Pythagorean Theorem and Special Right Triangles Practice Questions — ACT — Page 3

Question 41Question →

In quadrilateral $ABCD$ , $\angle B = 90^\circ$ and $\angle D = 90^\circ$ . If $AB = 12$ units, $BC = 16$ units, and $AD = 10$ units, what is the length, in units, of side $CD$ ?

10\sqrt{3}

10\sqrt{5}

6\sqrt{19}

18

10

Show answer & explanation

Answer: $10\sqrt{3}$

Answer

10\sqrt{3}

units

Connecting vertex

A

to vertex

C

creates two right-angled triangles sharing hypotenuse

AC

. In right triangle

ABC

, the Pythagorean theorem yields

AC^2 = 12^2 + 16^2 = 400

, so

AC = 20

. Next, in right triangle

ADC

,

AC

serves as the hypotenuse and

AD = 10

is one leg. Solving for leg

CD

gives

CD^2 = AC^2 - AD^2 = 20^2 - 10^2 = 300

, which simplifies to

CD = \sqrt{300} = 10\sqrt{3}

.

Step-by-Step Solution

1

Draw diagonal

AC

to divide quadrilateral

ABCD

into two right triangles,

\triangle ABC

and

\triangle ADC

, sharing hypotenuse

AC

.

Two right-angled triangles

\triangle ABC

(with right angle at

B

) and

\triangle ADC

(with right angle at

D

).

Diagonal

AC

acts as the hypotenuse for both right triangles.

2

Apply the Pythagorean Theorem to right triangle

\triangle ABC

to calculate the length of hypotenuse

AC

.

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

units.

In right triangle

\triangle ABC

,

AB

and

BC

are legs.

3

Apply the Pythagorean Theorem to right triangle

\triangle ADC

to solve for leg

CD

.

CD = \sqrt{AC^2 - AD^2} = \sqrt{20^2 - 10^2} = \sqrt{400 - 100} = \sqrt{300} = 10\sqrt{3}

units.

In right triangle

\triangle ADC

,

AC

is the hypotenuse (

20

) and

AD

is a leg (

10

).

Key Concept

Multi-step applications of the Pythagorean Theorem using shared boundary hypotenuses.

Estimated Time:1m 15s

Question 42Question →

In right triangle $ABC$ , the right angle is located at vertex $B$ . Point $D$ lies on leg $BC$ such that $AB = 12$ inches and $BD = 9$ inches. If segment $AD$ is equal in length to segment $DC$ , what is the length of leg $BC$ , in inches?

Show answer & explanation

Answer: 24

Answer

The length of leg

BC

is 24 inches.

Applying the Pythagorean theorem to right triangle

ABD

gives

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

inches. Because segment

AD

equals segment

DC

,

DC

is also 15 inches. Adding the lengths of segments

BD

and

DC

gives

BC = 9 + 15 = 24

inches.

Step-by-Step Solution

1

Calculate the length of hypotenuse

AD

in right triangle

ABD

AD = 15

inches

Apply the Pythagorean theorem:

AD = \sqrt{AB^2 + BD^2} = \sqrt{12^2 + 9^2} = 15

.

2

Determine the length of segment

DC

DC = 15

inches

It is given that segment

AD

is equal in length to segment

DC

.

3

Calculate the total length of leg

BC

BC = 24

inches

Add the adjacent segment lengths along leg

BC

:

BC = BD + DC = 9 + 15 = 24

.

Key Concept

Applying the Pythagorean theorem to adjacent right triangles within geometric figures.

Question 43Question →

A park planner is designing a triangular walking path $ABC$ where the corner at vertex $B$ forms a $90^\circ$ angle. A straight path $AD$ is constructed from vertex $A$ to point $D$ on side $BC$ , dividing angle $\angle BAC$ into two equal angles measuring $30^\circ$ each. If side $AB = 18$ meters, what is the length, in meters, of segment $DC$ ?

6\sqrt{3}

9\sqrt{3}

12\sqrt{3}

18\sqrt{3}

36

Show answer & explanation

Answer: $12\sqrt{3}$

Answer

The length of segment

DC

is

12\sqrt{3}

meters.

In right triangle

ABD

,

\angle BAD = 30^\circ

and

AB = 18

, so

BD = \frac{18}{\sqrt{3}} = 6\sqrt{3}

meters. In right triangle

ABC

,

\angle BAC = 60^\circ

, so

BC = 18\sqrt{3}

meters. Subtracting

BD

from

BC

gives

DC = 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3}

meters.

Step-by-Step Solution

1

Determine the angles in right triangle

ABD

and right triangle

ABC

.

In

\triangle ABD

,

\angle B = 90^\circ

and

\angle BAD = 30^\circ

. In

\triangle ABC

,

\angle B = 90^\circ

and

\angle BAC = 30^\circ + 30^\circ = 60^\circ

.

Path

AD

bisects

\angle BAC

into two

30^\circ

angles.

2

Calculate the length of segment

BD

using the

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

right triangle ratio in

\triangle ABD

.

BD = \frac{AB}{\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}

meters.

In a

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

triangle, the leg opposite the

30^\circ

angle is equal to the adjacent leg divided by

\sqrt{3}

.

3

Calculate the total length of leg

BC

using the

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

right triangle ratio in

\triangle ABC

.

BC = AB \cdot \sqrt{3} = 18\sqrt{3}

meters.

In

\triangle ABC

, the leg opposite the

60^\circ

angle (

BC

) is

\sqrt{3}

times the adjacent leg (

AB = 18

).

4

Subtract segment

BD

from total leg

BC

to find segment

DC

.

DC = BC - BD = 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3}

meters.

Segment addition postulate state that

BD + DC = BC

.

Key Concept

Properties of

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

Special Right Triangles

Estimated Time:1m 30s

Question 44Question →

A rhombus-shaped garden plot $ABCD$ has side lengths of $15$ feet each. The length of the shorter diagonal, $AC$ , is $18$ feet. A gardener places a straight divider line along the longer diagonal, $BD$ . What is the length, in feet, of the divider line along diagonal $BD$ ?

Show answer & explanation

Answer: 24

Answer

The length of the divider line along diagonal

BD

is 24 feet.

The diagonals of a rhombus are perpendicular bisectors of each other. The point of intersection

E

creates right triangle

AEB

, where the hypotenuse is rhombus side

AB = 15

feet and one leg is

AE = \frac{18}{2} = 9

feet. Applying the Pythagorean Theorem yields

9^2 + BE^2 = 15^2

, which simplifies to

81 + BE^2 = 225

, giving

BE = 12

feet. Doubling

BE

gives the complete length of diagonal

BD = 24

feet.

Step-by-Step Solution

1

Identify geometric properties of a rhombus regarding its diagonals.

The diagonals of rhombus

ABCD

are perpendicular to each other and bisect each other at intersection point

E

.

In any rhombus, the diagonals act as perpendicular bisectors, forming four right triangles.

2

Calculate the leg length

AE

in right triangle

AEB

.

AE = \frac{18}{2} = 9

feet.

Point

E

is the midpoint of diagonal

AC

.

3

Use the Pythagorean Theorem to calculate leg length

BE

.

9^2 + BE^2 = 15^2 \implies 81 + BE^2 = 225 \implies BE^2 = 144 \implies BE = 12

feet.

In right triangle

AEB

, side

AB = 15

is the hypotenuse, and

AE = 9

is one leg.

4

Find the total length of diagonal

BD

.

BD = 2 \times BE = 2 \times 12 = 24

feet.

Point

E

bisects diagonal

BD

, so

BD

is twice the length of

BE

.

Key Concept

Using the Pythagorean Theorem on right triangles formed by the perpendicular bisecting diagonals of a rhombus.

Question 45Question →

A rectangular billboard frame $ABCD$ has a length $AB = 16$ feet and width $BC = 12$ feet. A straight diagonal support brace connects vertex $A$ to vertex $C$ . To add structural stability, a secondary beam is installed perpendicular to diagonal $AC$ , extending from vertex $B$ to meet $AC$ at point $P$ . What is the length, in feet, of segment $BP$ ?

9.6

10.0

14.0

19.2

20.0

Show answer & explanation

Answer: 9.6

Answer

9.6 feet

The hypotenuse

AC

of right triangle

ABC

equals

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

feet. Since the area of triangle

ABC

can be calculated either as

\frac{1}{2} \times 16 \times 12 = 96

or as

\frac{1}{2} \times 20 \times BP

, solving

10 \times BP = 96

gives

BP = 9.6

feet.

Step-by-Step Solution

1

Calculate the length of diagonal

AC

using the Pythagorean theorem.

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

feet.

Triangle

ABC

is a right triangle with right angle at

B

and hypotenuse

AC

.

2

Express the area of triangle

ABC

using the two legs.

\text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 16 \times 12 = 96

square feet.

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

3

Express the area using hypotenuse

AC

as the base and

BP

as the height, then solve for

BP

.

\text{Area} = \frac{1}{2} \times AC \times BP \implies 96 = \frac{1}{2} \times 20 \times BP \implies 10 \times BP = 96 \implies BP = 9.6

feet.

Segment

BP

is given as perpendicular to base

AC

.

Key Concept

Altitude to the Hypotenuse in a Right Triangle

Estimated Time:1m 15s

Question 46Question →

In right triangle $ABC$ , the right angle is located at vertex $B$ , and the measure of angle $A$ is $30^\circ$ . The hypotenuse $AC$ has a length of 16 inches. Segment $BD$ is an altitude drawn from vertex $B$ perpendicular to hypotenuse $AC$ at point $D$ . What is the length, in inches, of segment $CD$ ?

Show answer & explanation

Answer: 4

Answer

The length of segment CD is 4 inches.

In right triangle ABC with angle A = 30°, the side opposite angle A (BC) is half the hypotenuse AC, so BC = 16 / 2 = 8 inches. Drawing altitude BD creates smaller right triangle BCD with right angle at D and angle C = 60°. This makes triangle BCD another 30°-60°-90° right triangle where segment BC = 8 inches is the hypotenuse. Segment CD lies opposite the 30° angle DBC, meaning CD is half of BC: 8 / 2 = 4 inches.

Step-by-Step Solution

1

Determine the length of leg BC in right triangle ABC

BC = 8 inches

In a 30°-60°-90° triangle, the length of the side opposite the 30° angle is equal to half the length of the hypotenuse. Since hypotenuse AC = 16 inches, BC = 16 / 2 = 8 inches.

2

Identify the angles of right triangle BCD

Angle C = 60°, Angle BDC = 90°, and Angle DBC = 30°

Since angle A = 30° in right triangle ABC, angle C must equal 90° - 30° = 60°. Altitude BD creates right angle BDC = 90°, leaving angle DBC = 180° - 90° - 60° = 30°.

3

Calculate the length of segment CD in 30°-60°-90° triangle BCD

CD = 4 inches

In triangle BCD, segment BC (8 inches) is the hypotenuse. Segment CD lies opposite the 30° angle DBC, so its length is half of the hypotenuse BC: 8 / 2 = 4 inches.

Key Concept

Altitude to Hypotenuse in Special 30°-60°-90° Right Triangles

Question 47Question →

In right triangle $XYZ$ , $\angle Y = 90^\circ$ and $\angle X = 45^\circ$ . The hypotenuse $XZ$ has a length of $12\sqrt{2}$ units. Point $W$ lies on leg $XY$ such that $XW = 7$ units. What is the length, in units, of segment $ZW$ ?

13

\sqrt{119}

17

\sqrt{193}

\sqrt{313}

Show answer & explanation

Answer: $13$

Answer

The length of segment

ZW

is

13

units.

Because

\triangle XYZ

is an isosceles right triangle (

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

), its leg lengths

XY

and

YZ

are equal to the hypotenuse divided by

\sqrt{2}

, giving

12

. Segment

YW

is

12 - 7 = 5

. In right triangle

\triangle ZYW

, the hypotenuse

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

.

Step-by-Step Solution

1

Determine the leg lengths of

\triangle XYZ

using special right triangle properties.

XY = YZ = 12

In a

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

right triangle, the hypotenuse is

\text{leg} \cdot \sqrt{2}

. Given

XZ = 12\sqrt{2}

, each leg length is

12

.

2

Calculate the length of segment

YW

.

YW = 5

Since point

W

lies on leg

XY

and

XW = 7

,

YW = XY - XW = 12 - 7 = 5

.

3

Apply the Pythagorean Theorem to right triangle

\triangle ZYW

to solve for hypotenuse

ZW

.

ZW = 13

ZW = \sqrt{YZ^2 + YW^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13

.

Key Concept

Properties of

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

special right triangles and multi-step applications of the Pythagorean Theorem.

Estimated Time:1m 0s

Question 48Question →

In right triangle $PQR$ , the measure of $\angle PQR$ is $90^\circ$ . Segment $QS$ is an altitude drawn perpendicular to hypotenuse $PR$ with point $S$ lying on $PR$ . If $PQ = 15$ centimeters and $PS = 9$ centimeters, what is the length, in centimeters, of hypotenuse $PR$ ?

Show answer & explanation

Answer: 25

Answer

The length of hypotenuse

PR

is 25 centimeters.

Using the leg-hypotenuse geometric mean theorem for right triangles (

PQ^2 = PS \cdot PR

), substituting

PQ = 15

and

PS = 9

yields

15^2 = 9 \cdot PR \implies 225 = 9 \cdot PR

, solving directly to

PR = 25

cm.

Step-by-Step Solution

1

Calculate the length of altitude

QS

using the Pythagorean theorem on right triangle

\triangle PQS

QS = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12

cm

Altitude

QS

forms right angle

\angle PSQ = 90^\circ

, making

\triangle PQS

a right triangle.

2

Determine the length of hypotenuse segment

SR

using the geometric mean relationship

QS^2 = PS \cdot SR

12^2 = 9 \cdot SR \implies 144 = 9 \cdot SR \implies SR = 16

cm

The altitude to the hypotenuse divides the original right triangle into two smaller similar right triangles.

3

Sum the segment lengths

PS

and

SR

to find the total length of hypotenuse

PR

PR = PS + SR = 9 + 16 = 25

cm

Point

S

lies directly on segment

PR

between endpoints

P

and

R

.

Key Concept

Geometric Mean Theorem and Pythagorean Theorem in Right Triangles

Estimated Time:1m 30s

Question 49Question →

A mountain zipline course consists of two connected ascending sections. The first section starts at base station $A$ and rises to intermediate platform $B$ at an angle of $30^\circ$ relative to the horizontal ground, covering a horizontal distance of $40\sqrt{3}$ meters. The second section rises from platform $B$ to peak platform $C$ at an angle of $45^\circ$ relative to the horizontal, covering a horizontal distance of $30$ meters. What is the total vertical height, in meters, of peak platform $C$ above base station $A$ ?

40

70

150

20\sqrt{3} + 30

40\sqrt{3} + 30

Show answer & explanation

Answer: $70$

Answer

The total vertical height of peak platform C above base station A is 70 meters.

To find the total vertical height of peak platform C above base station A, determine the vertical rise of each section individually using special right triangle rules and sum them together. For the first section, the

30^\circ

incline forms a

30^\circ

-

60^\circ

-

90^\circ

right triangle where the horizontal distance of

40\sqrt{3}

meters is adjacent to the

30^\circ

angle. The vertical rise is opposite the

30^\circ

angle, so dividing

40\sqrt{3}

by

\sqrt{3}

gives a vertical rise of

40

meters. For the second section, the

45^\circ

incline forms a

45^\circ

-

45^\circ

-

90^\circ

right triangle where the vertical leg equals the horizontal leg, giving a vertical rise of

30

meters. Adding these two vertical heights yields

40 + 30 = 70

meters.

Step-by-Step Solution

1

Calculate the vertical rise of the first section using

30^\circ

-

60^\circ

-

90^\circ

special right triangle properties.

The vertical height of the first section is

40

meters.

In a

30^\circ

-

60^\circ

-

90^\circ

triangle, the ratio of the side opposite the

30^\circ

angle (vertical rise) to the side opposite the

60^\circ

angle (horizontal distance) is

1 : \sqrt{3}

. Dividing the horizontal distance

40\sqrt{3}

by

\sqrt{3}

yields

40

meters.

2

Calculate the vertical rise of the second section using

45^\circ

-

45^\circ

-

90^\circ

special right triangle properties.

The vertical height of the second section is

30

meters.

In an isosceles right triangle (

45^\circ

-

45^\circ

-

90^\circ

), the two legs are equal in length. Since the horizontal distance is

30

meters, the vertical rise is also

30

meters.

3

Sum the vertical rises from both sections to find the total vertical height.

Total height =

40 + 30 = 70

meters.

The total vertical elevation of point C above point A is the sum of the vertical changes along each segment of the path.

Key Concept

Special Right Triangle Side Ratios (

30^\circ

-

60^\circ

-

90^\circ

and

45^\circ

-

45^\circ

-

90^\circ

)

Estimated Time:1m 30s

Question 50Question →

A drone begins at point $P$ and flies directly north for $24$ meters to point $Q$ . It then turns directly east and flies for $10$ meters to point $R$ . From point $R$ , the drone flies along a straight path directed $45^\circ$ south of east until it reaches point $S$ , which lies directly east of starting point $P$ . What is the distance, in meters, from point $P$ to point $S$ ?

Show answer & explanation

Answer: 34

Answer

The total distance from point P to point S is 34 meters.

Flying north 24 meters places point R at a height of 24 meters above the horizontal line passing east through P. Returning to this horizontal line at point S along a path 45° south of east forms a 45°-45°-90° right triangle with a vertical leg of 24 meters. Because the legs of a 45°-45°-90° right triangle are congruent, the horizontal leg is also 24 meters. Combining this with the initial 10 meters of eastward travel gives a total distance from P to S of 10 + 24 = 34 meters.

Step-by-Step Solution

1

Determine the vertical and horizontal position of point R relative to point P.

Point R is located 24 meters north and 10 meters east of point P.

The flight 24 meters north sets the vertical distance to 24 meters, and the flight 10 meters east sets the horizontal displacement to 10 meters.

2

Apply the properties of a 45°-45°-90° special right triangle to find the horizontal distance from point R to point S.

The horizontal distance traveled between point R and point S is 24 meters.

Since point S lies directly east of point P (at vertical height 0), the vertical drop from point R to point S is 24 meters. A line angled 45° south of east creates a 45°-45°-90° right triangle whose two leg lengths are equal, so the horizontal leg length equals the vertical leg length of 24 meters.

3

Calculate the total horizontal distance from point P to point S.

34 meters

Sum the initial east displacement of 10 meters from P to R with the additional horizontal displacement of 24 meters from R to S: 10 + 24 = 34 meters.

Key Concept

45°-45°-90° Special Right Triangle Properties and Planar Displacement

Estimated Time:1m 30s

Question 51Question →

In the figure below, quadrilateral $ABCD$ is composed of two adjacent right triangles, $\triangle ABD$ and $\triangle BCD$ , sharing side $BD$ . In $\triangle ABD$ , the right angle is at vertex $A$ , and the measure of $\angle ABD$ is $60^\circ$ . In $\triangle BCD$ , the right angle is at vertex $C$ , and the measure of $\angle BDC$ is $45^\circ$ . If the length of segment $AB$ is $6$ inches, what is the length, in inches, of segment $BC$ ?

6\sqrt{2}

6\sqrt{3}

12

12\sqrt{2}

4\sqrt{3}

Show answer & explanation

Answer: $6\sqrt{2}$

Answer

The length of segment

BC

is

6\sqrt{2}

inches.

In the

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

triangle

ABD

, short leg

AB = 6

gives hypotenuse

BD = 12

. In the

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

triangle

BCD

, hypotenuse

BD = 12

gives leg

BC = \frac{12}{\sqrt{2}} = 6\sqrt{2}

inches.

Step-by-Step Solution

1

Analyze

\triangle ABD

using

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

special right triangle ratios.

Since

\angle A = 90^\circ

and

\angle ABD = 60^\circ

,

\angle ADB = 30^\circ

. The side opposite

30^\circ

is

AB = 6

. Therefore, the hypotenuse

BD = 2 \times AB = 2(6) = 12

.

In a

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

triangle, the hypotenuse is twice the shorter leg.

2

Analyze

\triangle BCD

using

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

special right triangle ratios.

Triangle

BCD

is an isosceles right triangle with right angle at

C

and hypotenuse

BD = 12

. The legs are equal, so

BC = CD

.

In a

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

triangle, hypotenuse =

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

.

3

Solve for leg

BC

and rationalize the denominator.

BC = \frac{BD}{\sqrt{2}} = \frac{12}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2}

.

Dividing the hypotenuse by

\sqrt{2}

yields the leg length in standard simplified radical form.

Key Concept

Applying

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

and

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

special right triangle side ratio rules across multi-step figures.

Estimated Time:1m 15s

Question 52Question →

A regular hexagon $ABCDEF$ has a perpendicular distance of $12\sqrt{3}$ inches between its two parallel opposite sides. What is the perimeter, in inches, of hexagon $ABCDEF$ ?

Show answer & explanation

Answer: 72

Answer

The perimeter of the regular hexagon is 72 inches.

A regular hexagon with side length

s

can be partitioned from its center into 6 congruent equilateral triangles of side length

s

. Dropping an altitude from the center to any side creates two

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

right triangles. In each right triangle, the side opposite the

60^\circ

angle (the altitude) has length

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

. The total perpendicular distance between two parallel opposite sides of the hexagon equals twice this altitude,

s\sqrt{3}

. Setting

s\sqrt{3} = 12\sqrt{3}

gives

s = 12

inches. The perimeter of the regular hexagon is

6 \times 12 = 72

inches.

Step-by-Step Solution

1

Express the perpendicular distance between parallel opposite sides of a regular hexagon in terms of its side length

s

.

The distance between opposite sides is

s\sqrt{3}

.

A regular hexagon with side length

s

consists of 6 congruent equilateral triangles. The altitude of each equilateral triangle divides it into two

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

right triangles with legs

s/2

and

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

. The distance between opposite parallel sides spans two altitudes, which equals

2 \times \frac{s\sqrt{3}}{2} = s\sqrt{3}

.

2

Solve for the side length

s

.

s = 12

inches.

Equating the given distance

12\sqrt{3}

to

s\sqrt{3}

yields

s = 12

.

3

Calculate the total perimeter of the hexagon.

Perimeter

= 72

inches.

A regular hexagon has 6 equal sides, so its perimeter is

6 \times 12 = 72

inches.

Key Concept

Applying

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

special right triangle relationships to regular polygons

Question 53Question →

A vertical flagpole stands perpendicular to level ground. Two support cables are attached from the top of the flagpole to ground anchors located on opposite sides of the pole. The first cable makes a $60^\circ$ angle with the ground and its ground anchor is $10\sqrt{3}$ feet from the base of the pole. The second cable makes a $45^\circ$ angle with the ground. What is the sum of the lengths, in feet, of the two support cables?

20\sqrt{3} + 30\sqrt{2}

10\sqrt{3} + 30\sqrt{2}

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

20\sqrt{3} + 30

50\sqrt{5}

Show answer & explanation

Answer: $20\sqrt{3} + 30\sqrt{2}$

Answer

20\sqrt{3} + 30\sqrt{2}

feet

The first support cable forms a 30°-60°-90° right triangle with the flagpole and ground. Given the adjacent leg to the 60° angle is

10\sqrt{3}

feet, the opposite leg (the flagpole's height) is

10\sqrt{3} \times \sqrt{3} = 30

feet, and the hypotenuse (first cable) is

2 \times 10\sqrt{3} = 20\sqrt{3}

feet. The second cable forms a 45°-45°-90° right triangle sharing the 30-foot height as one leg. Thus, the hypotenuse (second cable) is

30\sqrt{2}

feet. Adding both cable lengths gives

20\sqrt{3} + 30\sqrt{2}

feet.

Step-by-Step Solution

1

Analyze the first right triangle formed by the pole and the first cable

The triangle is a

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

triangle with the ground angle equal to

60^\circ

. The side adjacent to

60^\circ

(shorter leg) is

10\sqrt{3}

ft.

The cable makes a

60^\circ

angle with the horizontal ground, making the angle at the top of the pole

30^\circ

.

2

Calculate the height of the flagpole and the length of the first cable

Flagpole height

= 10\sqrt{3} \cdot \sqrt{3} = 30

ft. First cable length (hypotenuse)

= 2 \cdot 10\sqrt{3} = 20\sqrt{3}

ft.

In a

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

triangle with shorter leg

x

, the longer leg is

x\sqrt{3}

and the hypotenuse is

2x

.

3

Calculate the length of the second cable using the second right triangle

Second cable length (hypotenuse)

= 30\sqrt{2}

ft.

The second triangle is a

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

right triangle with leg equal to the pole height (

30

ft). The hypotenuse is

\text{leg} \cdot \sqrt{2}

.

4

Add the lengths of the two support cables

Total length

= 20\sqrt{3} + 30\sqrt{2}

ft.

Summing the two hypotenuse lengths gives the total combined cable length.

Key Concept

Properties of

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

and

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

special right triangles

Estimated Time:1m 30s

Question 54Question →

In right triangle $ABC$ , the hypotenuse $AC$ has a length of 13 centimeters, and leg $AB$ has a length of 5 centimeters. What is the length, in centimeters, of leg $BC$ ?

Show answer & explanation

Answer: 12

Answer

The length of leg

BC

is 12 centimeters.

Applying the Pythagorean theorem, we have

5^2 + BC^2 = 13^2

, which simplifies to

25 + BC^2 = 169

. Subtracting 25 from both sides gives

BC^2 = 144

, and taking the square root of both sides gives

BC = 12

centimeters.

Step-by-Step Solution

1

Set up the Pythagorean Theorem equation for right triangle

ABC

.

AB^2 + BC^2 = AC^2

The Pythagorean Theorem states that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse.

2

Substitute the given values for

AB

and

AC

into the formula.

5^2 + BC^2 = 13^2

The length of leg

AB

is given as 5 centimeters, and the length of the hypotenuse

AC

is given as 13 centimeters.

3

Solve for the unknown leg length

BC

.

BC = 12

Squaring the values gives

25 + BC^2 = 169

. Subtracting 25 from both sides yields

BC^2 = 144

. Taking the square root of both sides gives

BC = 12

.

Key Concept

Pythagorean Theorem

Alternative Method

Recognize the triangle as a standard 5-12-13 Pythagorean triple, which immediately gives the missing leg length of 12 without needing calculations.

Estimated Time:30s

Question 55Question →

A surveyor is mapping a triangular park. Starting at point $A$ , she walks due east for $80$ meters to point $B$ . She then turns $120^\circ$ to her left and walks in a straight line to point $C$ , which is located directly north of point $A$ . What is the straight-line distance, in meters, from point $B$ to point $C$ ?

40

80\sqrt{2}

80\sqrt{3}

160

200

Show answer & explanation

Answer: $160$

Answer

The straight-line distance from point

B

to point

C

is

160

meters.

The surveyor's movement forms a

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

right triangle where the side opposite the

30^\circ

angle is

AB = 80

meters. The hypotenuse

BC

represents the distance from

B

to

C

. In a

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

triangle, the hypotenuse is exactly twice the length of the side opposite the

30^\circ

angle. Thus, the distance is

2 \times 80 = 160

meters.

Step-by-Step Solution

1

Determine the orientation and angles of the path.

A right triangle

ABC

with a right angle at vertex

A

.

Since the path from

A

to

B

goes due east, and

C

is directly north of

A

, the angle

\angle A

is exactly

90^\circ

.

2

Calculate the interior angle at vertex

B

.

\angle B = 60^\circ

and

\angle C = 30^\circ

.

The surveyor turns

120^\circ

to the left from the extension of the eastward segment

AB

. The interior angle is the supplement:

180^\circ - 120^\circ = 60^\circ

. The sum of angles in a triangle is

180^\circ

, so the angle at

C

is

180^\circ - (90^\circ + 60^\circ) = 30^\circ

.

3

Use special right triangle ratios to find the hypotenuse.

The distance

BC = 160

meters.

In a

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

triangle, the sides are in the ratio

1 : \sqrt{3} : 2

. The side opposite the

30^\circ

angle is

AB = 80

meters. The hypotenuse

BC

is twice the length of this side:

2 \times 80 = 160

meters.

Key Concept

Ratios of a

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

special right triangle

Estimated Time:1m 0s

Pythagorean Theorem and Special Right Triangles

Answer

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Key Concept

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Key Concept

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