All practice questions

Question 601Question →

Line segments $AC$ and $BD$ intersect at point $E$ such that segment $AB$ is parallel to segment $CD$ . If the length of $AE$ is $5$ , the length of $CE$ is $10$ , the length of $BE$ is $x - 2$ , and the length of $DE$ is $x + 4$ , what is the value of $x$ ?

Show answer & explanation

Answer: 8

Answer

8

Since segment

AB

is parallel to segment

CD

, the alternate interior angles

\angle EAB

and

\angle ECD

are congruent, and vertical angles

\angle AEB

and

\angle CED

are congruent. By the Angle-Angle (AA) similarity theorem, triangle

ABE

is similar to triangle

CDE

. The ratio of the lengths of corresponding sides is equal, so

\frac{AE}{CE} = \frac{BE}{DE}

. Substituting the given lengths gives

\frac{5}{10} = \frac{x - 2}{x + 4}

. Simplifying the left side to

\frac{1}{2}

and cross-multiplying gives

x + 4 = 2(x - 2)

, which expands to

x + 4 = 2x - 4

. Solving for

x

yields

x = 8

.

Step-by-Step Solution

1

Establish the similarity of triangles

ABE

and

CDE

.

\triangle ABE \sim \triangle CDE

Since segment

AB

is parallel to segment

CD

, alternate interior angles

\angle EAB

and

\angle ECD

are congruent, and vertical angles

\angle AEB

and

\angle CED

are congruent. Thus, the triangles are similar by AA similarity.

2

Set up a proportion using the ratio of corresponding sides.

\frac{AE}{CE} = \frac{BE}{DE}

In similar triangles, the ratio of corresponding side lengths is constant.

3

Substitute the given algebraic expressions and segment lengths into the proportion.

\frac{5}{10} = \frac{x - 2}{x + 4}

The given values are

AE = 5

,

CE = 10

,

BE = x - 2

, and

DE = x + 4

.

4

Simplify the fraction and solve the linear equation for

x

.

x = 8

Simplifying

\frac{5}{10}

yields

\frac{1}{2}

. Cross-multiplying gives

1 \cdot (x + 4) = 2 \cdot (x - 2)

, which simplifies to

x + 4 = 2x - 4

. Subtracting

x

from both sides and adding

4

to both sides gives

x = 8

.

Key Concept

Triangle similarity criteria (specifically AA similarity) and using proportions of corresponding sides in similar triangles to solve for unknown variables.

Question 602Question →

In triangle $ABC$ , the side lengths are $AB = AC = 5$ and $BC = 6$ . Point $D$ is the midpoint of side $BC$ , and point $E$ lies on side $AB$ such that segment $DE$ is perpendicular to side $AB$ . What is the length of segment $DE$ ?

Show answer & explanation

Answer: 2.4

Answer

The length of segment

DE

is

2.4

(or the equivalent fraction

12/5

).

The correct answer is

2.4

(or

12/5

). In the isosceles triangle

ABC

with

AB = AC = 5

, the median

AD

to the base

BC

is also an altitude. Since

D

is the midpoint of

BC

, we have

BD = 3

. Applying the Pythagorean theorem to right triangle

ABD

gives

AD = \sqrt{5^2 - 3^2} = 4

. Since

DE

is perpendicular to

AB

, triangle

AED

is a right triangle that shares angle

A

with right triangle

ADB

. Therefore, triangle

AED

is similar to triangle

ADB

. The ratio of the opposite side to the hypotenuse in both triangles must be equal:

\frac{DE}{BD} = \frac{AD}{AB}

, which gives

\frac{DE}{3} = \frac{4}{5}

, or

DE = 2.4

.

Step-by-Step Solution

1

Determine the properties of the altitude

AD

in the isosceles triangle

ABC

.

AD

is perpendicular to

BC

, and

BD = 3

.

In an isosceles triangle, the median to the base is also the altitude to the base. Since

D

is the midpoint of

BC

,

BD = \frac{BC}{2} = \frac{6}{2} = 3

, and

\angle ADB = 90^\circ

.

2

Calculate the length of segment

AD

using the Pythagorean theorem in right triangle

ABD

.

AD = 4

Applying the Pythagorean theorem to right triangle

ABD

gives

AD^2 + BD^2 = AB^2

. Substituting the known lengths yields

AD^2 + 3^2 = 5^2

, which simplifies to

AD^2 = 25 - 9 = 16

, so

AD = 4

.

3

Find the length of segment

DE

using triangle similarity.

DE = 2.4

Since segment

DE

is perpendicular to side

AB

,

\angle AED = 90^\circ

. The right triangles

AED

and

ADB

share the angle at

A

, so they are similar by AA similarity (

\triangle AED \sim \triangle ADB

). This allows us to set up the ratio of corresponding sides:

\frac{DE}{BD} = \frac{AD}{AB}

. Substituting the values gives

\frac{DE}{3} = \frac{4}{5}

, which results in

DE = \frac{12}{5} = 2.4

.

Key Concept

Properties of isosceles triangles, the Pythagorean theorem, and right triangle similarity theorems.

Alternative Method

Alternatively, the length of

DE

can be found using the area of right triangle

ABD

. The area of triangle

ABD

is

\frac{1}{2} \times BD \times AD = \frac{1}{2} \times 3 \times 4 = 6

. The area can also be expressed using the hypotenuse

AB

as the base and

DE

as the height:

\text{Area} = \frac{1}{2} \times AB \times DE = \frac{1}{2} \times 5 \times DE

. Setting these equal gives

\frac{5}{2} DE = 6

, which yields

DE = 2.4

.

Estimated Time:1m 30s

Question 603Question →

In triangle $ABC$ , point $D$ lies on side $AB$ and point $E$ lies on side $AC$ such that segment $DE$ is parallel to segment $BC$ . The length of segment $AD$ is $2x + 1$ , the length of segment $DB$ is $x + 1$ , the length of segment $AE$ is $10$ , and the length of segment $EC$ is $6$ . What is the length of segment $AB$ ?

Show answer & explanation

Answer: 8

Answer

The length of segment

AB

is 8.

By the Triangle Proportionality Theorem, since segment

DE

is parallel to segment

BC

, the segments on the transversal sides are proportional:

\frac{AD}{DB} = \frac{AE}{EC}

. Substituting the given expressions and values yields

\frac{2x + 1}{x + 1} = \frac{10}{6}

. Simplifying the fraction on the right side to

\frac{5}{3}

and cross-multiplying gives

3(2x + 1) = 5(x + 1)

. Solving this equation yields

x = 2

. The length of segment

AB

is the sum of

AD

and

DB

, which is

(2x + 1) + (x + 1) = 3x + 2

. Substituting

x = 2

gives

AB = 8

.

Step-by-Step Solution

1

Set up the proportion using the Triangle Proportionality Theorem.

\frac{AD}{DB} = \frac{AE}{EC}

Since segment

DE

is parallel to segment

BC

, it divides the sides of triangle

ABC

proportionally.

2

Substitute the given values and simplify the constant ratio.

\frac{2x + 1}{x + 1} = \frac{5}{3}

The length of segment

AE

is 10 and

EC

is 6, so

\frac{AE}{EC} = \frac{10}{6} = \frac{5}{3}

.

3

Cross-multiply and solve the linear equation for

x

.

x = 2

Cross-multiplying gives

3(2x + 1) = 5(x + 1)

, which simplifies to

6x + 3 = 5x + 5

. Subtracting

5x

and 3 from both sides yields

x = 2

.

4

Calculate the total length of segment

AB

.

AB = 8

The total length

AB

is the sum of

AD

and

DB

. Thus,

AB = (2x + 1) + (x + 1) = 3x + 2

. Substituting

x = 2

gives

3(2) + 2 = 8

.

Key Concept

Triangle Proportionality Theorem and Similar Triangles

Alternative Method

Instead of using the Triangle Proportionality Theorem directly, we can use the similarity of triangles

ADE

and

ABC

. Since

DE \parallel BC

, we have

\triangle ADE \sim \triangle ABC

by AA similarity. This gives the ratio of corresponding side lengths:

\frac{AD}{AB} = \frac{AE}{AC}

. Substituting the expressions yields

\frac{2x + 1}{3x + 2} = \frac{10}{16} = \frac{5}{8}

. Cross-multiplying gives

8(2x + 1) = 5(3x + 2) \implies 16x + 8 = 15x + 10 \implies x = 2

. Then,

AB = 3x + 2 = 3(2) + 2 = 8

.

Estimated Time:1m 30s

Question 604Question →

Points $A$ , $B$ , and $C$ lie on a circle with center $O$ . The length of the minor arc $AC$ is $\frac{4}{9}$ of the circumference of the circle. What is the measure, in degrees, of the inscribed angle $\angle ABC$ ?

Show answer & explanation

Answer: 80

Answer

The measure of the inscribed angle is 80 degrees.

The correct answer is 80. The minor arc

AC

constitutes

\frac{4}{9}

of the circle's circumference, which corresponds to an arc measure of

\frac{4}{9} \times 360^\circ = 160^\circ

. By the Inscribed Angle Theorem, the measure of the inscribed angle

\angle ABC

is half the measure of the intercepted arc, which is

\frac{1}{2} \times 160^\circ = 80^\circ

.

Step-by-Step Solution

1

Determine the degree measure of the minor arc

AC

.

The measure of minor arc

AC

is

160^\circ

.

Since a full circle has a circumference corresponding to

360^\circ

, minor arc

AC

has a degree measure of

\frac{4}{9} \times 360^\circ = 160^\circ

.

2

Calculate the measure of the inscribed angle

\angle ABC

.

The measure of

\angle ABC

is

80^\circ

.

According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the arc it intercepts. Thus, the measure of

\angle ABC

is

\frac{1}{2} \times 160^\circ = 80^\circ

.

Key Concept

Inscribed Angle Theorem and Arc Measure

Question 605Question →

In the $xy$ -plane, the graph of the equation $x^2 + y^2 - 10x - 24y + 69 = 0$ is a circle. What is the distance between the center of this circle and the origin?

Show answer & explanation

Answer: 13

Answer

The distance between the center of the circle and the origin is 13.

Completing the square on the given equation yields standard form

(x - 5)^2 + (y - 12)^2 = 100

, identifying the center of the circle as

(5, 12)

. The distance from

(5, 12)

to

(0,0)

is calculated using the distance formula:

\sqrt{5^2 + 12^2} = \sqrt{169} = 13

.

Step-by-Step Solution

1

Group the

x

and

y

terms and complete the square for each variable.

(x - 5)^2 + (y - 12)^2 = 100

By rewriting

x^2 - 10x

as

(x - 5)^2 - 25

and

y^2 - 24y

as

(y - 12)^2 - 144

, the equation becomes

(x - 5)^2 - 25 + (y - 12)^2 - 144 + 69 = 0

. Combining the constant terms gives

(x - 5)^2 + (y - 12)^2 - 100 = 0

, which simplifies to standard form.

2

Identify the center of the circle from the standard form

(x - h)^2 + (y - k)^2 = r^2

.

The center is

(5, 12)

.

Comparing

(x - 5)^2 + (y - 12)^2 = 100

to the standard form shows that

h = 5

and

k = 12

.

3

Use the distance formula to calculate the distance between the center

(5, 12)

and the origin

(0, 0)

.

13

The distance formula is

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

. Substituting the points gives

d = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13

.

Key Concept

Equations of Circles in the Coordinate Plane

Question 606Question →

A rectangular field has a diagonal path of length $40$ meters. The length of the field is $8$ meters greater than its width. What is the width, in meters, of the field?

Show answer & explanation

Answer: 24

Answer

The correct answer is 24.

Letting

w

represent the width of the rectangular field in meters, the length is

w + 8

meters. The diagonal forms the hypotenuse of a right triangle with the width and the length as legs. By the Pythagorean theorem,

w^2 + (w+8)^2 = 40^2

. Expanding this yields

2w^2 + 16w + 64 = 1600

, which simplifies to

w^2 + 8w - 768 = 0

. Factoring this equation gives

(w-24)(w+32) = 0

. Since width must be positive, the width is 24.

Step-by-Step Solution

1

Represent the dimensions of the rectangular field.

Let the width of the field be

w

meters, so the length is

w + 8

meters.

The problem states the length is 8 meters greater than the width.

2

Apply the Pythagorean theorem.

w^2 + (w + 8)^2 = 40^2

The diagonal of a rectangle forms a right triangle with the width and length as its legs.

3

Simplify the quadratic equation.

w^2 + 8w - 768 = 0

Expanding the equation gives

2w^2 + 16w - 1536 = 0

, which simplifies when divided by 2.

4

Solve for the variable

w

.

(w - 24)(w + 32) = 0

, so

w = 24

or

w = -32

.

Factoring the quadratic equation gives the possible values for the width.

5

Determine the valid physical width.

w = 24

Since the width of a field must be positive, we reject the negative solution.

Key Concept

Using the Pythagorean theorem to relate the sides of a right triangle in a geometric word problem.

Question 607Question →

In the $xy$ -plane, a circle with center at the origin passes through the point $(0, 8)$ . A line tangent to the circle at point $T$ passes through the point $P(15, 8)$ . What is the length of segment $PT$ ?

Show answer & explanation

Answer: 15

Answer

The length of segment

PT

is

15

.

The radius of the circle is

8

because the circle is centered at

(0,0)

and passes through

(0,8)

. The distance from the center

O(0,0)

to point

P(15,8)

is

OP = \sqrt{15^2 + 8^2} = 17

. The radius

OT

is perpendicular to the tangent segment

PT

at point

T

, forming a right triangle

\triangle OTP

with hypotenuse

OP

and legs

OT

and

PT

. Using the Pythagorean theorem,

PT = \sqrt{17^2 - 8^2} = 15

.

Step-by-Step Solution

1

Determine the radius of the circle

Radius

r = 8

The circle is centered at the origin

(0, 0)

and passes through

(0, 8)

, so the distance from the center to this point is the radius.

2

Calculate the distance from the origin

O(0,0)

to the point

P(15, 8)

Distance

OP = 17

Using the distance formula in the coordinate plane:

OP = \sqrt{(15-0)^2 + (8-0)^2} = \sqrt{225 + 64} = 17

.

3

Apply the Pythagorean theorem to the right triangle

\triangle OTP

Length

PT = 15

Since the tangent line

PT

is perpendicular to the radius

OT

at the point of tangency

T

,

\triangle OTP

is a right triangle with hypotenuse

OP = 17

and leg

OT = 8

. Thus,

PT = \sqrt{OP^2 - OT^2} = \sqrt{17^2 - 8^2} = 15

.

Key Concept

Tangent lines to circles and the Pythagorean theorem in the coordinate plane

Question 608Question →

In the $xy$ -plane, the graph of the equation $x^2 + y^2 - 8x + 6y - 11 = 0$ is a circle. If the center of the circle is $(h, k)$ and the radius is $r$ , what is the value of $h + k + r$ ?

Show answer & explanation

Answer: 7

Answer

7

To find

h + k + r

, we convert the general equation

x^2 + y^2 - 8x + 6y - 11 = 0

to standard form

(x - h)^2 + (y - k)^2 = r^2

by completing the square. Grouping terms gives

(x^2 - 8x) + (y^2 + 6y) = 11

. Adding 16 and 9 to both sides to complete the square yields

(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9

, which simplifies to

(x - 4)^2 + (y + 3)^2 = 36

. Comparing this to the standard form gives center

(h, k) = (4, -3)

and radius

r = \sqrt{36} = 6

. The sum of these values is

4 + (-3) + 6 = 7

.

Step-by-Step Solution

1

Group the variable terms and move the constant term to the right side of the equation.

(x^2 - 8x) + (y^2 + 6y) = 11

To arrange the equation for completing the square.

2

Complete the square for both variables by adding the square of half the coefficient of the linear terms to both sides.

(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9

This creates factorable perfect square trinomials on the left side while maintaining equality.

3

Factor the perfect square trinomials and add the constants on the right side.

(x - 4)^2 + (y + 3)^2 = 36

To write the equation in the standard form of a circle:

(x - h)^2 + (y - k)^2 = r^2

.

4

Determine the coordinates of the center

(h, k)

and the radius

r

from the standard form.

Center coordinates

h = 4

,

k = -3

, and radius

r = 6

Comparing

(x - 4)^2 + (y + 3)^2 = 36

to standard form reveals

h = 4

,

k = -3

, and

r = \sqrt{36} = 6

.

5

Sum the values of

h

,

k

, and

r

together.

4 + (-3) + 6 = 7

To find the final requested value.

Key Concept

Converting the general equation of a circle into standard form by completing the square to identify its center and radius.

Question 609Question →

In the $xy$ -plane, the graph of the equation $x^2 + y^2 - 12x + 10y + k = 0$ , where $k$ is a constant, is a circle. If this circle is tangent to the $y$ -axis, what is the value of $k$ ?

Show answer & explanation

Answer: 25

Answer

The value of

k

is 25.

Completing the square for the equation

x^2 + y^2 - 12x + 10y + k = 0

yields standard form

(x - 6)^2 + (y + 5)^2 = 61 - k

. The center of the circle is

(6, -5)

and the radius squared is

r^2 = 61 - k

. Since the circle is tangent to the

y

-axis (the line

x = 0

), its radius must be equal to the horizontal distance from the center to the

y

-axis, which is the absolute value of the

x

-coordinate of the center,

|6| = 6

. Therefore, the radius is

6

, and the radius squared is

r^2 = 36

. Setting

61 - k = 36

gives

k = 25

.

Step-by-Step Solution

1

Complete the square for

x

and

y

variables in the equation.

(x - 6)^2 + (y + 5)^2 = 61 - k

Completing the square allows the general form equation of the circle to be written in standard form, where the center and radius are easily identifiable.

2

Determine the center and the expression for the radius squared from the standard form.

Center is

(6, -5)

and

r^2 = 61 - k

In the standard circle equation

(x - h)^2 + (y - k_c)^2 = r^2

, the center is

(h, k_c)

and the right side represents the radius squared.

3

Relate the tangency condition to the radius of the circle.

r = 6

A circle tangent to the

y

-axis has its radius equal to the distance from its center to the line

x = 0

, which is the absolute value of the center's

x

-coordinate.

4

Calculate the value of the constant

k

.

k = 25

Since

r = 6

, then

r^2 = 36

. Setting

61 - k = 36

and solving for

k

gives

k = 25

.

Key Concept

Completing the square to find the standard form of a circle's equation and applying geometric tangency properties to determine coordinates or constants.

Question 610Question →

In right triangle $PQR$ , the measure of angle $PQR$ is $90^\circ$ . Point $S$ lies on side $PQ$ . If the length of side $QR$ is $15$ , the length of segment $SR$ is $17$ , and the length of segment $PS$ is $12$ , what is the length of side $PR$ ?

Show answer & explanation

Answer: 25

Answer

The length of side

PR

is

25

.

First, the Pythagorean theorem is applied to the smaller right triangle

SQR

to solve for the unknown leg

QS

:

QS = \sqrt{SR^2 - QR^2} = \sqrt{17^2 - 15^2} = 8

. Next, the total length of leg

PQ

is the sum of its parts,

PS

and

QS

, which gives

PQ = 12 + 8 = 20

. Finally, the Pythagorean theorem is applied to the main right triangle

PQR

to find the hypotenuse

PR

:

PR = \sqrt{PQ^2 + QR^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25

.

Step-by-Step Solution

1

Use the Pythagorean theorem on the right triangle

SQR

to find the length of segment

QS

.

QS = 8

Triangle

SQR

shares the right angle at

Q

with triangle

PQR

. Thus,

QS^2 + QR^2 = SR^2 \implies QS^2 + 15^2 = 17^2 \implies QS = 8

.

2

Determine the length of side

PQ

.

PQ = 20

Since

S

lies on side

PQ

, the length of

PQ

is the sum of

PS

and

QS

:

PQ = 12 + 8 = 20

.

3

Use the Pythagorean theorem on right triangle

PQR

to calculate the length of the hypotenuse

PR

.

PR = 25

PQ^2 + QR^2 = PR^2 \implies 20^2 + 15^2 = PR^2 \implies 400 + 225 = 625 \implies PR = 25

.

Key Concept

Multi-step application of the Pythagorean theorem in nested right triangles

Question 611Question →

An isosceles trapezoid has a shorter base of length $10$ , a longer base of length $22$ , and a height of $8$ . What is the perimeter of the trapezoid?

Show answer & explanation

Answer: 52

Answer

The perimeter of the trapezoid is

52

.

To find the perimeter of the isosceles trapezoid, we first determine the length of the two congruent slanted sides. By drawing altitudes from the endpoints of the shorter top base to the longer bottom base, we create two right triangles at the sides of the trapezoid. The vertical leg of each right triangle is equal to the height of the trapezoid (

8

). The horizontal leg of each right triangle is equal to half the difference between the two base lengths, which is

\frac{22 - 10}{2} = 6

. Applying the Pythagorean theorem, the length of each slanted side is

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

. Finally, the perimeter is the sum of all four sides:

10 + 22 + 10 + 10 = 52

.

Step-by-Step Solution

1

Calculate the horizontal base of the right triangles formed by drawing heights from the top base to the bottom base.

6

An isosceles trapezoid is symmetrical. Drawing vertical lines representing the height of

8

from the endpoints of the shorter base of length

10

down to the longer base of length

22

divides the longer base into a middle segment of length

10

and two equal end segments. The length of each end segment is

\frac{22 - 10}{2} = 6

.

2

Use the Pythagorean theorem to find the length of the slanted legs of the trapezoid.

10

Each slanted leg is the hypotenuse of a right triangle with legs of length

6

and

8

. According to the Pythagorean theorem, the hypotenuse length is

\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

.

3

Find the perimeter of the trapezoid by adding the lengths of all four sides.

52

The perimeter is the sum of the shorter base (

10

), the longer base (

22

), and the two slanted legs (each of length

10

). Therefore, the perimeter is

10 + 22 + 10 + 10 = 52

.

Key Concept

Applying the Pythagorean theorem to find missing side lengths in composite geometric figures

Question 612Question →

In the $xy$ -plane, triangle $ABC$ is a right triangle with the right angle at $B$ . The vertices of the triangle are $A(3, 11)$ , $B(15, 20)$ , and $C(27, y)$ , where $y$ is a constant. If the length of the hypotenuse $AC$ is 25 and $y < 20$ , what is the value of $y$ ?

Show answer & explanation

Answer: 4

Answer

The value of

y

is 4.

Applying the distance formula to vertices

A(3, 11)

and

B(15, 20)

yields a length of 15 for side

AB

. Since triangle

ABC

has its right angle at

B

, the hypotenuse is

AC = 25

. Using the Pythagorean theorem (

AB^2 + BC^2 = AC^2

), we find that

BC = 20

. Setting the distance between

B(15, 20)

and

C(27, y)

equal to 20 gives

(27-15)^2 + (y-20)^2 = 20^2

. This simplifies to

(y-20)^2 = 256

, which has solutions

y = 36

or

y = 4

. Given that

y < 20

, the value of

y

must be 4.

Step-by-Step Solution

1

Calculate the length of segment

AB

using the distance formula.

AB = 15

To find one of the leg lengths of the right triangle.

2

Apply the Pythagorean theorem to find the length of leg

BC

.

BC = 20

To determine the length of the other leg of the right triangle.

3

Use the distance formula for

BC

and the condition

y < 20

to solve for

y

.

y = 4

To find the vertical coordinate of vertex

C

.

Key Concept

Using the Pythagorean theorem in coordinate geometry to solve for unknown coordinates.

Answer

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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

Answer

Step-by-Step Solution

Key Concept

Common Mistakes