Geometry Practice Questions — GRE General Test — Page 7

Question 121Question →

In trapezoid $ABCD$ , side $AB$ is perpendicular to parallel bases $AD$ and $BC$ . Diagonal $AC$ is perpendicular to side $CD$ . If the measure of angle $CAD$ is $30^\circ$ and the length of $BC$ is $9$ , what is the length of side $CD$ ?

Show answer & explanation

Answer: 6

Answer

6

Because base

BC

and base

AD

are parallel, transversal

AC

creates equal alternate interior angles, making

\angle ACB = \angle CAD = 30^\circ

. Triangle

ABC

is a right triangle with

\angle B = 90^\circ

, so

BC

is adjacent to

30^\circ

. The ratio of the side adjacent to

30^\circ

to the hypotenuse

AC

is

\frac{\sqrt{3}}{2}

, which yields

AC = \frac{9}{\sqrt{3}/2} = 6\sqrt{3}

. Triangle

ACD

is also a

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

right triangle with right angle at

C

. Side

CD

is opposite the

30^\circ

angle and side

AC

is opposite the

60^\circ

angle. The ratio of the short leg to the long leg is

\frac{1}{\sqrt{3}}

, giving

CD = \frac{6\sqrt{3}}{\sqrt{3}} = 6

.

Step-by-Step Solution

1

Determine angle measures in right triangle ABC using parallel line properties

Angle ACB = 30 degrees and angle BAC = 60 degrees

Since base BC is parallel to base AD, alternate interior angles formed by transversal AC are equal: angle ACB = angle CAD = 30 degrees.

2

Calculate the length of diagonal AC using special right triangle ratios

AC = 6*sqrt(3)

In 30-60-90 right triangle ABC, the ratio of the side adjacent to 30 degrees (BC) to the hypotenuse (AC) is sqrt(3)/2. Therefore, 9 / AC = sqrt(3)/2, which gives AC = 18 / sqrt(3) = 6*sqrt(3).

3

Calculate the length of side CD using right triangle ACD

CD = 6

In 30-60-90 right triangle ACD, angle ACD = 90 degrees and angle CAD = 30 degrees. The ratio of the short leg opposite 30 degrees (CD) to the long leg opposite 60 degrees (AC) is 1/sqrt(3). Therefore, CD = AC / sqrt(3) = (6*sqrt(3)) / sqrt(3) = 6.

Key Concept

Side ratios of 30-60-90 special right triangles and alternate interior angles

Question 122Question →

In convex quadrilateral $ABCD$ , diagonals $AC$ and $BD$ intersect at point $E$ . The area of $\triangle ABE$ is $4$ , the area of $\triangle BCE$ is $8$ , and the area of $\triangle CDE$ is $16$ . Which of the following statements must be true? Select all that apply.

Select all that apply

The area of

\triangle DAE

is

8

.

Quadrilateral

ABCD

is a trapezoid with

AB

parallel to

CD

.

The length of segment

CD

is twice the length of segment

AB

.

The total area of quadrilateral

ABCD

is

32

.

The midpoint of diagonal

AC

lies on segment

DE

.

Show answer & explanation

Answer: The area of $\triangle DAE$ is $8$ .; Quadrilateral $ABCD$ is a trapezoid with $AB$ parallel to $CD$ .; The length of segment $CD$ is twice the length of segment $AB$ .

Answer

The correct statements are that the area of triangle DAE is 8, quadrilateral ABCD is a trapezoid with AB parallel to CD, and the length of segment CD is twice the length of segment AB.

The statements confirming that the area of triangle DAE is 8, that ABCD is a trapezoid with AB parallel to CD, and that CD is twice length AB are all derived using area ratio properties of intersecting diagonals and triangle similarity.

Step-by-Step Solution

1

Determine the area of triangle DAE using diagonal segment ratios.

Area of triangle DAE = 8.

Triangles sharing a base line have areas proportional to the segments into which the intersecting line divides that base line: AE/EC = Area(ABE)/Area(BCE) = 4/8 = 1/2, so Area(DAE) = (1/2) * 16 = 8.

2

Check parallelism of opposite sides AB and CD.

AB is parallel to CD, making ABCD a trapezoid.

Area(ABC) = 4 + 8 = 12 and Area(ABD) = 4 + 8 = 12. Triangles with equal areas on the common base AB must have equal heights, implying line CD is parallel to line AB.

3

Calculate side ratio CD / AB using similar triangles.

CD = 2 * AB.

Since AB || CD, triangle ABE is similar to triangle CDE. The ratio of their areas is 16/4 = 4, so the side length ratio CD/AB = sqrt(4) = 2.

4

Compute total area of quadrilateral ABCD and evaluate diagonal midpoint position.

Total area is 36 (not 32), and the midpoint of AC is at (1/2)AC from A, which is distinct from E at (1/3)AC.

Total area = 4 + 8 + 16 + 8 = 36. Since E divides AC in a 1:2 ratio, E is not the midpoint of AC.

Key Concept

Properties of convex quadrilaterals, area decomposition via diagonal ratios, trapezoid parallelism criteria, and triangle similarity.

Question 123Question →

In $\triangle ABC$ , angle $B$ is a right angle, and line segment $BD$ is an altitude drawn to side $AC$ with point $D$ lying on $AC$ . If $AD = 4$ units and $DC = 16$ units, what is the area, in square units, of $\triangle ABC$ ?

Show answer & explanation

Answer: 80

Answer

The area of triangle ABC is 80 square units.

In right triangle

ABC

with right angle at

B

, altitude

BD

drawn to hypotenuse

AC

divides the hypotenuse into segments

AD

and

DC

. By the geometric mean theorem,

BD^2 = AD \times DC = 4 \times 16 = 64

, which gives

BD = 8

units. The length of hypotenuse

AC

is

AD + DC = 4 + 16 = 20

units. The area of triangle

ABC

is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8 = 80

square units.

Step-by-Step Solution

1

Calculate the height (altitude)

BD

of the triangle.

BD = 8

units.

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that

BD^2 = AD \times DC

. Substituting the given values yields

BD^2 = 4 \times 16 = 64

, so

BD = \sqrt{64} = 8

.

2

Calculate the total length of hypotenuse

AC

.

AC = 20

units.

Since point

D

lies on segment

AC

, the total length is the sum of its parts:

AC = AD + DC = 4 + 16 = 20

.

3

Calculate the area of

\triangle ABC

.

Area = 80 square units.

Using the triangle area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, with base

AC = 20

and altitude

BD = 8

, we get

\text{Area} = \frac{1}{2} \times 20 \times 8 = 80

.

Key Concept

Geometric mean theorem for right triangle altitude and area of a triangle

Question 124Question →

In the $xy$ -plane, line $k$ has a slope of $-\frac{3}{4}$ and intersects the positive $x$ -axis at $(a, 0)$ and the positive $y$ -axis at $(0, b)$ . If the distance between the two intercept points $(a, 0)$ and $(0, b)$ is $15$ , what is the value of $a$ ?

9

12

15

16

20

Show answer & explanation

Answer: $12$

Answer

The value of

a

is

12

.

Using the slope formula between

(a,0)

and

(0,b)

, we find

m = -\frac{b}{a} = -\frac{3}{4}

, giving

b = \frac{3}{4}a

. Applying the distance formula yields

\sqrt{a^2 + b^2} = 15

, or

a^2 + b^2 = 225

. Substituting

b = \frac{3}{4}a

produces

a^2 + \frac{9}{16}a^2 = 225

, which simplifies to

\frac{25}{16}a^2 = 225

. Multiplying by 16 and dividing by 25 yields

a^2 = 144

, giving

a = 12

since

a > 0

.

Step-by-Step Solution

1

Express

b

in terms of

a

using the slope formula.

The slope of line

k

passing through

(a, 0)

and

(0, b)

is

m = \frac{b - 0}{0 - a} = -\frac{b}{a} = -\frac{3}{4}

, which simplifies to

b = \frac{3}{4}a

.

The slope of a line through two points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

.

2

Set up the distance equation between

(a, 0)

and

(0, b)

.

\sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} = 15

, so

a^2 + b^2 = 225

.

The distance formula between two points in the coordinate plane is derived from the Pythagorean theorem.

3

Substitute

b = \frac{3}{4}a

into the distance equation and solve for

a

.

a^2 + \left(\frac{3}{4}a\right)^2 = 225 \implies a^2 + \frac{9}{16}a^2 = 225 \implies \frac{25}{16}a^2 = 225 \implies a^2 = 144 \implies a = 12

.

Since

(a, 0)

is on the positive

x

-axis,

a

must be positive.

Key Concept

Slope and Distance in Coordinate Geometry

Estimated Time:1m 30s

Question 125Question →

In $\triangle ABC$ , the length of side $AB$ is $13$ units, the length of side $BC$ is $21$ units, and the area of $\triangle ABC$ is $126$ square units. If $\angle ABC$ is an acute angle, what is the length of side $AC$ ?

15

18

20

22

25

Show answer & explanation

Answer: 20

Answer

20

The area of triangle

ABC

is

\frac{1}{2} \times 21 \times h = 126

, which yields an altitude

AH = 12

perpendicular to side

BC

. In right triangle

ABH

, the base segment

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

. Because angle

ABC

is acute,

H

falls between

B

and

C

, making

HC = 21 - 5 = 16

. Finally, in right triangle

AHC

,

AC = \sqrt{12^2 + 16^2} = \sqrt{400} = 20

. Therefore, 20 is the correct answer.

Step-by-Step Solution

1

Calculate the altitude

h

from vertex

A

to base

BC

.

h = 12

units.

The area formula for a triangle is

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

. Substituting

BC = 21

and

\text{Area} = 126

yields

126 = \frac{1}{2} \times 21 \times h \implies h = 12

.

2

Find the length of segment

BH

where

H

is the foot of the altitude on

BC

.

BH = 5

units.

In right triangle

\triangle ABH

,

AB = 13

and

AH = 12

. By the Pythagorean theorem,

BH = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5

.

3

Determine the length of segment

HC

.

HC = 16

units.

Since

\angle ABC

is an acute angle, point

H

lies on segment

BC

. Therefore,

HC = BC - BH = 21 - 5 = 16

.

4

Calculate the length of side

AC

.

AC = 20

units.

In right triangle

\triangle AHC

,

AH = 12

and

HC = 16

. By the Pythagorean theorem,

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

.

Key Concept

Triangles: Altitude, Area, and Pythagorean Theorem

Estimated Time:1m 30s

Question 126Question →

In the $xy$ -coordinate plane, point $P$ lies in the first quadrant such that the distance from the origin $O(0,0)$ to $P$ is $4\sqrt{3}$ , and line segment $OP$ forms a $30^\circ$ angle with the positive $x$ -axis. A circle centered at point $P$ with radius $10$ intersects the $y$ -axis at two distinct points. Which of the following statements are true? Select all such statements.

Select all that apply

The

x

-coordinate of point

P

is

6

.

The length of the chord formed by the circle on the

y

-axis is

16

.

The

x

-coordinate of point

P

is

2\sqrt{3}

.

The length of the chord formed by the circle on the

y

-axis is

4\sqrt{34}

.

The perpendicular distance from point

P

to the

y

-axis is

2\sqrt{3}

.

Show answer & explanation

Answer: The $x$ -coordinate of point $P$ is $6$ .; The length of the chord formed by the circle on the $y$ -axis is $16$ .

Answer

The correct statements are that the

x

-coordinate of point

P

is

6

and the length of the chord formed by the circle on the

y

-axis is

16

.

The horizontal leg of the

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

triangle gives the

x

-coordinate

x_P = 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6

. The perpendicular distance from center

P

to the

y

-axis is

6

. Using the Pythagorean theorem with circle radius

10

and distance

6

, half the chord length is

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

, yielding a full chord length of

16

.

Step-by-Step Solution

1

Determine the coordinates of point P using special right triangle ratio (30-60-90)

Point

P = (6, 2\sqrt{3})

Segment

OP

forms the hypotenuse of length

4\sqrt{3}

in a right triangle with the

x

-axis. The horizontal component is

x_P = 4\sqrt{3} \cos(30^\circ) = 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6

. The vertical component is

y_P = 4\sqrt{3} \sin(30^\circ) = 4\sqrt{3} \cdot \frac{1}{2} = 2\sqrt{3}

.

2

Find the perpendicular distance from P to the y-axis

Perpendicular distance

= 6

The

y

-axis is defined by the line equation

x = 0

. The perpendicular distance from

P(6, 2\sqrt{3})

to

x = 0

is

|6 - 0| = 6

.

3

Apply the Pythagorean theorem to calculate the chord length on the y-axis

Total chord length

= 16

A perpendicular segment from center

P

to the chord bisects the chord at midpoint

M(0, 2\sqrt{3})

. In right triangle

\triangle PMQ

formed by radius

PQ = 10

, distance

PM = 6

, and half-chord

MQ = d

:

d^2 + PM^2 = PQ^2 \implies d^2 + 6^2 = 10^2 \implies d^2 = 64 \implies d = 8

. The full chord length is

2d = 16

.

Key Concept

Properties of 30-60-90 special right triangles and application of the Pythagorean theorem to circle chord geometry

Question 127Question →

In the $xy$ -plane, line segment $AB$ has endpoints $A(-3, 1)$ and $B(1, 9)$ . Line $N$ is the perpendicular bisector of segment $AB$ . Which of the following statements must be true? Select all that apply.

Select all that apply

Line

N

has a slope of

-\frac{1}{2}

.

Line

N

passes through the point

(3, 3)

.

Line

N

has a

y

-intercept of

(0, 5)

.

The length of line segment

AB

is

10

.

The midpoint of line segment

AB

lies in Quadrant I.

Show answer & explanation

Answer: Line $N$ has a slope of $-\frac{1}{2}$ .; Line $N$ passes through the point $(3, 3)$ .

Answer

The statement that Line

N

has a slope of

-\frac{1}{2}

and the statement that Line

N

passes through the point

(3, 3)

are both true.

The slope of segment

AB

is

2

, making the perpendicular slope

-\frac{1}{2}

. The midpoint of

AB

is

(-1, 5)

. Using point-slope form with

M(-1, 5)

and slope

-\frac{1}{2}

, Line

N

has equation

y = -\frac{1}{2}x + \frac{9}{2}

. Evaluating at

x = 3

gives

y = 3

, so Line

N

passes through

(3, 3)

.

Step-by-Step Solution

1

Calculate the slope of line segment

AB

slope

m_{AB} = \frac{9 - 1}{1 - (-3)} = \frac{8}{4} = 2

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is

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

.

2

Determine the slope of the perpendicular bisector (Line

N

)

slope

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

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

3

Find the midpoint of line segment

AB

Midpoint

M = \left(\frac{-3 + 1}{2}, \frac{1 + 9}{2}\right) = (-1, 5)

A perpendicular bisector must pass through the midpoint of the segment it bisects.

4

Find the equation of Line

N

and test given points

Line

N

:

y - 5 = -\frac{1}{2}(x - (-1)) \implies y = -\frac{1}{2}x + \frac{9}{2}

. Testing

(3, 3)

:

3 = -\frac{1}{2}(3) + \frac{9}{2} = \frac{6}{2} = 3

(True). Testing

y

-intercept (

x=0

):

y = 4.5 \neq 5

.

Point-slope form

y - y_1 = m(x - x_1)

defines the line.

5

Calculate the length of segment

AB

AB = \sqrt{(1 - (-3))^2 + (9 - 1)^2} = \sqrt{4^2 + 8^2} = \sqrt{80} = 4\sqrt{5} \approx 8.94

The distance formula between two points is

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

.

Key Concept

Coordinate Geometry: Perpendicular Bisectors, Slopes, Midpoints, and Distance

Estimated Time:1m 45s

Question 128Question →

In the coordinate plane, triangle $JKL$ has vertices $J(0,0)$ , $K(14,0)$ , and $L(x, 12)$ , where $x > 0$ . If the perimeter of triangle $JKL$ is $42$ units and the length of side $JL$ is less than the length of side $KL$ , what is the value of $x$ ?

4

5

7

9

12

Show answer & explanation

Answer: 5

Answer

5

The correct answer is 5. Using the distance formula, the base length

JK = 14

. Expressing the side lengths as

JL = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + 144}

, setting the perimeter

JK + JL + KL = 42

leads to the quadratic equation

x^2 - 14x + 45 = 0

. This yields

x = 5

or

x = 9

. Evaluating the sides for

x = 5

gives

JL = 13

and

KL = 15

, which satisfies the problem condition

JL < KL

.

Step-by-Step Solution

1

Calculate the length of base

JK

using the distance formula.

The distance between

J(0,0)

and

K(14,0)

is

14 - 0 = 14

units.

Base

JK

lies along the horizontal

x

-axis.

2

Express side lengths

JL

and

KL

in terms of

x

.

JL = \sqrt{(x-0)^2 + (12-0)^2} = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + (12-0)^2} = \sqrt{(14-x)^2 + 144}

.

Apply the distance formula between coordinates

L(x,12)

and vertices

J

and

K

.

3

Set up and solve the perimeter equation.

14 + \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 42 \implies \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 28

. Squaring both sides systematically yields

x^2 - 14x + 45 = 0

, giving roots

x = 5

and

x = 9

.

The total perimeter is given as 42 units.

4

Apply the constraint

JL < KL

to choose the valid root.

For

x = 5

,

JL = \sqrt{25 + 144} = 13

and

KL = \sqrt{81 + 144} = 15

, satisfying

JL < KL

.

For

x = 9

,

JL = 15

and

KL = 13

, which violates

JL < KL

.

Key Concept

Coordinate Geometry and Triangle Side Length Constraints

Estimated Time:1m 30s

Question 129Question →

In the $xy$ -plane, line $k$ passes through the points $(-1, 2)$ and $(3, 10)$ . Line $m$ is perpendicular to line $k$ and passes through the point $(2, 4)$ . Line $m$ intersects the $x$ -axis at point $P$ and the $y$ -axis at point $Q$ . What is the distance between point $P$ and point $Q$ ?

3\sqrt{5}

4\sqrt{5}

5\sqrt{3}

5\sqrt{5}

15

Show answer & explanation

Answer: $5\sqrt{5}$

Answer

5\sqrt{5}

The answer

5\sqrt{5}

is obtained by taking the negative reciprocal of the slope of line

k

(

m = 2

) to find line

m

's slope (

m = -1/2

), establishing line

m

's equation as

y = -1/2 x + 5

, finding the intercepts

P(10, 0)

and

Q(0, 5)

, and applying the distance formula.

Step-by-Step Solution

1

Calculate the slope of line

k

.

The slope of line

k

is

m_k = \frac{10 - 2}{3 - (-1)} = \frac{8}{4} = 2

.

The slope formula is

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

.

2

Find the slope of line

m

, which is perpendicular to line

k

.

The slope of line

m

is

m_m = -\frac{1}{2}

.

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

m_1 \cdot m_2 = -1

).

3

Determine the equation of line

m

using point-slope form with point

(2, 4)

.

y - 4 = -\frac{1}{2}(x - 2) \implies y = -\frac{1}{2}x + 5

.

Substituting slope

m_m = -\frac{1}{2}

and point

(x_1, y_1) = (2, 4)

yields line

m

's slope-intercept equation.

4

Find the coordinates of the

x

-intercept

P

and

y

-intercept

Q

.

P = (10, 0)

and

Q = (0, 5)

.

Setting

y = 0

gives

0 = -\frac{1}{2}x + 5 \implies x = 10

. Setting

x = 0

gives

y = 5

.

5

Calculate the distance between

P(10, 0)

and

Q(0, 5)

.

d = \sqrt{(10 - 0)^2 + (0 - 5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}

.

Apply the distance formula

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

.

Key Concept

Perpendicular Line Slopes and Coordinate Distance

Estimated Time:2m 0s

Question 130Question →

In the $xy$ -plane, line $k$ has a slope of $\frac{3}{4}$ and passes through the point $(2, 1)$ . Line $k$ intersects the line $x = 10$ at point $P$ . What is the distance between point $P$ and the point $(2, 1)$ ?

Show answer & explanation

Answer: 10

Answer

10

Point P lies on the line x = 10, so its x-coordinate is 10. The horizontal change from x = 2 to x = 10 is 8 units. Given that line k has slope 3/4, the corresponding vertical change is (3/4)(8) = 6 units. Therefore, point P has coordinates (10, 7). The distance between (2, 1) and (10, 7) is sqrt((10 - 2)^2 + (7 - 1)^2) = sqrt(64 + 36) = sqrt(100) = 10.

Step-by-Step Solution

1

Determine the coordinates of point P

P is located at (10, 7)

Since P lies on the line x = 10, its x-coordinate is 10. The horizontal distance from (2, 1) to P is 10 - 2 = 8. Using the slope m = 3/4, the vertical change is (3/4) * 8 = 6, so the y-coordinate of P is 1 + 6 = 7.

2

Calculate the distance between (2, 1) and (10, 7)

The distance is 10

Using the distance formula sqrt((10 - 2)^2 + (7 - 1)^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10.

Key Concept

Slope definition and distance formula in coordinate geometry

Estimated Time:1m 30s

Question 131Question →

In the $xy$ -coordinate plane, point $A$ has coordinates $(-9, 0)$ and point $C$ has coordinates $(0, 12)$ . Point $B$ lies on the positive $x$ -axis such that line segment $BD$ is perpendicular to segment $AC$ , with point $D$ lying on segment $AC$ . If the area of right triangle $ABD$ is $54$ , what is the length of segment $OB$ , where $O$ is the origin $(0,0)$ ?

Show answer & explanation

Answer: 6

Answer

The length of segment

OB

is

6

.

The length of segment

OB

is

6

. Using the Pythagorean theorem on

\triangle AOC

, hypotenuse

AC = 15

, establishing a

3:4:5

side ratio for

\triangle AOC

. Because

\triangle ABD

shares acute angle

\angle A

with

\triangle AOC

and has a right angle at

D

,

\triangle ABD

is also a

3:4:5

right triangle with hypotenuse

AB

. Expressing the area

\frac{1}{2} \times \left(\frac{3}{5}AB\right) \times \left(\frac{4}{5}AB\right) = 54

yields

AB = 15

. Since

A

is at

(-9,0)

, point

B

is at

(6,0)

, making

OB = 6

.

Step-by-Step Solution

1

Find the side lengths and hypotenuse of right triangle

AOC

.

Leg

AO = 9

, leg

OC = 12

, and by the Pythagorean theorem, hypotenuse

AC = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

.

The coordinates of

A(-9,0)

and

C(0,12)

form a right triangle at the origin

O(0,0)

.

2

Use angle similarity to determine the side ratio of right triangle

ABD

.

Triangle

ABD

is similar to triangle

AOC

because both contain a right angle and share

\angle DAO

. Thus, the sides of

\triangle ABD

maintain the ratio

AD : BD : AB = 3 : 4 : 5

.

Right triangles with a shared acute angle are similar.

3

Express legs

AD

and

BD

in terms of hypotenuse

AB

and set up the area equation.

AD = \frac{3}{5}AB

and

BD = \frac{4}{5}AB

. The area of

\triangle ABD = \frac{1}{2} \times AD \times BD = \frac{1}{2} \times \frac{3}{5}AB \times \frac{4}{5}AB = \frac{6}{25}AB^2

. Setting

\frac{6}{25}AB^2 = 54

yields

AB^2 = 225

, so

AB = 15

.

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

4

Calculate the length of segment

OB

.

Since point

A

is at

(-9,0)

and

B

lies on the positive

x

-axis,

AB = x_B - (-9) = 15 \implies x_B = 6

. Therefore, the length of

OB

is

6

.

The distance from the origin

(0,0)

to

(6,0)

on the

x

-axis is equal to the

x

-coordinate

6

.

Key Concept

Applying Pythagorean triples (3-4-5 right triangle family) and similar right triangles in coordinate geometry.

Question 132Question →

In the $xy$ -plane, line $L$ passes through the points $(-2, 5)$ and $(4, 1)$ . Line $M$ is the perpendicular bisector of the line segment connecting these two points. What is the $y$ -intercept of line $M$ ?

-\frac{3}{2}

\frac{1}{2}

\frac{3}{2}

\frac{7}{3}

\frac{11}{3}

Show answer & explanation

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

Answer

\frac{3}{2}

The midpoint of the segment with endpoints

(-2, 5)

and

(4, 1)

is

\left(\frac{-2+4}{2}, \frac{5+1}{2}\right) = (1, 3)

. The slope of the segment is

\frac{1-5}{4-(-2)} = -\frac{2}{3}

. Therefore, the perpendicular bisector (line

M

) has a slope equal to the negative reciprocal,

\frac{3}{2}

. Using point-slope form with point

(1, 3)

, line

M

has the equation

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

, which simplifies to

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

. The

y

-intercept is

\frac{3}{2}

.

Step-by-Step Solution

1

Calculate the midpoint of the line segment with endpoints

(-2, 5)

and

(4, 1)

Midpoint

(x_m, y_m) = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3)

The perpendicular bisector passes through the midpoint of the segment.

2

Determine the slope of line

L

m_L = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}

Slope is calculated as change in

y

divided by change in

x

.

3

Determine the slope of line

M

(perpendicular bisector)

m_M = -\frac{1}{m_L} = \frac{3}{2}

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

4

Find the equation of line

M

using point-slope form through

(1, 3)

and evaluate its

y

-intercept

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

. Setting

x = 0

gives

y = \frac{3}{2}

.

The

y

-intercept is the value of

y

when

x = 0

.

Key Concept

Perpendicular Bisectors and Slope-Intercept Form

Question 133Question →

A chord $AB$ of length $24$ is drawn in a circle with center $O$ and radius $13$ . A point $P$ lies on the circle such that the area of triangle $ABP$ is maximized. What is the perimeter of triangle $ABP$ ?

24 + 8\sqrt{13}

12 + 12\sqrt{13}

24 + 12\sqrt{13}

24 + 6\sqrt{13}

24 + 12\sqrt{17}

Show answer & explanation

Answer: $24 + 12\sqrt{13}$

Answer

The perimeter of triangle

ABP

is

24 + 12\sqrt{13}

.

The distance from center O to chord AB forms a 5-12-13 right triangle with half the chord length (12) and the radius (13), giving OM = 5. To maximize triangle area, P must be on the major arc, making the altitude PM = 13 + 5 = 18. Using the Pythagorean theorem in right triangle AMP with legs 12 and 18 gives AP = sqrt(12^2 + 18^2) = 6sqrt(13). The total perimeter is base AB plus twice AP, which equals 24 + 12sqrt(13).

Step-by-Step Solution

1

Find the distance from the center

O

to the chord

AB

.

Let

M

be the midpoint of chord

AB

. Since

AB = 24

,

AM = 12

. Triangle

OMA

is a right triangle with hypotenuse

OA = 13

and leg

AM = 12

. By the Pythagorean theorem:

OM = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5

.

The perpendicular from the center of a circle to a chord bisects the chord and forms a right triangle with the radius.

2

Determine the location of

P

that maximizes the area of

\triangle ABP

and calculate the altitude.

The area of

\triangle ABP

is

\frac{1}{2} \times AB \times h

, where

h

is the perpendicular distance from

P

to segment

AB

. Area is maximized when

h

is maximized. Point

P

must lie on the major arc along the diameter perpendicular to

AB

. Thus, the maximum height is

PM = PO + OM = 13 + 5 = 18

.

Since the base

AB

is fixed, maximizing the area requires maximizing the height perpendicular to

AB

.

3

Calculate the length of side

AP

(and

BP

) using the Pythagorean theorem.

In right triangle

AMP

,

AM = 12

and

PM = 18

. Therefore:

AP = \sqrt{12^2 + 18^2} = \sqrt{144 + 324} = \sqrt{468} = 6\sqrt{13}

. By symmetry,

BP = AP = 6\sqrt{13}

.

The perpendicular bisector of a chord creates two congruent right triangles for any point

P

lying on it.

4

Calculate the total perimeter of

\triangle ABP

.

\text{Perimeter} = AB + AP + BP = 24 + 6\sqrt{13} + 6\sqrt{13} = 24 + 12\sqrt{13}.

The perimeter of a triangle is the sum of all three side lengths.

Key Concept

Pythagorean Theorem and Special Right Triangles in Circle Geometry

Question 134Question →

In the $xy$ -plane, the graph of the function $g(x)$ is obtained by taking the graph of $f(x) = (x + 2)^3 - 1$ , reflecting it across the $y$ -axis, translating it $4$ units to the right, and then translating it $3$ units upward. What is the $y$ -intercept of the graph of $y = g(x)$ ?

-214

-6

12

212

218

Show answer & explanation

Answer: $218$

Answer

The

y

-intercept of the graph of

y = g(x)

is

218

.

Applying the transformations step-by-step to

f(x) = (x + 2)^3 - 1

gives

g(x) = (-x + 6)^3 + 2

. Substituting

x = 0

yields

g(0) = 6^3 + 2 = 218

, making

218

the correct

y

-intercept.

Step-by-Step Solution

1

Reflect the function

f(x) = (x + 2)^3 - 1

across the

y

-axis.

Replacing

x

with

-x

yields

y_1 = f(-x) = (-x + 2)^3 - 1

.

Reflecting a graph across the

y

-axis corresponds to replacing

x

with

-x

in the function rule.

2

Translate the reflected graph

4

units to the right.

Replacing

x

with

x - 4

yields

y_2 = (-(x - 4) + 2)^3 - 1 = (-x + 4 + 2)^3 - 1 = (-x + 6)^3 - 1

.

Translating a graph

h

units to the right replaces

x

with

x - h

.

3

Translate the graph

3

units upward to form

g(x)

.

g(x) = (-x + 6)^3 - 1 + 3 = (-x + 6)^3 + 2

.

Translating a graph

k

units upward adds

k

to the expression.

4

Find the

y

-intercept by evaluating

g(0)

.

g(0) = (-0 + 6)^3 + 2 = 6^3 + 2 = 216 + 2 = 218

.

The

y

-intercept occurs where

x = 0

.

Key Concept

Function Transformations in Coordinate Geometry

Estimated Time:1m 30s

Question 135Question →

The ratio of the measure of each interior angle of a regular $n$ -sided polygon $P$ to the measure of each interior angle of a regular $(n+2)$ -sided polygon is $\frac{24}{25}$ . Which of the following statements about polygon $P$ must be true? Select all such statements.

Select all that apply

Polygon

P

has

10

sides.

Polygon

P

has

35

diagonals.

The measure of each exterior angle of polygon

P

is

40^\circ

.

The sum of the measures of the interior angles of polygon

P

is

1{}800^\circ

.

The measure of each interior angle of polygon

P

is

144^\circ

.

Show answer & explanation

Answer: Polygon $P$ has $10$ sides.; Polygon $P$ has $35$ diagonals.; The measure of each interior angle of polygon $P$ is $144^\circ$ .

Answer

The statements asserting that polygon P has 10 sides, polygon P has 35 diagonals, and each interior angle of polygon P measures 144 degrees are all correct.

To determine which statements are true, we set up the ratio of the interior angle of a regular

n

-sided polygon to that of a regular

(n+2)

-sided polygon:

\frac{(n-2)(n+2)}{n^2} = \frac{24}{25}

. Simplifying gives

1 - \frac{4}{n^2} = \frac{24}{25}

, which leads to

n^2 = 100

, so

n = 10

. Therefore, polygon

P

is a regular decagon (10 sides). Evaluating the properties of a regular 10-gon shows that the number of diagonals is

\frac{10(10-3)}{2} = 35

, each interior angle measures

\frac{8 \cdot 180^\circ}{10} = 144^\circ

, each exterior angle measures

\frac{360^\circ}{10} = 36^\circ

, and the sum of the interior angles is

8 \cdot 180^\circ = 1{}440^\circ

. Consequently, the options stating that the polygon has 10 sides, has 35 diagonals, and has interior angles measuring

144^\circ

are correct.

Step-by-Step Solution

1

Set up the algebraic equation comparing the interior angles of an

n

-gon and an

(n+2)

-gon.

\frac{\frac{(n-2) \cdot 180^\circ}{n}}{\frac{n \cdot 180^\circ}{n+2}} = \frac{24}{25} \implies \frac{(n-2)(n+2)}{n^2} = \frac{24}{25}

The formula for each interior angle of a regular polygon with

k

sides is

\frac{(k-2) \cdot 180^\circ}{k}

.

2

Solve for

n

.

1 - \frac{4}{n^2} = \frac{24}{25} \implies \frac{4}{n^2} = \frac{1}{25} \implies n^2 = 100 \implies n = 10

Expanding

(n-2)(n+2) = n^2 - 4

allows simplifying the algebraic ratio.

3

Evaluate polygon properties for

n = 10

.

Diagonals:

\frac{10(10-3)}{2} = 35

; Interior angle:

\frac{8 \cdot 180^\circ}{10} = 144^\circ

; Exterior angle:

\frac{360^\circ}{10} = 36^\circ

; Interior angle sum:

8 \cdot 180^\circ = 1{}440^\circ

.

Apply standard formulas for diagonal count, exterior angle measure, and interior angle sum for a regular decagon.

Key Concept

Interior and exterior angle formulas of regular polygons and diagonal counting formulas

Question 136Question →

In the $xy$ -coordinate plane, point $P$ has coordinates $(0, 0)$ and point $Q$ has coordinates $(6, 8)$ . Point $R$ is positioned such that triangle $PQR$ is a right triangle with hypotenuse $PQ$ . Which of the following could be the coordinates of point $R$ ? Select all such points.

Select all that apply

(8, 4)

(-2, 4)

(3, -1)

(4, 8)

(6, 6)

Show answer & explanation

Answer: $(8, 4)$ ; $(-2, 4)$ ; $(3, -1)$

Answer

The valid coordinates for point

R

are

(8, 4)

,

(-2, 4)

, and

(3, -1)

.

According to the Pythagorean theorem and Thales's theorem, any point forming a right angle with segment

PQ

as hypotenuse must satisfy

PR^2 + RQ^2 = PQ^2 = 100

, placing it on a circle centered at

(3, 4)

with radius

5

. The points

(8, 4)

,

(-2, 4)

, and

(3, -1)

each lie on this circle because their squared distances to

P

and

Q

sum to

100

.

Step-by-Step Solution

1

Calculate the square of hypotenuse

PQ

using the distance formula.

PQ^2 = (6 - 0)^2 + (8 - 0)^2 = 36 + 64 = 100

.

Since

PQ

is given as the hypotenuse of right triangle

PQR

, the Pythagorean theorem requires

PR^2 + RQ^2 = PQ^2 = 100

.

2

Express the condition

PR^2 + RQ^2 = 100

in terms of coordinates

(x, y)

of point

R

.

(x - 0)^2 + (y - 0)^2 + (x - 6)^2 + (y - 8)^2 = 100 \implies (x - 3)^2 + (y - 4)^2 = 25

.

By Thales's Theorem, any point

R

that forms a right triangle with hypotenuse

PQ

lies on a circle whose diameter is

PQ

, centered at the midpoint

(3, 4)

with radius

R_{circle} = 5

.

3

Test each candidate coordinate pair to verify if it satisfies

(x - 3)^2 + (y - 4)^2 = 25

.

Points

(8,4)

,

(-2,4)

, and

(3,-1)

satisfy

(5)^2 + 0^2 = 25

,

(-5)^2 + 0^2 = 25

, and

0^2 + (-5)^2 = 25

, respectively. Points

(4,8)

and

(6,6)

yield

(4-3)^2 + (8-4)^2 = 17 \neq 25

and

(6-3)^2 + (6-4)^2 = 13 \neq 25

.

Only coordinates located on the circle of diameter

PQ

form a right angle

\angle PRQ = 90^\circ

.

Key Concept

Pythagorean Theorem and Right Triangles in Coordinate Geometry

Question 137Question →

In the $xy$ -plane, triangle $ABC$ has vertices $A(1, 2)$ , $B(4, 2)$ , and $C(1, 6)$ . The triangle is reflected across the line $y = x$ , and then translated $2$ units to the left and $3$ units downward. What is the $y$ -coordinate of the image of vertex $C$ ?

Show answer & explanation

Answer: -2

Answer

The y-coordinate of the image of vertex C is -2.

Reflecting a point

(x, y)

across the line

y = x

swaps the coordinates, transforming

C(1, 6)

into

(6, 1)

. Subsequently, translating the point

2

units to the left and

3

units downward subtracts

2

from the

x

-coordinate and

3

from the

y

-coordinate, resulting in

(6 - 2, 1 - 3) = (4, -2)

. The

y

-coordinate of this image point is

-2

.

Step-by-Step Solution

1

Apply reflection across the line

y = x

to point

C(1, 6)

.

The transformed point is

C'(6, 1)

.

Reflecting a point

(x, y)

across the line

y = x

swaps its coordinates to

(y, x)

.

2

Apply translation left by

2

units and down by

3

units to

C'(6, 1)

.

The final point is

C''(6 - 2, 1 - 3) = C''(4, -2)

.

Translating left subtracts from the

x

-coordinate, and translating downward subtracts from the

y

-coordinate.

3

Extract the

y

-coordinate of

C''(4, -2)

.

-2

The

y

-coordinate is the second entry in the coordinate pair

(x, y)

.

Key Concept

Coordinate Geometry Transformations: Reflection across y = x and Translation

Estimated Time:1m 15s

Question 138Question →

In quadrilateral $ABCD$ , diagonal $AC$ divides the figure into two right triangles, $\triangle ABC$ and $\triangle ACD$ . It is given that $\angle ABC = 90^\circ$ , $\angle ACD = 90^\circ$ , $\angle BAC = 30^\circ$ , and $\angle CAD = 45^\circ$ . If $AB = 6\sqrt{3}$ , what is the perimeter of quadrilateral $ABCD$ ?

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

18 + 6\sqrt{3} + 12\sqrt{2}

24 + 6\sqrt{3} + 12\sqrt{2}

30 + 6\sqrt{3} + 12\sqrt{2}

18 + 18\sqrt{3} + 12\sqrt{6}

Show answer & explanation

Answer: $18 + 6\sqrt{3} + 12\sqrt{2}$

Answer

18 + 6\sqrt{3} + 12\sqrt{2}

The correct answer is derived by sequentially calculating the side lengths of the two special right triangles that share segment

AC

. In

\triangle ABC

, the given side

AB = 6\sqrt{3}

is adjacent to the

30^\circ

angle, making it the side opposite

60^\circ

. Using the

1 : \sqrt{3} : 2

ratio yields

BC = 6

and hypotenuse

AC = 12

. In

\triangle ACD

, leg

AC = 12

is adjacent to a

45^\circ

angle, making

\triangle ACD

a

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

triangle with equal leg

CD = 12

and hypotenuse

AD = 12\sqrt{2}

. Summing the outer edges

AB + BC + CD + AD = 6\sqrt{3} + 6 + 12 + 12\sqrt{2} = 18 + 6\sqrt{3} + 12\sqrt{2}

.

Step-by-Step Solution

1

Analyze right triangle

\triangle ABC

using

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

special right triangle ratios.

BC = 6

and

AC = 12

In a

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

triangle, the ratio of sides opposite to

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

is

1 : \sqrt{3} : 2

. Since

AB = 6\sqrt{3}

is opposite

60^\circ

, the shorter leg

BC = \frac{6\sqrt{3}}{\sqrt{3}} = 6

. The hypotenuse

AC = 2 \times BC = 12

.

2

Analyze right triangle

\triangle ACD

using

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

special right triangle ratios.

CD = 12

and

AD = 12\sqrt{2}

In right triangle

\triangle ACD

with

\angle ACD = 90^\circ

and

\angle CAD = 45^\circ

,

\triangle ACD

is an isosceles right triangle with side ratio

1 : 1 : \sqrt{2}

. Since leg

AC = 12

, leg

CD = 12

and hypotenuse

AD = 12\sqrt{2}

.

3

Sum the lengths of the four outer boundary segments to compute the perimeter of quadrilateral

ABCD

.

Perimeter

= 18 + 6\sqrt{3} + 12\sqrt{2}

Perimeter

= AB + BC + CD + AD = 6\sqrt{3} + 6 + 12 + 12\sqrt{2} = 18 + 6\sqrt{3} + 12\sqrt{2}

.

Key Concept

Special Right Triangles (

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

and

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

side ratios)

Question 139Question →

In the $xy$ -plane, the graph of a function $g(x)$ is obtained by shifting the graph of $f(x) = x^2 - 6x + 2$ horizontally to the right by $3$ units and then reflecting the resulting graph across the $x$ -axis. What is the value of $g(1)$ ?

-18

-10

0

6

18

Show answer & explanation

Answer: $-18$

Answer

The value of

g(1)

is

-18

.

Shifting the graph of

f(x)

horizontally to the right by

3

units gives the function

h(x) = f(x - 3)

. Reflecting this graph across the

x

-axis negates the entire function output, yielding

g(x) = -h(x) = -f(x - 3)

. To find

g(1)

, evaluate

-f(1 - 3) = -f(-2)

. Substituting

x = -2

into

f(x) = x^2 - 6x + 2

gives

f(-2) = (-2)^2 - 6(-2) + 2 = 4 + 12 + 2 = 18

. Negating this gives

-18

.

Step-by-Step Solution

1

Determine the algebraic expression for the transformed function

g(x)

in terms of

f(x)

.

g(x) = -f(x - 3)

Shifting a graph horizontally to the right by

3

units replaces

x

with

x - 3

, giving

f(x - 3)

. Reflecting across the

x

-axis negates the entire function, resulting in

-f(x - 3)

.

2

Substitute

x = 1

into the expression for

g(x)

.

g(1) = -f(1 - 3) = -f(-2)

Evaluating

g(1)

requires finding the value of

-f(x - 3)

when

x = 1

.

3

Evaluate

f(-2)

using the original definition

f(x) = x^2 - 6x + 2

.

f(-2) = (-2)^2 - 6(-2) + 2 = 4 + 12 + 2 = 18

Substituting

-2

into

f(x)

follows standard algebraic order of operations.

4

Apply the outer negative sign to calculate

g(1)

.

g(1) = -18

Since

g(1) = -f(-2)

and

f(-2) = 18

,

g(1) = -18

.

Key Concept

Graph Transformations of Functions

Question 140Question →

In right triangle $ABC$ , the measure of angle $ACB$ is $90^\circ$ . Altitude $CD$ is drawn from vertex $C$ to hypotenuse $AB$ , with point $D$ lying on line segment $AB$ . If $AD = 9$ and $DB = 16$ , what is the perimeter of triangle $ABC$ ?

Show answer & explanation

Answer: 60

Answer

The perimeter of triangle

ABC

is 60.

By the Geometric Mean Theorem for right triangles, the altitude

CD

to hypotenuse

AB

satisfies

CD^2 = AD \cdot DB = 9 \cdot 16 = 144

, giving

CD = 12

. Applying the Pythagorean Theorem to the smaller right triangles

\triangle ADC

and

\triangle BDC

yields

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

and

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

. The hypotenuse

AB = 9 + 16 = 25

. Summing the side lengths gives the perimeter:

15 + 20 + 25 = 60

.

Step-by-Step Solution

1

Calculate the length of altitude

CD

using the Geometric Mean Theorem.

CD = \sqrt{AD \cdot DB} = \sqrt{9 \cdot 16} = \sqrt{144} = 12

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that the altitude is the geometric mean of the two segment lengths.

2

Calculate leg

AC

using the Pythagorean Theorem in right triangle

ADC

.

AC = \sqrt{AD^2 + CD^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

Triangle

ADC

is a right triangle with right angle at

D

(

3-4-5

triple scaled by

3

).

3

Calculate leg

BC

using the Pythagorean Theorem in right triangle

BDC

.

BC = \sqrt{BD^2 + CD^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

Triangle

BDC

is a right triangle with right angle at

D

(

3-4-5

triple scaled by

4

).

4

Calculate the total perimeter of triangle

ABC

.

Perimeter =

AC + BC + AB = 15 + 20 + (9 + 16) = 15 + 20 + 25 = 60

The perimeter is the sum of the three outer sides of triangle

ABC

.

Key Concept

Right Triangle Altitude Relationships and Pythagorean Triples

Geometry

Answer

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