Pythagorean Theorem and Special Right Triangles Practice Questions — ACT — Page 2

Question 21Question →

In the standard $(x, y)$ coordinate plane, a circle is centered at the origin $(0,0)$ and has a radius of $8$ . A horizontal chord $AB$ lies entirely in the first and second quadrants at a distance of $4$ units from the $x$ -axis. A point $P$ is located on the circle such that $\triangle ABP$ is a right triangle. If the hypotenuse of $\triangle ABP$ is a diameter of the circle, what is the area of $\triangle ABP$ ?

16\sqrt{3}

32

32\sqrt{3}

32\sqrt{5}

48\sqrt{3}

Show answer & explanation

Answer: $32\sqrt{3}$

Answer

The area of the right triangle is

32\sqrt{3}

.

The correct answer is

32\sqrt{3}

. The horizontal chord

AB

has

y

-coordinate

4

, and its endpoints lie on the circle

x^2 + y^2 = 64

. Solving for

x

gives

x = \pm 4\sqrt{3}

, so the length of the chord is

8\sqrt{3}

. Because the triangle is inscribed in the circle and is a right triangle, its hypotenuse must be a diameter of the circle (length

16

). Since

AB < 16

,

AB

is a leg, and the hypotenuse is one of the other sides (e.g.,

AP

). The remaining leg

BP

is found using the Pythagorean theorem:

BP = \sqrt{16^2 - (8\sqrt{3})^2} = \sqrt{256 - 192} = \sqrt{64} = 8

. The area of the right triangle is

\frac{1}{2} \times 8\sqrt{3} \times 8 = 32\sqrt{3}

.

Step-by-Step Solution

1

Determine the length of chord

AB

.

The length of chord

AB

is

8\sqrt{3}

.

The equation of the circle is

x^2 + y^2 = 64

. Since the chord is horizontal and at a distance of

4

units from the

x

-axis, its

y

-coordinate is

4

. Substituting

y = 4

gives

x^2 + 16 = 64 \Rightarrow x^2 = 48 \Rightarrow x = \pm 4\sqrt{3}

. The distance between

A(-4\sqrt{3}, 4)

and

B(4\sqrt{3}, 4)

is

8\sqrt{3}

.

2

Apply the rule for a right triangle inscribed in a circle to identify the hypotenuse.

The hypotenuse must be a diameter of length

16

, so the right angle is at

B

(or

A

).

Any right triangle inscribed in a circle must have a diameter as its hypotenuse. The diameter of this circle is

2 \times 8 = 16

. Since the chord

AB = 8\sqrt{3} \approx 13.86

is shorter than the diameter, it cannot be the hypotenuse. Therefore, either

AP

or

BP

is the hypotenuse (a diameter), making the angle opposite to it (either

\angle ABP

or

\angle BAP

) the

90^\circ

angle.

3

Calculate the length of the remaining leg of the right triangle.

The length of leg

BP

is

8

.

Using the Pythagorean theorem for right triangle

ABP

with hypotenuse

AP = 16

and leg

AB = 8\sqrt{3}

:

AB^2 + BP^2 = AP^2 \Rightarrow (8\sqrt{3})^2 + BP^2 = 16^2 \Rightarrow 192 + BP^2 = 256 \Rightarrow BP^2 = 64 \Rightarrow BP = 8

.

4

Compute the area of right triangle

ABP

.

The area is

32\sqrt{3}

.

The area of a right triangle is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BP = \frac{1}{2} \times 8\sqrt{3} \times 8 = 32\sqrt{3}

.

Key Concept

Applying the Pythagorean theorem and Thales's theorem (inscribed right triangles) to solve multi-step geometric problems on the coordinate plane.

Question 22Question →

A regular hexagon $ABCDEF$ has a side length of $8$ inches. Point $M$ lies on side $CD$ such that the length of segment $CM$ is $2$ inches. What is the length, in inches, of segment $AM$ ?

Show answer & explanation

Answer: 14

Answer

14

The correct answer is 14 because

\triangle ACM

is a right triangle with legs

AC = 8\sqrt{3}

and

CM = 2

. Applying the Pythagorean Theorem yields

AM^2 = (8\sqrt{3})^2 + 2^2 = 192 + 4 = 196

, so

AM = \sqrt{196} = 14

.

Step-by-Step Solution

1

Find the properties of the regular hexagon and the diagonal

AC

.

The interior angle at vertex

B

is

120^\circ

. Since

AB = BC = 8

, the triangle

\triangle ABC

is an isosceles triangle with angles

\angle BAC = \angle BCA = 30^\circ

. Using the properties of

30^\circ

-

60^\circ

-

90^\circ

triangles, the diagonal length is

AC = 8\sqrt{3}

.

To find the length of the leg

AC

for the right triangle

\triangle ACM

.

2

Determine the angle

\angle ACD

to show

\triangle ACM

is a right triangle.

Since the interior angle

\angle BCD = 120^\circ

and

\angle BCA = 30^\circ

, the remaining angle is

\angle ACD = 120^\circ - 30^\circ = 90^\circ

. Thus,

\triangle ACM

is a right triangle with the right angle at vertex

C

.

To establish the right-angle relationship between the legs

AC

and

CM

.

3

Apply the Pythagorean Theorem to calculate the hypotenuse

AM

.

AM^2 = AC^2 + CM^2 = (8\sqrt{3})^2 + 2^2 = 192 + 4 = 196

. Taking the square root gives

AM = 14

.

To find the final length of segment

AM

.

Key Concept

Using properties of regular hexagons, special right triangles, and the Pythagorean Theorem to find lengths in multi-step plane geometry configurations.

Question 23Question →

A security camera is mounted on a vertical wall at point $C$ , exactly $15$ feet above the flat ground. The camera is programmed to monitor two objects, $A$ and $B$ , on the ground. The line of sight from the camera to object $A$ makes a $30^\circ$ angle with the wall, and the line of sight from the camera to object $B$ makes a $45^\circ$ angle with the wall. On the ground, the path from the base of the wall directly below the camera to object $A$ is perpendicular to the path from the base of the wall to object $B$ . What is the straight-line distance, in feet, between object $A$ and object $B$ ?

5\sqrt{6}

4\sqrt{15}

10\sqrt{3}

15 + 5\sqrt{3}

30

Show answer & explanation

Answer: $10\sqrt{3}$

Answer

The distance between object

A

and object

B

is

10\sqrt{3}

feet.

The horizontal distances from the base of the wall to objects

A

and

B

are calculated using the trigonometric ratios of the

30^\circ

-

60^\circ

-

90^\circ

and

45^\circ

-

45^\circ

-

90^\circ

triangles formed by the vertical wall. This gives legs of

5\sqrt{3}

feet and

15

feet. Applying the Pythagorean theorem to the right triangle on the ground yields a hypotenuse of

10\sqrt{3}

feet.

Step-by-Step Solution

1

Find the horizontal distance from the base of the wall to object

A

.

Let

O

be the base of the wall directly below the camera

C

, so

OC = 15

feet. The line of sight

CA

makes a

30^\circ

angle with the wall, so

\triangle OCA

is a

30^\circ

-

60^\circ

-

90^\circ

right triangle with leg

OA = 15 \tan(30^\circ) = \frac{15}{\sqrt{3}} = 5\sqrt{3}

feet.

We need to determine the length of one of the perpendicular legs on the ground.

2

Find the horizontal distance from the base of the wall to object

B

.

The line of sight

CB

makes a

45^\circ

angle with the wall, so

\triangle OCB

is a

45^\circ

-

45^\circ

-

90^\circ

right triangle with leg

OB = 15 \tan(45^\circ) = 15

feet.

We need to determine the length of the other perpendicular leg on the ground.

3

Use the Pythagorean theorem to calculate the straight-line distance

AB

on the ground.

Since the paths

OA

and

OB

are perpendicular,

\triangle AOB

is a right triangle with legs

OA = 5\sqrt{3}

and

OB = 15

. The hypotenuse

AB

is

\sqrt{(5\sqrt{3})^2 + 15^2} = \sqrt{75 + 225} = \sqrt{300} = 10\sqrt{3}

feet.

The straight-line distance between the two objects corresponds to the hypotenuse of the right triangle formed by their ground distances.

Key Concept

Applying special right triangle ratios and the Pythagorean theorem to solve multi-step problems in three-dimensional contexts.

Question 24Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $60^\circ$ , and the length of $AC$ is $16$ units. Point $D$ lies on side $BC$ such that the measure of $\angle ADB$ is $45^\circ$ . What is the length of segment $CD$ , rounded to the nearest tenth?

Show answer & explanation

Answer: 5.9

Answer

The length of segment CD is approximately 5.9 units.

By recognizing that triangle ABC is a 30-60-90 right triangle, the shorter leg AB is found to be 8 (half of the hypotenuse 16), and the longer leg BC is 8*sqrt(3). Because triangle ABD is a 45-45-90 right triangle, the leg BD is equal to the leg AB, which is 8. Subtracting BD from BC yields CD = 8*sqrt(3) - 8, which is approximately 5.9.

Step-by-Step Solution

1

Determine the type of triangle ABC

Triangle ABC is a 30-60-90 special right triangle.

The triangle has a right angle (90 degrees) at B and an angle of 60 degrees at A, which leaves 30 degrees for angle C.

2

Calculate the lengths of sides AB and BC

AB = 8 units and BC = 8*sqrt(3) units.

In a 30-60-90 triangle with hypotenuse AC = 16, the side opposite 30 degrees (AB) is half the hypotenuse, and the side opposite 60 degrees (BC) is the shorter leg multiplied by sqrt(3).

3

Determine the type of triangle ABD

Triangle ABD is a 45-45-90 special right triangle.

Since D lies on BC, angle ABD is a right angle (90 degrees). Given that angle ADB is 45 degrees, the remaining angle BAD must also be 45 degrees.

4

Calculate the length of side BD

BD = 8 units.

In a 45-45-90 right triangle, the two legs opposite the 45-degree angles are equal in length, so BD = AB.

5

Calculate the length of segment CD and round to the nearest tenth

CD ≈ 5.9 units.

Since D lies on side BC, CD = BC - BD = 8*sqrt(3) - 8 ≈ 8(1.732) - 8 = 13.856 - 8 = 5.856, which rounds to 5.9.

Key Concept

Applying properties of 30-60-90 and 45-45-90 special right triangles to find segment lengths within nested figures.

Question 25Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , and the measure of $\angle C$ is $30^\circ$ . Point $D$ lies on segment $BC$ such that the measure of $\angle ADC$ is $135^\circ$ . If the length of segment $AD$ is $8$ units, what is the length, in units, of segment $AC$ ?

4\sqrt{6}

8

8\sqrt{2}

16

16\sqrt{2}

Show answer & explanation

Answer: $8\sqrt{2}$

Answer

8\sqrt{2}

The correct answer is

8\sqrt{2}

. Since points

B

,

D

, and

C

lie on a straight line, the adjacent angles

\angle ADB

and

\angle ADC

must sum to

180^\circ

. Subtracting the given angle measure shows that

\angle ADB = 180^\circ - 135^\circ = 45^\circ

. In the right triangle

ABD

, since

\angle B = 90^\circ

and

\angle ADB = 45^\circ

, the triangle is a

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

special right triangle. Using the ratio of side lengths for this triangle type, the leg

AB

is equal to the hypotenuse

AD

divided by

\sqrt{2}

, which simplifies to

AB = \frac{8}{\sqrt{2}} = 4\sqrt{2}

. Next, looking at the larger right triangle

ABC

, the angle at

C

is given as

30^\circ

, which makes

\triangle ABC

a

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

special right triangle. In this type of triangle, the hypotenuse

AC

is twice the length of the shorter leg

AB

, which is opposite the

30^\circ

angle. Multiplying the length of

AB

by

2

yields

AC = 2 \times 4\sqrt{2} = 8\sqrt{2}

units.

Step-by-Step Solution

1

Find the measure of angle

ADB

using the supplementary angle relationship along the line segment

BC

.

\angle ADB = 180^\circ - 135^\circ = 45^\circ

Points

B

,

D

, and

C

are collinear, meaning

\angle ADB

and

\angle ADC

form a linear pair and must sum to

180^\circ

.

2

Determine the properties of right triangle

ABD

and solve for the length of side

AB

.

\triangle ABD

is a

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

right triangle, where leg

AB = \frac{AD}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}

units.

Since

\angle B = 90^\circ

and

\angle ADB = 45^\circ

, the remaining angle

\angle DAB

is also

45^\circ

. In a

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

triangle, the leg length equals the hypotenuse divided by

\sqrt{2}

.

3

Use the properties of the larger

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

right triangle

ABC

to find the length of hypotenuse

AC

.

AC = 2 \times AB = 2 \times 4\sqrt{2} = 8\sqrt{2}

units.

In right triangle

ABC

, the angle opposite leg

AB

is

\angle C = 30^\circ

. In any

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

right triangle, the hypotenuse is exactly twice the length of the leg opposite the

30^\circ

angle.

Key Concept

Using multi-step properties of special right triangles (

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

and

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

) sharing a common boundary line.

Question 26Question →

A wheelchair ramp is constructed in two consecutive straight segments. The first segment rises at a $30^\circ$ angle relative to the flat ground and has a length of $12$ feet. The second segment starts at the end of the first segment and rises at a $45^\circ$ angle relative to the horizontal, with a length of $8\sqrt{2}$ feet. What is the total vertical rise, in feet, from the start of the first segment to the end of the second segment?

22

6 + 8\sqrt{2}

14

8 + 6\sqrt{3}

16 + 6\sqrt{3}

Show answer & explanation

Answer: $14$

Answer

The total vertical rise is

14

feet.

The total vertical rise is the sum of the vertical rises of the two individual segments. The first segment forms a

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

right triangle where the hypotenuse is

12

feet, so the vertical rise (the leg opposite the

30^\circ

angle) is half of the hypotenuse, which is

6

feet. The second segment forms a

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

right triangle where the hypotenuse is

8\sqrt{2}

feet, so the vertical rise (the leg opposite the

45^\circ

angle) is the hypotenuse divided by

\sqrt{2}

, which is

8

feet. Adding these two values gives a total vertical rise of

14

feet.

Step-by-Step Solution

1

Calculate the vertical rise of the first segment.

The first segment has a length of

12

feet and rises at a

30^\circ

angle. It forms a

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

right triangle where the vertical rise is the side opposite the

30^\circ

angle. Since the leg opposite the

30^\circ

angle is half the hypotenuse, the vertical rise is

\frac{12}{2} = 6

feet.

To find the vertical component of the first ramp segment.

2

Calculate the vertical rise of the second segment.

The second segment has a length of

8\sqrt{2}

feet and rises at a

45^\circ

angle. It forms a

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

right triangle where the leg length is the hypotenuse divided by

\sqrt{2}

. Thus, the vertical rise is

\frac{8\sqrt{2}}{\sqrt{2}} = 8

feet.

To find the vertical component of the second ramp segment.

3

Sum the vertical rises of both segments.

The total vertical rise is

6\text{ feet} + 8\text{ feet} = 14

feet.

To find the combined vertical height gained over the entire ramp.

Key Concept

Using special right triangle ratios (

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

and

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

) to find missing side lengths.

Estimated Time:1m 30s

Question 27Question →

An equilateral triangle $ABC$ has a side length of $8$ inches. Point $D$ lies on side $BC$ such that the distance from $B$ to $D$ is $3$ inches. What is the length, in inches, of the segment $AD$ ?

Show answer & explanation

Answer: 7

Answer

The length of segment

AD

is

7

inches.

Dropping altitude

AM

from

A

to

BC

divides the equilateral triangle into two

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

right triangles. Since

M

is the midpoint of

BC

,

BM = 4

inches. In

\triangle ABM

, the hypotenuse is

8

and the shorter leg is

4

, so the altitude

AM = 4\sqrt{3}

inches. Since

BD = 3

inches, the segment

DM

has length

BM - BD = 4 - 3 = 1

inch. Applying the Pythagorean Theorem to right triangle

\triangle ADM

gives

AD^2 = AM^2 + DM^2 = (4\sqrt{3})^2 + 1^2 = 48 + 1 = 49

, which simplifies to

AD = 7

inches.

Step-by-Step Solution

1

Find the midpoint of side

BC

by dropping altitude

AM

.

BM = 4

inches

In an equilateral triangle, the altitude to a side bisects that side.

2

Calculate the length of the altitude

AM

.

AM = 4\sqrt{3}

inches

The altitude forms a

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

triangle with the hypotenuse of

8

inches, making the altitude length equal to

8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}

.

3

Determine the length of the segment

DM

.

DM = 1

inch

Since

D

is

3

inches from

B

and

M

is

4

inches from

B

, the remaining distance is

4 - 3 = 1

.

4

Apply the Pythagorean Theorem on right triangle

\triangle ADM

to find

AD

.

AD = 7

inches

The hypotenuse squared is the sum of the squares of the legs:

AD^2 = (4\sqrt{3})^2 + 1^2 = 48 + 1 = 49

, which gives

AD = 7

.

Key Concept

Using the altitude of an equilateral triangle to create special right triangles and applying the Pythagorean Theorem.

Question 28Question →

A rectangular park $ABCD$ has a length of $20$ meters and a width of $15$ meters. A straight walking path is built from corner $A$ to a point $P$ on the diagonal path $BD$ such that the path $AP$ is perpendicular to $BD$ . What is the length, in meters, of the path $AP$ ?

Show answer & explanation

Answer: 12

Answer

The length of the path

AP

is

12

meters.

The diagonal

BD

divides the rectangular park into two congruent right triangles. For right triangle

ABD

, the legs are

AB = 20

and

AD = 15

. Using the Pythagorean theorem, the hypotenuse

BD = \sqrt{20^2 + 15^2} = 25

meters. The area of triangle

ABD

is

\frac{1}{2} \times 20 \times 15 = 150

square meters. Since

AP

is perpendicular to

BD

,

AP

is the altitude to base

BD

, so the area can also be written as

\frac{1}{2} \times 25 \times AP

. Equating the two areas,

12.5 \times AP = 150

, which simplifies to

AP = 12

meters.

Step-by-Step Solution

1

Calculate the length of the diagonal

BD

using the Pythagorean theorem on right triangle

ABD

.

BD = \sqrt{20^2 + 15^2} = 25

meters.

The diagonal forms the hypotenuse of the right triangle

ABD

, which is needed to calculate the altitude

AP

.

2

Express the area of right triangle

ABD

using the two perpendicular legs,

AB

and

AD

.

\text{Area} = \frac{1}{2} \times 20 \times 15 = 150

square meters.

This establishes the total area of the triangle.

3

Express the area of the same triangle using the diagonal

BD

as the base and the perpendicular path

AP

as the height.

\text{Area} = \frac{1}{2} \times 25 \times AP

.

This sets up an equation using the unknown path length

AP

.

4

Equate the two area expressions and solve for

AP

.

12.5 \times AP = 150 \implies AP = 12

meters.

Since both expressions represent the area of the same triangle, they must be equal.

Key Concept

Using the Pythagorean theorem to find the hypotenuse of a right triangle, and then using the area formula to find the altitude to the hypotenuse.

Alternative Method

Alternatively, you can use similar right triangles. Since triangle

ABP

is similar to triangle

DBA

, the ratio of their corresponding sides is equal:

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

. Substituting the known values gives

\frac{AP}{15} = \frac{20}{25}

, which simplifies to

AP = 15 \times \frac{4}{5} = 12

meters.

Estimated Time:1m 30s

Question 29Question →

In the figure below, quadrilateral $ABCD$ is divided by diagonal $BD$ into two right triangles. In right triangle $ABD$ , the angle at $B$ is a right angle, $\angle ADB = 30^\circ$ , and the hypotenuse $AD = 12$ centimeters. In right triangle $BCD$ , the angle at $C$ is a right angle, and $BC = CD$ . What is the perimeter, in centimeters, of quadrilateral $ABCD$ ?

18 + 6\sqrt{3}

18 + 12\sqrt{3}

18 + 6\sqrt{6}

18 + 12\sqrt{6}

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

Show answer & explanation

Answer: $18 + 6\sqrt{6}$

Answer

The perimeter of the quadrilateral is

18 + 6\sqrt{6}

centimeters.

The perimeter of quadrilateral

ABCD

is the sum of the lengths of its four outer boundary sides:

AB

,

BC

,

CD

, and

DA

. In the

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

right triangle

ABD

, the shorter leg

AB

is half the hypotenuse

AD

, so

AB = 6\text{ cm}

. The longer leg

BD

is

AB\sqrt{3} = 6\sqrt{3}\text{ cm}

. In the isosceles right triangle

BCD

, the hypotenuse is

BD = 6\sqrt{3}\text{ cm}

. The legs

BC

and

CD

are congruent, with each length equal to the hypotenuse divided by

\sqrt{2}

, which simplifies to

3\sqrt{6}\text{ cm}

. Adding the four outer side lengths (

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

) yields a perimeter of

18 + 6\sqrt{6}\text{ cm}

.

Step-by-Step Solution

1

Find the lengths of the legs of right triangle

ABD

using the properties of a

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

right triangle.

AB = 6\text{ cm}

and

BD = 6\sqrt{3}\text{ cm}

In a

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

triangle, the leg opposite the

30^\circ

angle is half the length of the hypotenuse (

AB = \frac{12}{2} = 6

), and the leg opposite the

60^\circ

angle is

\sqrt{3}

times the shorter leg (

BD = 6\sqrt{3}

).

2

Find the lengths of the legs of the isosceles right triangle

BCD

(

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

) using the hypotenuse

BD

.

BC = CD = 3\sqrt{6}\text{ cm}

In a

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

triangle, the length of each leg is the hypotenuse divided by

\sqrt{2}

. Thus,

BC = CD = \frac{6\sqrt{3}}{\sqrt{2}} = \frac{6\sqrt{3}\cdot\sqrt{2}}{2} = 3\sqrt{6}

.

3

Calculate the perimeter of quadrilateral

ABCD

by summing the lengths of its four outer boundary sides:

AB

,

BC

,

CD

, and

DA

.

Perimeter

= 6 + 3\sqrt{6} + 3\sqrt{6} + 12 = 18 + 6\sqrt{6}\text{ cm}

The perimeter is the sum of the outer boundary sides of the quadrilateral, which are

AB

,

BC

,

CD

, and

DA

.

Key Concept

Pythagorean Theorem and Special Right Triangles

Alternative Method

Instead of using special right triangle formulas, the Pythagorean theorem can be used with variables:

AB^2 + BD^2 = AD^2

, where

AB = \frac{1}{2}AD = 6

, so

36 + BD^2 = 144 \implies BD = \sqrt{108} = 6\sqrt{3}

. Then

BC^2 + CD^2 = BD^2 \implies 2BC^2 = 108 \implies BC = \sqrt{54} = 3\sqrt{6}

.

Estimated Time:1m 30s

Question 30Question →

In the figure below, quadrilateral $ABCD$ is composed of two right triangles, $\triangle ABC$ and $\triangle ACD$ . The measure of $\angle ABC$ is $90^\circ$ , and the measure of $\angle ACD$ is $90^\circ$ . The side lengths are $AB = 3$ units and $BC = 4$ units. If the measure of $\angle CAD$ is $60^\circ$ , what is the length, in units, of segment $CD$ ?

\sqrt{21}

5\sqrt{3}

7\sqrt{3}

10

\sqrt{39}

Show answer & explanation

Answer: $5\sqrt{3}$

Answer

The length of segment

CD

is

5\sqrt{3}

units.

The correct answer is the length of

5\sqrt{3}

units. By using the Pythagorean theorem on the first right triangle

\triangle ABC

, the length of the hypotenuse is

AC = \sqrt{3^2 + 4^2} = 5

. Since

\triangle ACD

is a 30-60-90 right triangle with a right angle at

C

and

\angle CAD = 60^\circ

, the side

AC

is the shorter leg (opposite the

30^\circ

angle). The length of the longer leg

CD

(opposite the

60^\circ

angle) is therefore

AC\sqrt{3} = 5\sqrt{3}

.

Step-by-Step Solution

1

Use the Pythagorean theorem in right triangle

\triangle ABC

to find the length of the hypotenuse

AC

.

AC = 5

Since

\triangle ABC

is a right triangle with legs

AB = 3

and

BC = 4

, the hypotenuse is

AC = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5

.

2

Identify the type of right triangle for

\triangle ACD

.

\triangle ACD

is a 30-60-90 right triangle.

The triangle has a right angle at

C

(measure of

\angle ACD = 90^\circ

) and an acute angle at

A

(measure of

\angle CAD = 60^\circ

), which leaves the remaining angle

\angle ADC = 30^\circ

.

3

Apply the special right triangle ratios to find the length of leg

CD

.

CD = 5\sqrt{3}

In a 30-60-90 triangle, the leg opposite the

60^\circ

angle is

\sqrt{3}

times the leg opposite the

30^\circ

angle. Here,

AC = 5

is opposite the

30^\circ

angle, so the longer leg

CD = AC\sqrt{3} = 5\sqrt{3}

.

Key Concept

Using the Pythagorean theorem to find a shared side and then applying special right triangle ratios (30-60-90) to solve for an unknown length.

Alternative Method

Instead of using the special right triangle ratios directly, right triangle trigonometry can be applied:

\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CD}{AC}

. Since

\tan(60^\circ) = \sqrt{3}

and

AC = 5

, we have

\sqrt{3} = \frac{CD}{5}

, which yields

CD = 5\sqrt{3}

.

Estimated Time:1m 0s

Question 31Question →

A hiker starts at a trailhead, point $A$ , and walks $9$ miles due east, then $12$ miles due north to reach a campsite, point $C$ . A lookout tower, point $T$ , is located due west of the campsite $C$ . The straight-line distance from the starting point $A$ to the tower $T$ is $13$ miles. If the tower $T$ is located east of the north-south line passing through point $A$ , what is the distance, in miles, between the campsite $C$ and the tower $T$ ?

Show answer & explanation

Answer: 4

Answer

The distance between the campsite and the tower is 4 miles.

The correct answer is 4 miles. By modeling the hiker's path, the vertical height of both the campsite and the tower is 12 miles north of the starting point's east-west line. A right triangle is formed by the trailhead, the projection of the tower onto the east-west axis, and the tower itself. The hypotenuse is 13 miles and the vertical leg is 12 miles. By the Pythagorean theorem, the horizontal leg is 5 miles. Since the tower is east of the trailhead's north-south line, it is 5 miles east of the trailhead. The campsite is 9 miles east of the trailhead. The horizontal distance between the campsite and the tower is the difference: 9 - 5 = 4 miles.

Step-by-Step Solution

1

Determine the vertical height of the campsite and the tower.

The campsite

C

is

12

miles north of the trailhead

A

's east-west line. Since the tower

T

is located due west of

C

, it lies on the same horizontal line. Therefore, the vertical distance from the east-west line to the tower

T

is also

12

miles.

Points on the same due east-west line share the same vertical offset (latitude) from the reference axis.

2

Use the Pythagorean theorem to find the horizontal distance from the trailhead to the tower.

Let

D

be the point on the east-west line directly below the tower

T

. A right triangle

\triangle ADT

is formed where the vertical leg

DT = 12

miles, the hypotenuse

AT = 13

miles, and the horizontal leg is

AD

. Using the Pythagorean theorem:

AD^2 + DT^2 = AT^2 \implies AD^2 + 12^2 = 13^2 \implies AD^2 + 144 = 169 \implies AD^2 = 25 \implies AD = 5

miles.

Calculating the horizontal offset of the tower from the trailhead's north-south line.

3

Calculate the horizontal distance between the campsite and the tower.

The campsite

C

is

9

miles east of the trailhead

A

's north-south line, and the tower

T

is

5

miles east of it. The horizontal distance between them is the difference:

9 - 5 = 4

miles.

Since both points have the same vertical coordinate, the direct distance is simply the difference in their horizontal coordinates.

Key Concept

Using the Pythagorean theorem to solve multi-step geometry problems on a coordinate-like plane.

Estimated Time:1m 30s

Question 32Question →

In the trapezoid $ABCD$ , bases $AB$ and $CD$ are parallel, and side $AD$ is perpendicular to base $AB$ . The length of $AB$ is $10$ inches, the length of $CD$ is $4$ inches, and the measure of angle $B$ is $60^\circ$ . What is the perimeter, in inches, of the trapezoid?

20 + 3\sqrt{3}

20 + 6\sqrt{3}

26 + 6\sqrt{3}

14 + 6\sqrt{3}

32\sqrt{3}

Show answer & explanation

Answer: $26 + 6\sqrt{3}$

Answer

The perimeter of the trapezoid is

26 + 6\sqrt{3}

inches.

Drawing an altitude from vertex

C

to base

AB

splits the trapezoid into a rectangle

AECD

and a right triangle

\triangle CEB

. Since

AE = CD = 4

, the leg

EB = 10 - 4 = 6

. The right triangle

\triangle CEB

is a

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

triangle where

EB

is the leg opposite the

30^\circ

angle. The hypotenuse

BC = 2 \times 6 = 12

and the height

CE = AD = 6\sqrt{3}

. Summing all four outer sides of the trapezoid (

10 + 12 + 4 + 6\sqrt{3}

) yields the correct perimeter of

26 + 6\sqrt{3}

inches.

Step-by-Step Solution

1

Draw an altitude from vertex

C

perpendicular to base

AB

at point

E

.

A rectangle

AECD

and a right triangle

\triangle CEB

are formed, with

AE = CD = 4

inches and

CE = AD

.

To break down the trapezoid into a rectangle and a right triangle so we can find the unknown side lengths.

2

Find the length of segment

EB

.

EB = AB - AE = 10 - 4 = 6

inches.

To find the length of the base of the right triangle

\triangle CEB

.

3

Use the properties of a

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

right triangle to determine the lengths of sides

CE

(which is

AD

) and

BC

.

Since

\angle B = 60^\circ

is opposite to

CE

, and

EB = 6

is the shorter leg adjacent to

60^\circ

, the hypotenuse is

BC = 2 \times 6 = 12

inches and the longer leg is

CE = AD = 6\sqrt{3}

inches.

To compute the remaining unknown outer side lengths of the trapezoid.

4

Sum the four outer sides of the trapezoid to find the perimeter.

Perimeter =

AB + BC + CD + DA = 10 + 12 + 4 + 6\sqrt{3} = 26 + 6\sqrt{3}

inches.

The perimeter is the total boundary length of the shape.

Key Concept

The perimeter of a right trapezoid can be found by drawing an altitude to create a rectangle and a

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

special right triangle, then determining the missing side lengths using special right triangle ratios.

Estimated Time:1m 30s

Question 33Question →

A vertical flagpole is secured by two straight guide wires anchored to the flat ground on opposite sides of the pole. The first guide wire is $13$ feet long and its anchor is $5$ feet from the base of the pole. The second guide wire is anchored such that it makes a $30^\circ$ angle of elevation with the ground. If both guide wires are attached to the flagpole at the same height, what is the length, in feet, of the second guide wire?

Show answer & explanation

Answer: 24

Answer

The length of the second guide wire is

24

feet.

The correct answer is

24

. First, the height of the attachment point is found using the Pythagorean theorem:

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

feet. Since the second wire makes a

30^\circ

angle of elevation with the ground, it forms a 30-60-90 right triangle where the flagpole height of

12

feet is the leg opposite the

30^\circ

angle. The length of the wire is the hypotenuse of this triangle, which is twice the length of the opposite leg:

2 \times 12 = 24

feet.

Step-by-Step Solution

1

Use the Pythagorean theorem to calculate the height of the flagpole where the guide wires are attached.

The flagpole height is

12

feet.

The first guide wire, the flagpole, and the ground form a right triangle with a hypotenuse of

13

feet and a horizontal leg of

5

feet. Thus,

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

.

2

Apply the properties of a 30-60-90 special right triangle to find the length of the second guide wire.

The length of the second guide wire is

24

feet.

The second wire forms a 30-60-90 right triangle with the flagpole and the ground. The angle of elevation is

30^\circ

, which means the side opposite this angle is the vertical height of the flagpole (

12

feet). In a 30-60-90 triangle, the hypotenuse (the wire length) is twice the length of the shorter leg (opposite the

30^\circ

angle), so the length is

2 \times 12 = 24

.

Key Concept

Applying the Pythagorean Theorem and the ratio properties of 30-60-90 special right triangles to solve multi-step geometry problems.

Question 34Question →

A right triangle has a hypotenuse of length $13\text{ inches}$ . One of the legs is $7\text{ inches}$ longer than the other leg. What is the length, in inches, of the shorter leg?

3.0

5.0

6.0

8.6

11.0

Show answer & explanation

Answer: 5.0

Answer

The length of the shorter leg is 5.0 inches.

The correct answer is the option representing 5.0. By setting the shorter leg as

x

, the longer leg is

x + 7

. Applying the Pythagorean theorem gives

x^2 + (x + 7)^2 = 13^2

, which simplifies to

2x^2 + 14x - 120 = 0

. Factoring the divided equation

x^2 + 7x - 60 = 0

gives

(x - 5)(x + 12) = 0

. Since length must be positive, the shorter leg is 5.0 inches.

Step-by-Step Solution

1

Define variables for the side lengths of the right triangle based on the problem statement.

Let the length of the shorter leg be

x

inches. The length of the longer leg is

x + 7

inches, and the hypotenuse is

13

inches.

This translates the verbal descriptions into mathematical expressions using a single variable.

2

Apply the Pythagorean theorem to set up an equation relating the side lengths.

x^2 + (x + 7)^2 = 13^2

For any right triangle, the sum of the squares of the legs equals the square of the hypotenuse.

3

Expand the squared binomial and simplify the equation into standard quadratic form.

x^2 + (x^2 + 14x + 49) = 169 \Rightarrow 2x^2 + 14x - 120 = 0

Expanding allows us to group like terms and solve for the variable

x

.

4

Divide the quadratic equation by 2 and factor the resulting expression.

x^2 + 7x - 60 = 0 \Rightarrow (x + 12)(x - 5) = 0

Factoring is the most efficient way to find the roots of this quadratic equation.

5

Solve for

x

and select the mathematically and physically valid solution.

x = 5

or

x = -12

. Since a side length must be positive,

x = 5

inches.

Lengths in geometry must be positive, so we discard the negative root.

Key Concept

Applying the Pythagorean theorem to solve for unknown side lengths of a right triangle given algebraic relationships between the sides.

Estimated Time:1m 30s

Question 35Question →

An architect is designing a triangular roof truss, $\triangle ABC$ , where side $AB$ is equal in length to side $AC$ . The height of the truss, represented by the altitude from vertex $A$ to the base $BC$ , is $12$ feet. If the measure of the base angle $\angle ABC$ is $30^\circ$ , what is the length, in feet, of the base $BC$ ? (Round your answer to the nearest tenth.)

Show answer & explanation

Answer: 41.6

Answer

The correct answer is 41.6

The correct answer is obtained by recognizing that the altitude of the isosceles triangle bisects the base into two congruent

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

right triangles. The leg opposite the

30^\circ

angle is

12

feet, so the leg adjacent (which is half the base) is

12\sqrt{3}

feet. Doubling this gives a total base length of

24\sqrt{3}

feet, which is approximately

41.6

feet when rounded to the nearest tenth.

Step-by-Step Solution

1

Identify the right triangle formed by the altitude.

An altitude

AD

perpendicular to base

BC

, creating two right triangles,

\triangle ABD

and

\triangle ACD

, with

AD = 12

feet.

In an isosceles triangle, the altitude to the base bisects the base and is perpendicular to it.

2

Determine the angles of the right triangle

\triangle ABD

.

Triangle

\triangle ABD

is a

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

special right triangle.

Angle

B

is

30^\circ

and angle

ADB

is

90^\circ

, leaving

60^\circ

for angle

BAD

.

3

Calculate the length of the segment

BD

.

BD = 12\sqrt{3}

feet

In a

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

triangle, the longer leg is

\sqrt{3}

times the shorter leg (which is opposite the

30^\circ

angle).

4

Find the total length of the base

BC

.

BC = 24\sqrt{3}

feet

Since

D

is the midpoint of

BC

, the total length

BC

is

2 \times BD

.

5

Convert the exact value to a decimal rounded to the nearest tenth.

BC \approx 41.6

24 \times 1.73205 = 41.569

, which rounds to

41.6

.

Key Concept

Properties of special 30-60-90 right triangles and altitudes of isosceles triangles

Question 36Question →

In the figure, point $B$ lies on the line segment $AC$ , and segment $BD$ is perpendicular to $AC$ . Triangle $ABD$ is a right triangle with hypotenuse $AD = 8$ and $\angle ADB = 30^\circ$ . If the length of segment $BC$ is $11$ , what is the length of segment $CD$ ?

Show answer & explanation

Answer: 13

Answer

The length of segment

CD

is

13

.

The length of segment

CD

is

13

. Since

BD

is perpendicular to segment

AC

at point

B

,

\triangle ABD

and

\triangle DBC

are both right triangles. In the

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

right triangle

\triangle ABD

, the hypotenuse is

AD = 8

, so the longer leg opposite the

60^\circ

angle is

BD = 4\sqrt{3}

. In right triangle

\triangle DBC

, using the Pythagorean Theorem:

CD^2 = BD^2 + BC^2 = (4\sqrt{3})^2 + 11^2 = 48 + 121 = 169

. Taking the square root gives

CD = 13

.

Step-by-Step Solution

1

Find the length of the shared perpendicular segment

BD

using the properties of the special

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

right triangle

\triangle ABD

.

BD = 4\sqrt{3}

In a

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

right triangle, the side opposite the

60^\circ

angle is

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

times the hypotenuse.

2

Apply the Pythagorean Theorem to right triangle

\triangle DBC

to calculate the length of hypotenuse

CD

.

CD = 13

The Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the squares of the legs (

CD^2 = BD^2 + BC^2

).

Key Concept

Solving for unknown sides in adjacent right triangles by combining special right triangle ratios (

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

) and the Pythagorean Theorem.

Question 37Question →

A maintenance worker leans a ladder against a vertical wall such that the ladder makes a $60^\circ$ angle with the horizontal ground, reaching a height of $15\sqrt{3}\text{ feet}$ up the wall. If the base of the ladder is then pulled further away from the wall until the ladder makes a $45^\circ$ angle with the horizontal ground, how many feet further from the wall is the base of the ladder?

15\sqrt{2} - 15

15\sqrt{3} - 15

15\sqrt{2}

30\sqrt{2} - 30

30 - 15\sqrt{3}

Show answer & explanation

Answer: $15\sqrt{2} - 15$

Answer

The base of the ladder is

15\sqrt{2} - 15

feet further from the wall.

In the initial position, the ladder forms a 30°-60°-90° right triangle with the wall and ground. The side opposite the 60° angle (height on the wall) is

15\sqrt{3}\text{ ft}

. Using the ratio

1:\sqrt{3}:2

, the shorter leg (initial distance from the wall) is

15\text{ ft}

, and the hypotenuse (ladder length) is

30\text{ ft}

. In the second position, the ladder forms a 45°-45°-90° triangle with hypotenuse

30\text{ ft}

. The leg length (new distance from the wall) is

\frac{30}{\sqrt{2}} = 15\sqrt{2}\text{ ft}

. The additional distance the ladder base was pulled is

15\sqrt{2} - 15\text{ ft}

.

Step-by-Step Solution

1

Determine the initial base distance and ladder length using 30°-60°-90° triangle relationships.

Initial base distance =

15\text{ ft}

, ladder length =

30\text{ ft}

.

In a 30°-60°-90° right triangle, the side opposite the 60° angle is

x\sqrt{3}

. Given

x\sqrt{3} = 15\sqrt{3}

, the shorter leg (initial base distance) is

x = 15\text{ ft}

and the hypotenuse (ladder length) is

2x = 30\text{ ft}

.

2

Determine the new base distance using 45°-45°-90° triangle relationships.

New base distance =

15\sqrt{2}\text{ ft}

.

When the ladder (hypotenuse of 30 ft) makes a 45° angle with the ground, it forms a 45°-45°-90° right triangle where hypotenuse =

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

. Thus, the new base distance is

\frac{30}{\sqrt{2}} = 15\sqrt{2}\text{ ft}

.

3

Calculate how much further the base was pulled from the wall.

15\sqrt{2} - 15\text{ ft}

.

Subtract the initial base distance (

15\text{ ft}

) from the new base distance (

15\sqrt{2}\text{ ft}

).

Key Concept

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

Estimated Time:1m 30s

Question 38Question →

In rectangle $ABCD$ , the length of side $AD$ is $12$ units and the length of side $CD$ is $17$ units. Point $E$ lies on side $CD$ such that $\triangle ADE$ is an isosceles right triangle with the right angle at vertex $D$ . What is the length, in units, of segment $BE$ ?

Show answer & explanation

Answer: 13

Answer

The length of segment

BE

is

13

units.

Because

\triangle ADE

is an isosceles right triangle with the right angle at vertex

D

, leg

DE

equals leg

AD = 12

. Subtracting

DE

from total side length

CD = 17

gives segment

EC = 5

. Since

ABCD

is a rectangle, angle

C

is a right angle (

90^\circ

) and

BC = AD = 12

. Applying the Pythagorean Theorem to right triangle

\triangle BCE

gives hypotenuse

BE = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13

.

Step-by-Step Solution

1

Find the length of segment

DE

using the properties of an isosceles right triangle.

DE = 12

units

In isosceles right triangle

\triangle ADE

with right angle at

D

, legs

AD

and

DE

are equal in length. Given

AD = 12

,

DE

must also be

12

.

2

Determine the length of segment

EC

.

EC = 5

units

Since point

E

lies on side

CD

,

EC = CD - DE = 17 - 12 = 5

.

3

Use the Pythagorean Theorem in right triangle

\triangle BCE

to find

BE

.

BE = 13

units

Because

ABCD

is a rectangle, angle

C

is

90^\circ

and

BC = AD = 12

. Applying the Pythagorean Theorem with legs

BC = 12

and

EC = 5

gives

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

.

Key Concept

Applying properties of isosceles right triangles (

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

) and the Pythagorean Theorem in composite figures.

Question 39Question →

A delivery drone departs from a central launch pad and travels due north for $12$ miles, then turns and travels due east for $16$ miles to reach Drop Point P. A secondary relay station is located $18$ miles due south of the central launch pad. What is the straight-line distance, in miles, from Drop Point P to the secondary relay station?

20

28

30

34

46

Show answer & explanation

Answer: $34$

Answer

The straight-line distance from Drop Point P to the secondary relay station is

34

miles.

To find the straight-line distance from Drop Point P to the secondary relay station, model the displacement as a right triangle. The horizontal distance is

16

miles (east). The vertical distance is the sum of

12

miles north and

18

miles south, giving

30

miles. Using the Pythagorean Theorem,

d = \sqrt{16^2 + 30^2} = \sqrt{256 + 900} = \sqrt{1156} = 34

miles.

Step-by-Step Solution

1

Set up a coordinate grid relative to the central launch pad.

Central launch pad is at

(0, 0)

. Drop Point P is at

(16, 12)

. Secondary relay station is at

(0, -18)

.

Establishing coordinates converts the movement into horizontal and vertical components.

2

Calculate the horizontal and vertical distances between Drop Point P and the secondary relay station.

Horizontal distance =

|16 - 0| = 16

miles. Vertical distance =

|12 - (-18)| = 12 + 18 = 30

miles.

These distances represent the two perpendicular legs of a right triangle.

3

Apply the Pythagorean Theorem to find the hypotenuse.

Distance

= \\sqrt{16^2 + 30^2} = \\sqrt{256 + 900} = \\sqrt{1156} = 34

miles.

The straight-line distance is the hypotenuse of the right triangle formed by the horizontal and vertical legs.

Key Concept

Pythagorean Theorem for distance in perpendicular directions

Question 40Question →

In isosceles trapezoid $ABCD$ , the shorter base $AB$ measures $7$ units and the longer base $CD$ measures $17$ units. The congruent legs $AD$ and $BC$ each form a $45^\circ$ angle with base $CD$ . What is the length of diagonal $AC$ ?

13

17

5\sqrt{2}

\sqrt{119}

169

Show answer & explanation

Answer: $13$

Answer

13 units

Dropping altitude

AP

perpendicular to base

CD

divides base

CD

into

DP = 5

units and

PC = 12

units. Since triangle

APD

is a

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

right triangle, height

AP = DP = 5

. Right triangle

APC

has legs

AP = 5

and

PC = 12

, making hypotenuse

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

.

Step-by-Step Solution

1

Find the length of the base projection segment for the isosceles trapezoid.

Segment

DP = 5

units.

Draw altitude

AP

perpendicular to

CD

. Because trapezoid

ABCD

is isosceles, the projection

DP = \frac{CD - AB}{2} = \frac{17 - 7}{2} = 5

.

2

Determine the altitude of the trapezoid using special right triangle properties.

Altitude

AP = 5

units.

Triangle

APD

is a

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

right triangle, so its legs are congruent (

AP = DP = 5

).

3

Calculate the length of the remaining base segment in right triangle

APC

.

Segment

PC = 12

units.

Segment

PC = CD - DP = 17 - 5 = 12

.

4

Apply the Pythagorean Theorem to right triangle

APC

to solve for diagonal

AC

.

Diagonal

AC = 13

units.

AC = \sqrt{AP^2 + PC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

.

Key Concept

Pythagorean Theorem and Special Right Triangles

Estimated Time:1m 15s

Pythagorean Theorem and Special Right Triangles

Answer

Step-by-Step Solution

Key Concept

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