 0

gives

0 = -2x + 14 \implies x = 7

.

Key Concept

Midpoint Formula, Perpendicular Slopes, and Line Intercepts

Question 119Question →

A line with a slope of $-2$ passes through the point $(1, 7)$ and contains the point $P(a, 3)$ . What is the distance between point $P$ and the point $Q(11, -3)$ ?

2\sqrt{10}

6\sqrt{5}

10

100

14

Show answer & explanation

Answer: $10$

Answer

The distance between point

P

and point

Q

is

10

.

Using the slope definition

m = \frac{\Delta y}{\Delta x}

, we set up

\frac{3 - 7}{a - 1} = -2

, which solves to

a = 3

, giving point

P(3, 3)

. Applying the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

between

P(3, 3)

and

Q(11, -3)

yields

\sqrt{(11 - 3)^2 + (-3 - 3)^2} = \sqrt{64 + 36} = 10

.

Step-by-Step Solution

1

Use the slope formula to find the missing coordinate

a

of point

P(a, 3)

.

The slope

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 7}{a - 1} = -2

, which simplifies to

\frac{-4}{a - 1} = -2

, yielding

a - 1 = 2

, so

a = 3

.

The slope between any two points on a straight line must equal the given slope of

-2

.

2

Identify the coordinates of point

P

.

Point

P

has coordinates

(3, 3)

.

Substituting

a = 3

into

P(a, 3)

gives the exact position of

P

.

3

Apply the distance formula between

P(3, 3)

and

Q(11, -3)

.

d = \sqrt{(11 - 3)^2 + (-3 - 3)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10

.

The distance formula calculates the Euclidean distance between two points in the coordinate plane.

Key Concept

Line slope equation and coordinate distance formula

Alternative Method

Find the line equation directly in slope-intercept form:

y - 7 = -2(x - 1) \implies y = -2x + 9

. Substitute

y = 3

to get

3 = -2a + 9 \implies a = 3

. Then compute the distance between

(3, 3)

and

(11, -3)

using the standard distance formula.

Estimated Time:1m 30s

Question 120Question →

In triangle $PQR$ , point $S$ lies on segment $QR$ such that segment $PS$ is perpendicular to $QR$ . The length of altitude $PS$ is $8$ units. If the area of triangle $PQR$ is $56$ square units and the ratio of the area of triangle $PQS$ to the area of triangle $PSR$ is $3 : 4$ , what is the length of side $PQ$ ?

10

4\sqrt{13}

\sqrt{174.25}

4\sqrt{5}

14

Show answer & explanation

Answer: $10$

Answer

10

The area of triangle

PQR

is given as 56 square units and altitude

PS = 8

. Using

\text{Area} = \frac{1}{2} \times QR \times 8

, we find

QR = 14

. Because triangles

PQS

and

PSR

share altitude

PS

, the ratio of their areas equals the ratio of their bases

QS : SR = 3 : 4

. Dividing

QR = 14

into 7 equal parts yields

QS = 6

. In right-angled triangle

PQS

, legs are

PS = 8

and

QS = 6

, giving hypotenuse

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

.

Step-by-Step Solution

1

Calculate the total length of base

QR

using the area of triangle

PQR

.

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 56 = \frac{1}{2} \times QR \times 8 \implies 56 = 4 \times QR \implies QR = 14

.

The area formula for any triangle relates base, height, and total area.

2

Determine the length of segment

QS

using the given area ratio.

Since triangles

PQS

and

PSR

share the same altitude

PS

, their areas are proportional to their base lengths

QS

and

SR

. Therefore,

QS : SR = 3 : 4

. The total parts are

3 + 4 = 7

. Thus,

QS = 14 \times \frac{3}{7} = 6

.

Triangles sharing a common altitude have areas proportional to their respective bases.

3

Apply the Pythagorean theorem in right triangle

PQS

to find hypotenuse

PQ

.

PQ^2 = PS^2 + QS^2 = 8^2 + 6^2 = 64 + 36 = 100 \implies PQ = 10

.

Segment

PS

is perpendicular to

QR

, forming right-angled triangle

PQS

with legs of length 8 and 6.

Key Concept

Area of triangles, common altitude area ratio, and the Pythagorean theorem.

Geometry

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Common Mistakes

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Answer

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution