All practice questions

Question 461Question →

In rectangle $ABCD$ , the length of side $AB$ is $16$ centimeters and the length of side $BC$ is $12$ centimeters. Point $P$ lies on diagonal $AC$ such that segment $DP$ is perpendicular to $AC$ . What is the length, in centimeters, of segment $DP$ ?

Show answer & explanation

Answer: 9.6

Answer

The length of segment

DP

is

9.6

centimeters.

In rectangle

ABCD

, opposite sides are equal (

AD = BC = 12

cm and

DC = AB = 16

cm) and all interior angles are

90^\circ

. Right triangle

ADC

has legs

12

cm and

16

cm, making hypotenuse

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

cm. Calculating the area of triangle

ADC

using the legs gives

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

cm². Using hypotenuse

AC

as the base and perpendicular line segment

DP

as the altitude, the area is

\frac{1}{2} \times 20 \times DP = 10 \times DP

. Setting

10 \times DP = 96

yields

DP = 9.6

cm.

Step-by-Step Solution

1

Determine the length of diagonal

AC

Diagonal

AC = 20

cm

Since

ABCD

is a rectangle, angle

ADC

is a right angle with legs

AD = 12

cm and

DC = 16

cm. By the Pythagorean theorem,

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

cm.

2

Calculate the area of right triangle

ADC

Area of triangle

ADC = 96

cm²

The area of a right triangle equals half the product of its legs:

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

cm².

3

Solve for altitude

DP

Length

DP = 9.6

cm

The area can also be expressed using hypotenuse

AC

as the base and

DP

as the altitude:

\text{Area} = \frac{1}{2} \times AC \times DP = \frac{1}{2} \times 20 \times DP = 10 \times DP

. Equating the two area expressions yields

10 \times DP = 96

, giving

DP = 9.6

cm.

Key Concept

Properties of Rectangles and Altitudes in Right Triangles

Question 462Question →

In triangular plot $ABC$ , the boundary lengths are $AB = 13$ meters, $BC = 8$ meters, and $AC = 15$ meters. A straight drainage pipe is laid from vertex $B$ perpendicular to side $AC$ , meeting side $AC$ at point $D$ . What is the distance, in meters, from point $A$ to point $D$ ?

Show answer & explanation

Answer: 11

Answer

The distance from point A to point D is 11 meters.

Applying the Law of Cosines

a^2 = b^2 + c^2 - 2bc \cos A

with

a=8

,

b=15

, and

c=13

yields

64 = 225 + 169 - 390 \cos A

, which simplifies to

390 \cos A = 330

, giving

\cos A = \frac{11}{13}

. In right triangle

ABD

,

\cos A = \frac{AD}{AB}

, so

AD = 13 \cdot \frac{11}{13} = 11

meters.

Step-by-Step Solution

1

Apply the Law of Cosines to triangle ABC to solve for the cosine of angle A.

cos A = 11/13

The Law of Cosines relates all three side lengths of a triangle to the cosine of one of its interior angles.

2

Use right triangle trigonometry in right triangle ABD to calculate the length of AD.

AD = 11 meters

Since BD is perpendicular to AC, triangle ABD is a right triangle with hypotenuse AB and adjacent side AD relative to angle A.

Key Concept

Law of Cosines and Right Triangle Trigonometry

Estimated Time:2m 0s

Question 463Question →

For an angle $\theta$ in the third quadrant satisfying $\pi < \theta < \frac{3\pi}{2}$ , the tangent value is $\tan \theta = \frac{4}{3}$ . What is the exact value of the expression $\sin^4 \theta - \cos^4 \theta$ ?

Show answer & explanation

Answer: 0.28

Answer

The exact numerical value of the expression is 0.28 (or 7/25).

By factoring

\sin^4 \theta - \cos^4 \theta

as

(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)

, we can apply the fundamental Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

. The expression simplifies cleanly to

\sin^2 \theta - \cos^2 \theta

. Given

\tan \theta = \frac{4}{3}

in Quadrant III, the reference triangle has opposite side 4, adjacent side 3, and hypotenuse 5. Thus,

\sin \theta = -\frac{4}{5}

and

\cos \theta = -\frac{3}{5}

. Substituting these values yields

\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28

.

Step-by-Step Solution

1

Factor the fourth-degree trigonometric expression using difference of squares.

\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)

The difference of two squares

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

applies directly to

a = \sin^2 \theta

and

b = \cos^2 \theta

.

2

Simplify using the fundamental Pythagorean trigonometric identity.

\sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta

By the Pythagorean identity,

\sin^2 \theta + \cos^2 \theta = 1

for any angle

\theta

.

3

Determine

\sin \theta

and

\cos \theta

from the given quadrant and tangent value.

\sin \theta = -\frac{4}{5}

and

\cos \theta = -\frac{3}{5}

In Quadrant III (

\pi < \theta < \frac{3\pi}{2}

), both sine and cosine are negative. A standard 3-4-5 right triangle yields

\sin \theta = -\frac{4}{5}

and

\cos \theta = -\frac{3}{5}

.

4

Substitute the values into the simplified expression and compute the result.

\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28

Squaring each trigonometric ratio yields positive values, giving a final simplified decimal value of 0.28.

Key Concept

Pythagorean Identity and Difference of Squares

Estimated Time:1m 30s

Question 464Question →

In kite $ABCD$ , diagonals $AC$ and $BD$ intersect perpendicularly at point $P$ . If $AP = 9$ centimeters, $PC = 16$ centimeters, and $BP = PD = 12$ centimeters, what is the perimeter, in centimeters, of kite $ABCD$ ?

Show answer & explanation

Answer: 70

Answer

The perimeter of kite

ABCD

is 70 centimeters.

The diagonals of a kite intersect at right angles (

90^\circ

). Applying the Pythagorean theorem to right triangle

APB

with legs

9

cm and

12

cm gives hypotenuse

AB = 15

cm. Applying the Pythagorean theorem to right triangle

BPC

with legs

16

cm and

12

cm gives hypotenuse

BC = 20

cm. Since a kite has two pairs of equal adjacent sides (

AB = AD = 15

cm and

BC = CD = 20

cm), the total perimeter is

15 + 15 + 20 + 20 = 70

cm.

Step-by-Step Solution

1

Identify right triangles formed by the perpendicular diagonals

Four right triangles are formed:

\triangle APB

,

\triangle BPC

,

\triangle CPD

, and

\triangle DPA

.

Diagonals of a kite are perpendicular to each other.

2

Calculate upper side length

AB

AB = \sqrt{9^2 + 12^2} = \sqrt{225} = 15

cm

Apply the Pythagorean theorem

a^2 + b^2 = c^2

to right triangle

APB

.

3

Calculate lower side length

BC

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

cm

Apply the Pythagorean theorem

a^2 + b^2 = c^2

to right triangle

BPC

.

4

Compute the total perimeter

Perimeter

= 2(15) + 2(20) = 30 + 40 = 70

cm

A kite has two pairs of congruent adjacent sides (

AD = AB

and

CD = BC

).

Key Concept

Perpendicular diagonals and side length properties of a kite

Alternative Method

Instead of calculating all four sides individually, calculate one side from each distinct right triangle (

15

cm and

20

cm) and multiply their sum by

2

, using the property that a kite has two symmetric pairs of congruent adjacent sides:

2 \times (15 + 20) = 70

cm.

Estimated Time:1m 15s

Question 465Question →

If $\sin \theta + \cos \theta = \sqrt{1.5}$ , what is the value of $\tan \theta + \cot \theta$ ?

Show answer & explanation

Answer: 4

Answer

4

Squaring both sides of

\sin \theta + \cos \theta = \sqrt{1.5}

yields

\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5

. Using the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

, we get

1 + 2\sin \theta \cos \theta = 1.5

, which simplifies to

\sin \theta \cos \theta = 0.25

. Rewriting

\tan \theta + \cot \theta

using quotient identities gives

\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}

. Substituting

\sin \theta \cos \theta = 0.25

yields

\frac{1}{0.25} = 4

.

Step-by-Step Solution

1

Square both sides of the given equation

\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5

Squaring both sides allows the expansion of the binomial

(\sin \theta + \cos \theta)^2

to reveal the product term

\sin \theta \cos \theta

.

2

Apply the Pythagorean identity to solve for

\sin \theta \cos \theta

\sin \theta \cos \theta = 0.25

Since

\sin^2 \theta + \cos^2 \theta = 1

, substituting

1

into

1 + 2\sin \theta \cos \theta = 1.5

gives

2\sin \theta \cos \theta = 0.5

, so

\sin \theta \cos \theta = 0.25

.

3

Rewrite

\tan \theta + \cot \theta

using quotient identities

\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}

Using

\tan \theta = \frac{\sin \theta}{\cos \theta}

and

\cot \theta = \frac{\cos \theta}{\sin \theta}

, finding a common denominator yields

\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}

. Applying

\sin^2 \theta + \cos^2 \theta = 1

simplifies the numerator to

1

.

4

Substitute the numerical value of

\sin \theta \cos \theta

to find the answer

\tan \theta + \cot \theta = \frac{1}{0.25} = 4

Dividing

1

by

0.25

evaluates to the exact integer

4

.

Key Concept

Fundamental Trigonometric Identities (Pythagorean and Quotient Identities)

Estimated Time:1m 30s

Question 466Question →

What is the exact value of $\cos\left(2\arctan(-3)\right)$ expressed as a decimal?

Show answer & explanation

Answer: -0.8

Answer

The exact decimal value of

\cos\left(2\arctan(-3)\right)

is

-0.8

.

Letting

\theta = \arctan(-3)

gives

\tan(\theta) = -3

. Utilizing the double-angle formula

\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}

, we substitute

\tan(\theta) = -3

to obtain

\frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8

.

Step-by-Step Solution

1

Define an angle variable for the inverse trigonometric expression.

Let

\theta = \arctan(-3)

, which means

\tan(\theta) = -3

in Quadrant IV where

-\frac{\pi}{2} < \theta < 0

.

Applying the standard definition and principal range of the arctangent function.

2

Apply the double-angle identity for cosine expressed in terms of tangent.

\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}

This identity directly connects

\cos(2\theta)

to the known value of

\tan(\theta)

without needing radicals.

3

Substitute

\tan(\theta) = -3

into the identity and evaluate.

\cos(2\theta) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8

Simplifying the numerical expression produces the exact decimal value.

Key Concept

Inverse Trigonometric Functions and Double-Angle Identities

Estimated Time:1m 30s

Question 467Question →

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 468Question →

If $\tan \theta = 2$ , what is the value of $\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta}$ ?

Show answer & explanation

Answer: 1.25

Answer

The value of the expression is 1.25.

Using the trigonometric quotient identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

, dividing both the numerator and denominator of

\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta}

by

\cos \theta

transforms the fraction into

\frac{3\tan \theta + 4}{5\tan \theta - 2}

. Substituting

\tan \theta = 2

gives

\frac{3(2) + 4}{5(2) - 2} = \frac{10}{8} = 1.25

. Alternatively, constructing a right triangle with opposite side 2 and adjacent side 1 gives a hypotenuse of

\sqrt{5}

, yielding

\sin \theta = \frac{2}{\sqrt{5}}

and

\cos \theta = \frac{1}{\sqrt{5}}

, which produces the exact same ratio of

\frac{10/\sqrt{5}}{8/\sqrt{5}} = 1.25

.

Step-by-Step Solution

1

Divide every term in both the numerator and the denominator by

\cos \theta

.

The expression becomes

\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 4\left(\frac{\cos \theta}{\cos \theta}\right)}{5\left(\frac{\sin \theta}{\cos \theta}\right) - 2\left(\frac{\cos \theta}{\cos \theta}\right)}

.

Dividing by

\cos \theta

allows us to convert sine-and-cosine terms into tangent terms using the quotient identity.

2

Apply the quotient identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

.

The expression simplifies to

\frac{3\tan \theta + 4}{5\tan \theta - 2}

.

This reduces the trigonometric expression to an algebraic expression containing only

\tan \theta

.

3

Substitute

\tan \theta = 2

into the simplified expression and compute the result.

\frac{3(2) + 4}{5(2) - 2} = \frac{6 + 4}{10 - 2} = \frac{10}{8} = 1.25.

Performing basic arithmetic yields the exact numeric answer.

Key Concept

Quotient Identity of Tangent

Question 469Question →

A tangent line segment $\overline{PT}$ touches a circle at point $T$ . A secant line from external point $P$ intersects the circle at points $A$ and $B$ , such that point $A$ lies on segment $\overline{PB}$ . If $m\angle P = 35^\circ$ and the measure of minor arc $AT$ is $50^\circ$ , what is the degree measure of inscribed angle $\angle TAB$ ?

Show answer & explanation

Answer: 60

Answer

The degree measure of inscribed angle

\angle TAB

is 60 degrees.

According to the exterior angle theorem for circles, the angle formed by a tangent and a secant meeting at an external point

P

is equal to half the difference of the intercepted arcs:

m\angle P = \frac{1}{2}(m\widehat{BT} - m\widehat{AT})

. Substituting

m\angle P = 35^\circ

and

m\widehat{AT} = 50^\circ

into the equation gives

35^\circ = \frac{1}{2}(m\widehat{BT} - 50^\circ)

, which simplifies to

m\widehat{BT} = 120^\circ

. The inscribed angle

\angle TAB

intercepts arc

BT

. By the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc, giving

m\angle TAB = \frac{1}{2}(120^\circ) = 60^\circ

. Alternatively, inside triangle

PAT

, the tangent-chord angle

PTA

intercepts arc

AT

, so

m\angle PTA = \frac{1}{2}(50^\circ) = 25^\circ

. Since the angles in triangle

PAT

sum to

180^\circ

,

m\angle PAT = 180^\circ - (35^\circ + 25^\circ) = 120^\circ

. Angle

TAB

is supplementary to angle

PAT

, so

m\angle TAB = 180^\circ - 120^\circ = 60^\circ

.

Step-by-Step Solution

1

Use the exterior angle relationship for the secant and tangent to find the measure of arc

BT

.

m\widehat{BT} = 120^\circ

The exterior angle measure equals half the difference of intercepted arcs

BT

and

AT

:

35^\circ = \frac{1}{2}(m\widehat{BT} - 50^\circ)

.

2

Use the Inscribed Angle Theorem to find

m\angle TAB

.

m\angle TAB = 60^\circ

An inscribed angle measure is equal to half the measure of its intercepted arc:

m\angle TAB = \frac{1}{2}(120^\circ) = 60^\circ

.

Key Concept

Secant-Tangent Angle Theorem and Inscribed Angle Theorem

Estimated Time:1m 30s

Question 470Question →

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 471Question →

In parallelogram $ABCD$ , the diagonals $AC$ and $BD$ intersect at point $E$ . If $AE = 2x + 5$ , $EC = 5x - 7$ , $BE = 3y - 1$ , and $ED = y + 9$ , what is the length of diagonal $BD$ ?

Show answer & explanation

Answer: 28

Answer

The length of diagonal

BD

is 28.

Because the diagonals of a parallelogram bisect each other, the intersection point

E

divides diagonal

BD

into two equal segments (

BE = ED

). Equating the expressions gives

3y - 1 = y + 9

, which simplifies to

2y = 10

, so

y = 5

. Substituting

y = 5

into the expression for

BE

yields

BE = 14

. Since

BD

consists of

BE + ED

, the full length of diagonal

BD

is

14 + 14 = 28

.

Step-by-Step Solution

1

Apply the diagonal bisection property of parallelograms.

BE = ED

, so

3y - 1 = y + 9

.

The diagonals of any parallelogram bisect each other at their intersection point.

2

Solve the linear equation for

y

.

2y = 10 \implies y = 5

.

Subtract

y

from both sides and add 1 to both sides.

3

Calculate segment

BE

and total length

BD

.

BE = 3(5) - 1 = 14

, so

BD = 2 \times 14 = 28

.

Substitute

y = 5

into the segment length expression and double it for the full diagonal length.

Key Concept

Diagonals of a parallelogram bisect each other.

Question 472Question →

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 473Question →

In the standard $(x,y)$ coordinate plane, a sinusoidal function $f(x) = A \sin(Bx - C) + D$ reaches a maximum value of $9$ at $x = 1$ and its immediate next minimum value of $-1$ at $x = 4$ . What is the period of $f(x)$ ?

Show answer & explanation

Answer: 6

Answer

The period of the sinusoidal function is 6.

For any sinusoidal graph, the horizontal distance between a maximum point and the immediate next minimum point corresponds to one-half of the period. Given that the maximum occurs at

x = 1

and the next minimum occurs at

x = 4

, the half-period is

4 - 1 = 3

. Multiplying this half-period by

2

gives the full period of

6

.

Step-by-Step Solution

1

Identify the horizontal distance between the consecutive maximum and minimum points.

The horizontal distance is

4 - 1 = 3

.

The maximum occurs at

x = 1

and the consecutive minimum occurs at

x = 4

.

2

Relate the horizontal distance between consecutive extrema to the period of the function.

Half of the period is equal to

3

.

In any sinusoidal function, the horizontal distance between a peak (maximum) and the adjacent trough (minimum) represents exactly one-half of a complete cycle.

3

Calculate the full period of the function.

Period =

2 \times 3 = 6

.

Multiplying the half-period by 2 yields the full period of the function.

Key Concept

The horizontal distance between consecutive maximum and minimum points of a sinusoidal graph is half of the function's period.

Estimated Time:1m 15s

Question 474Question →

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 475Question →

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 476Question →

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 477Question →

A telecommunications company connects two remote cell towers, Tower X and Tower Y, to a central relay station R. The cable path from Relay Station R to Tower X is $7$ kilometers long, and the cable path from Relay Station R to Tower Y is $8$ kilometers long. The angle formed between the two cables at Relay Station R, $\angle XRY$ , measures $120^\circ$ . A straight wireless backup link is established directly between Tower X and Tower Y. What is the distance, in kilometers, of this direct wireless link?

Show answer & explanation

Answer: 13

Answer

The distance of the direct wireless link between Tower X and Tower Y is 13 kilometers.

Using the Law of Cosines with two side lengths of 7 km and 8 km and an included angle of 120° gives

XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ) = 49 + 64 - 112(-0.5) = 169

. Taking the square root gives 13 km.

Step-by-Step Solution

1

Identify known components of the triangle formed by the relay station and towers

Side

RX = 7\text{ km}

, side

RY = 8\text{ km}

, and included angle

\angle R = 120^\circ

Two sides and the included angle (SAS) are known, indicating that the Law of Cosines must be used to find the third side.

2

Apply the Law of Cosines formula for side

XY

XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ)

The Law of Cosines relates three sides of a triangle to the cosine of one of its angles:

c^2 = a^2 + b^2 - 2ab\cos(C)

.

3

Calculate the numeric value of

XY^2

XY^2 = 49 + 64 + 56 = 169

Since

\cos(120^\circ) = -0.5

, the term

-2(56)(-0.5)

evaluates to

+56

.

4

Take the principal square root of

169

XY = 13\text{ km}

Distance must be a positive length.

Key Concept

Law of Cosines (Side-Angle-Side configuration)

Question 478Question →

In rectangle $ABCD$ , diagonals $AC$ and $BD$ intersect at point $E$ . If the measure of $\angle AEB$ is $120^\circ$ and $AC = 16$ , what is the length of side $BC$ ?

Show answer & explanation

Answer: 8

Answer

The length of side

BC

is

8

.

In any rectangle, the diagonals are congruent and bisect each other. Given

AC = 16

, the distance from the intersection point

E

to any vertex is

8

, so

BE = EC = 8

. Because

\angle AEB

and

\angle BEC

lie along the straight diagonal line

AC

, they are supplementary, giving

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

. Triangle

BEC

is an isosceles triangle with

BE = EC = 8

and a vertex angle of

60^\circ

, which forces it to be equilateral. Consequently, all sides of

\triangle BEC

are equal, so

BC = 8

.

Step-by-Step Solution

1

Calculate the lengths of the diagonal segments from the center point

E

.

BE = EC = 8

The diagonals of a rectangle are congruent and bisect each other, so each half-diagonal equals half of

AC

.

2

Determine the measure of adjacent angle

\angle BEC

.

\angle BEC = 60^\circ

Angles

\angle AEB

and

\angle BEC

form a straight line (linear pair), so their sum is

180^\circ

.

3

Analyze

\triangle BEC

to find the length of side

BC

.

BC = 8

An isosceles triangle with a

60^\circ

vertex angle has base angles of

60^\circ

as well, making it an equilateral triangle where all sides are equal to

8

.

Key Concept

Diagonals of a rectangle are equal in length and bisect each other, dividing the rectangle into two pairs of congruent isosceles triangles.

Estimated Time:1m 15s

Question 479Question →

A park planner is designing a triangular walking trail connecting three landmarks: a fountain at point $F$ , a gazebo at point $G$ , and a bridge at point $B$ . The distance from the fountain to the gazebo is $800$ meters, and the distance from the gazebo to the bridge is $1{}500$ meters. If the angle formed at the gazebo ( $\angle FGB$ ) measures $60^\circ$ , what is the direct distance, in meters, from the fountain at point $F$ to the bridge at point $B$ ?

Show answer & explanation

Answer: 1300

Answer

1300 meters

Applying the Law of Cosines

c^2 = a^2 + b^2 - 2ab\cos(C)

with

a = 800

,

b = 1500

, and

C = 60^\circ

gives

c^2 = 800^2 + 1500^2 - 2(800)(1500)(0.5) = 1{}690{}000

. Taking the square root yields

c = 1300

meters.

Step-by-Step Solution

1

Identify given measurements and formula

FG = 800

m,

GB = 1500

m,

\angle FGB = 60^\circ

, using Law of Cosines

c^2 = a^2 + b^2 - 2ab\cos(C)

Two side lengths and the included angle (SAS) are known, requiring the Law of Cosines to solve for the opposite side.

2

Substitute values into the Law of Cosines equation

FB^2 = 800^2 + 1500^2 - 2(800)(1500)\cos(60^\circ)

Direct replacement of known side lengths and angle measure into the formula.

3

Evaluate the arithmetic terms

FB^2 = 640{}000 + 2{}250{}000 - 1{}200{}000 = 1{}690{}000

Squaring the side lengths and calculating the product

2(800)(1500)(0.5)

.

4

Take the square root to solve for

FB

FB = \sqrt{1{}690{}000} = 1300

Taking the principal square root yields the distance in meters.

Key Concept

Law of Cosines (SAS Triangle Solving)

Question 480Question →

On a standard number line, point $R$ has coordinate $-18$ and point $T$ has coordinate $22$ . Point $S$ lies on the line segment $RT$ such that the distance from $R$ to $S$ is $\frac{3}{5}$ of the distance from $R$ to $T$ . What is the coordinate of point $S$ ?

Show answer & explanation

Answer: 6

Answer

6

The distance between

R(-18)

and

T(22)

is calculated using absolute value:

|22 - (-18)| = 40

. Taking

\frac{3}{5}

of 40 yields 24 units. Adding 24 to the starting coordinate

-18

gives the coordinate of point

S

as 6.

Step-by-Step Solution

1

Calculate the total distance between points

R

and

T

Distance

= |22 - (-18)| = |40| = 40

units

The distance between two points on a number line is given by the absolute difference of their coordinates.

2

Determine the distance from

R

to

S

Distance

= \frac{3}{5} \times 40 = 24

units

Point

S

is located at a distance equal to

\frac{3}{5}

of the total segment length starting from

R

.

3

Calculate the coordinate of point

S

Coordinate of

S = -18 + 24 = 6

Since

S

lies between

R(-18)

and

T(22)

, moving from

R

toward

T

corresponds to increasing the coordinate value by 24.

Key Concept

Calculating distance between signed integers using absolute value and finding a point dividing a number line segment in a given ratio

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