Plane Geometry Practice Questions — ACT — Page 8

Question 141Question →

Is the statement that a convex quadrilateral with perpendicular and equal-length diagonals must be a square true or false?

True

False

Show answer & explanation

Answer: False

Answer

The statement is false because perpendicular and equal-length diagonals do not guarantee a quadrilateral is a square; they must also bisect each other.

The statement is false because having perpendicular and equal-length diagonals is a necessary condition for a square, but not a sufficient one. A quadrilateral must also have diagonals that bisect each other to be a square.

Step-by-Step Solution

1

Identify the properties of a square's diagonals.

In a square, the diagonals are perpendicular, equal in length, and bisect each other.

To evaluate the statement, we must compare the given diagonal conditions with the complete set of diagonal properties of a square.

2

Analyze whether perpendicularity and equality alone are sufficient to define a square.

Without the bisection property, we cannot guarantee the quadrilateral is a parallelogram, which is a prerequisite for being a square.

A square is a specific type of parallelogram, so any set of sufficient conditions must first satisfy the definition of a parallelogram.

3

Construct a counterexample where the diagonals are perpendicular and equal in length, but do not bisect each other.

Consider a quadrilateral with vertices

A(0, 3)

,

B(2, 0)

,

C(0, -1)

, and

D(-2, 0)

on a standard coordinate plane. The diagonal

AC

has a length of

4

along the

y

-axis, and the diagonal

BD

has a length of

4

along the

x

-axis. They intersect at the origin

(0, 0)

at a right angle.

Providing a single counterexample is sufficient to prove that the statement is false.

4

Verify if the constructed quadrilateral is a square.

The side lengths are

AB = \sqrt{13}

and

BC = \sqrt{5}

. Since the sides are not equal, this quadrilateral is a kite, not a square.

This confirms that a quadrilateral can have perpendicular and equal-length diagonals without being a square.

Key Concept

Diagonal properties of quadrilaterals

Question 142Question →

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 143Question →

The measures of the six interior angles of a convex hexagon are in the ratio $3:4:5:5:6:7$ . What is the measure, in degrees, of the largest interior angle of this hexagon?

72

84

126

144

168

Show answer & explanation

Answer: $168$

Answer

168

The sum of the interior angles of any convex

n

-gon is given by

(n-2) \times 180^\circ

. For a hexagon (

n=6

), this sum is

(6-2) \times 180^\circ = 720^\circ

. The angles are in the ratio

3:4:5:5:6:7

, which sum to

3+4+5+5+6+7 = 30

parts. Each part corresponds to

720^\circ \div 30 = 24^\circ

. The largest angle is represented by the largest part of the ratio, 7, which equals

7 \times 24^\circ = 168^\circ

.

Step-by-Step Solution

1

Calculate the sum of the interior angles of a convex hexagon.

720^\circ

The sum of the interior angles of a polygon with

n

sides is

(n-2) \times 180^\circ

. For a hexagon (

n=6

), the sum is

(6-2) \times 180^\circ = 4 \times 180^\circ = 720^\circ

.

2

Determine the total number of parts in the given ratio.

30 parts

Sum the components of the ratio:

3 + 4 + 5 + 5 + 6 + 7 = 30

.

3

Find the value of one part in degrees.

24^\circ

per part

Divide the total interior angle sum by the total number of ratio parts:

720^\circ \div 30 = 24^\circ

.

4

Calculate the measure of the largest interior angle.

168^\circ

The largest angle corresponds to the largest component in the ratio, which is 7. Multiply the value of one part by 7:

7 \times 24^\circ = 168^\circ

.

Key Concept

Calculating the interior angle measures of an irregular convex polygon using the polygon interior angle sum formula and a given ratio.

Estimated Time:1m 30s

Question 144Question →

In the standard $(x, y)$ coordinate plane, an isosceles trapezoid $ABCD$ has vertices at $A(0, 0)$ , $B(8, 0)$ , $C(6, 4)$ , and $D(2, 4)$ . The diagonals $AC$ and $BD$ intersect at point $P$ . What is the area of triangle $APB$ ?

8

\frac{32}{3}

12

\frac{64}{3}

16

Show answer & explanation

Answer: $\frac{32}{3}$

Answer

\frac{32}{3}

The correct answer is obtained by first finding the equations of the lines representing the diagonals

AC

and

BD

, which intersect at

P(4, \frac{8}{3})

. Since

AB

lies on the x-axis from

x = 0

to

x = 8

, the length of the base of triangle

APB

is

8

. The height of the triangle is the y-coordinate of

P

, which is

\frac{8}{3}

. Using the formula for the area of a triangle, we get

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

.

Step-by-Step Solution

1

Find the equations of the lines containing the diagonals

AC

and

BD

.

Line

AC

passes through

(0,0)

and

(6,4)

, so its equation is

y = \frac{2}{3}x

. Line

BD

passes through

(8,0)

and

(2,4)

, so its slope is

\frac{4-0}{2-8} = -\frac{2}{3}

and its equation is

y = -\frac{2}{3}(x-8) = -\frac{2}{3}x + \frac{16}{3}

.

Determining the equations of the lines allows us to find their intersection point, which is the vertex

P

of triangle

APB

.

2

Determine the coordinates of the intersection point

P

by solving the system of equations.

Equating the two expressions for

y

gives

\frac{2}{3}x = -\frac{2}{3}x + \frac{16}{3}

, which simplifies to

\frac{4}{3}x = \frac{16}{3}

, so

x = 4

. Substituting

x = 4

back into the equation for line

AC

gives

y = \frac{2}{3}(4) = \frac{8}{3}

. The intersection point

P

is

(4, \frac{8}{3})

.

The

y

-coordinate of point

P

represents the height of triangle

APB

relative to the base

AB

along the x-axis.

3

Calculate the area of triangle

APB

using the area formula.

The base of triangle

APB

is the segment

AB

, which has length

8 - 0 = 8

. The height is the

y

-coordinate of

P

, which is

\frac{8}{3}

. The area is

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

.

Applying the triangle area formula with the correct base and height yields the final answer.

Key Concept

Properties of Quadrilaterals (specifically, trapezoids and their diagonals on the coordinate plane)

Question 145Question →

Chords $AB$ and $CD$ intersect at point $E$ inside a circle. If $AE = 6$ , $EB = 8$ , and the total length of chord $CD$ is $16$ , what is the length of the shorter segment of chord $CD$ ?

Show answer & explanation

Answer: 4

Answer

The length of the shorter segment of chord

CD

is

4

.

According to the Intersecting Chords Theorem, when two chords intersect inside a circle, the product of the segments of one chord is equal to the product of the segments of the other. For chords

AB

and

CD

intersecting at point

E

, this relationship is expressed as

AE \cdot EB = CE \cdot ED

. Substituting the given values yields

6 \cdot 8 = CE \cdot ED

, so

CE \cdot ED = 48

. Since the total length of chord

CD

is

16

, we can define

CE = x

and

ED = 16 - x

. The equation becomes

x(16 - x) = 48

, which simplifies to the quadratic equation

x^2 - 16x + 48 = 0

. Factoring this equation gives

(x - 12)(x - 4) = 0

, meaning the two segments of chord

CD

have lengths of

12

and

4

. The length of the shorter segment is

4

.

Step-by-Step Solution

1

State the relationship between intersecting chord segments.

AE \cdot EB = CE \cdot ED

By the Intersecting Chords Theorem, the product of the segments of one chord equals the product of the segments of the other.

2

Substitute the known lengths and define the segments of

CD

using a variable

x

.

6 \cdot 8 = x(16 - x)

, which simplifies to

48 = 16x - x^2

.

We are given

AE = 6

and

EB = 8

. Since the total length of chord

CD

is

16

, if one segment is

x

, the remaining segment must be

16 - x

.

3

Solve the quadratic equation for

x

by factoring.

x^2 - 16x + 48 = 0 \implies (x - 12)(x - 4) = 0

, so

x = 12

or

x = 4

.

Rearranging the equation into standard quadratic form allows us to find the two possible segment lengths.

4

Identify the shorter segment length from the two solutions.

4

The two segment lengths are

12

and

4

. The problem asks for the shorter segment, which is

4

.

Key Concept

Intersecting Chords Theorem

Question 146Question →

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 147Question →

A sector of a circle has a central angle of $\frac{3\pi}{4}$ radians and an area of $3\pi$ square inches. What is the perimeter, in inches, of the sector?

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

2\sqrt{2} + \frac{3\pi\sqrt{2}}{2}

4\sqrt{2} + \frac{3\pi\sqrt{2}}{2}

4 + \frac{3\pi}{2}

8 + 3\pi

Show answer & explanation

Answer: $4\sqrt{2} + \frac{3\pi\sqrt{2}}{2}$

Answer

The perimeter of the sector is

4\sqrt{2} + \frac{3\pi\sqrt{2}}{2}

inches.

The area of a sector is given by

A = \frac{1}{2} r^2 \theta

. Substituting the given area

3\pi

and the central angle

\frac{3\pi}{4}

, we get

3\pi = \frac{3\pi}{8} r^2

, which simplifies to

r^2 = 8

and

r = 2\sqrt{2}

. The perimeter of the sector consists of the curved arc length

s = r\theta

plus the two straight radii of the sector (

2r

). Substituting the values yields a perimeter of

2(2\sqrt{2}) + (2\sqrt{2})\left(\frac{3\pi}{4}\right) = 4\sqrt{2} + \frac{3\pi\sqrt{2}}{2}

.

Step-by-Step Solution

1

Use the sector area formula in radians,

A = \frac{1}{2} r^2 \theta

, to solve for the radius

r

of the circle.

3\pi = \frac{1}{2} r^2 \left(\frac{3\pi}{4}\right) \implies 3\pi = \frac{3\pi}{8} r^2 \implies r^2 = 8 \implies r = 2\sqrt{2}

inches.

Finding the radius of the circle is necessary to calculate both the arc length and the straight boundary segments of the sector.

2

Calculate the arc length

s

of the sector using the formula

s = r\theta

.

s = (2\sqrt{2})\left(\frac{3\pi}{4}\right) = \frac{3\pi\sqrt{2}}{2}

inches.

The arc length represents the curved boundary of the sector.

3

Compute the total perimeter of the sector by adding the arc length to the two boundary radii:

P = 2r + s

.

P = 2(2\sqrt{2}) + \frac{3\pi\sqrt{2}}{2} = 4\sqrt{2} + \frac{3\pi\sqrt{2}}{2}

inches.

The perimeter of a sector is the sum of the curved arc length and the two straight radial boundary segments.

Key Concept

Calculating the perimeter of a circular sector using its area and central angle in radians.

Estimated Time:1m 35s

Question 148Question →

A sector of a circle with radius $r$ inches has a central angle measuring $60^\circ$ . The ratio of the area of the sector (in square inches) to the perimeter of the sector (in inches) is $3:1$ . What is the radius, $r$ , of the circle, in inches?

\frac{36 + 6\pi}{\pi}

6

\frac{12 + 2\pi}{3\pi}

\frac{18 + 6\pi}{\pi}

\frac{18 + 3\pi}{\pi}

Show answer & explanation

Answer: $\frac{36 + 6\pi}{\pi}$

Answer

The radius of the circle is

\frac{36 + 6\pi}{\pi}

inches.

The sector area is

\frac{1}{6}\pi r^2

and the total sector perimeter (arc length plus two radii) is

2r + \frac{\pi r}{3}

. Setting the area equal to 3 times the perimeter gives

\frac{1}{6}\pi r^2 = 6r + \pi r

. Dividing by non-zero

r

gives

\frac{1}{6}\pi r = 6 + \pi

, which yields

r = \frac{36 + 6\pi}{\pi}

.

Step-by-Step Solution

1

Express the sector area in terms of radius

r

.

Sector Area

= \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6}\pi r^2

The area of a sector with central angle

\theta

in degrees is

\frac{\theta}{360^\circ}\pi r^2

.

2

Express the sector arc length and total sector perimeter in terms of

r

.

Arc Length

= \frac{60^\circ}{360^\circ} \times 2\pi r = \frac{\pi r}{3}

; Sector Perimeter

= 2r + \frac{\pi r}{3}

The total perimeter of a sector includes the curved arc length plus the two straight radii that enclose it.

3

Set up the equation using the given ratio of Area to Perimeter (

3:1

).

\frac{\frac{1}{6}\pi r^2}{2r + \frac{\pi r}{3}} = 3 \implies \frac{1}{6}\pi r^2 = 3\left(2r + \frac{\pi r}{3}\right)

A ratio of

3:1

means Area

= 3 \times \text{Perimeter}

.

4

Solve the equation for

r

.

\frac{1}{6}\pi r^2 = 6r + \pi r \implies \frac{1}{6}\pi r = 6 + \pi \implies r = \frac{6(6 + \pi)}{\pi} = \frac{36 + 6\pi}{\pi}

Dividing both sides by

r

(since

r > 0

) simplifies the quadratic relationship to a linear equation in

r

.

Key Concept

Calculating sector area, arc length, and sector perimeter using central angle ratios.

Estimated Time:2m 0s

Question 149Question →

The tip of a mechanical pendulum swings along a circular arc, sweeping out a sector of a circle. The arc length traveled by the tip of the pendulum is $8\pi$ inches, and the area of the circular sector swept out is $48\pi$ square inches. What is the total perimeter, in inches, of this circular sector?

12 + 8\pi

24 + 8\pi

48 + 8\pi

8\pi

24 + 4\pi

Show answer & explanation

Answer: $24 + 8\pi$

Answer

The total perimeter of the circular sector is

24 + 8\pi

inches.

The area of a circular sector with radius

r

and arc length

s

is given by

A = \frac{1}{2} r s

. Substituting the given values

A = 48\pi

and

s = 8\pi

gives

48\pi = \frac{1}{2} r (8\pi)

, which simplifies to

48\pi = 4\pi r

, so

r = 12

inches. The total perimeter of the sector consists of the arc length plus two straight radii:

P = 2r + s = 2(12) + 8\pi = 24 + 8\pi

inches.

Step-by-Step Solution

1

Relate sector area, radius, and arc length

Use the formula

\text{Area} = \frac{1}{2} r s

, where

s

is the arc length and

r

is the radius.

This direct relationship allows solving for the radius without needing to calculate the central angle explicitly.

2

Solve for the radius

r

48\pi = \frac{1}{2} \cdot r \cdot (8\pi) \implies 48\pi = 4\pi r \implies r = 12

inches.

Dividing both sides by

4\pi

determines the length of the pendulum arm (the radius).

3

Calculate the total sector perimeter

\text{Perimeter} = 2r + s = 2(12) + 8\pi = 24 + 8\pi$ inches.

The boundary of a circular sector consists of two straight radii and the curved arc.

Key Concept

Relationship between Sector Area, Arc Length, Radius, and Sector Perimeter

Estimated Time:1m 30s

Question 150Question →

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 151Question →

In any parallelogram $ABCD$ , if consecutive angles $\angle A$ and $\angle B$ are supplementary, then quadrilateral $ABCD$ must be a rectangle.

True

False

Show answer & explanation

Answer: False

Answer

False. Consecutive interior angles are supplementary in all parallelograms, not only in rectangles.

The statement is false because consecutive interior angles are supplementary in every parallelogram due to parallel opposite sides. This property does not imply that the angles are right angles (

90^\circ

); for instance, a parallelogram with angles measuring

60^\circ

and

120^\circ

has supplementary consecutive angles but is clearly not a rectangle.

Step-by-Step Solution

1

Recall the consecutive angle property for any general parallelogram.

In any parallelogram

ABCD

, opposite sides are parallel (

AD \parallel BC

). When parallel lines are intersected by a transversal

AB

, consecutive interior angles are supplementary:

\angle A + \angle B = 180^\circ

.

Consecutive interior angles formed by parallel lines and a transversal always sum to

180^\circ

.

2

Compare this general property with the specific condition that defines a rectangle.

A parallelogram is a rectangle if and only if all four interior angles are right angles (

90^\circ

), which requires consecutive angles to be congruent (equal in measure), not merely supplementary.

Two supplementary angles can have measures such as

60^\circ

and

120^\circ

, which form an oblique parallelogram rather than a rectangle.

3

Determine the truth value of the statement.

Because the condition of having supplementary consecutive angles is satisfied by every parallelogram and does not guarantee

90^\circ

angles, the statement is false.

A property shared by all members of a general class (parallelograms) cannot be used as a sufficient condition to classify a figure into a restrictive subclass (rectangles).

Key Concept

Properties of Quadrilaterals: Parallelogram vs. Rectangle Angle Rules

Estimated Time:1m 0s

Question 152Question →

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 153Question →

A decorative emblem is shaped as the region bounded by two concentric circular sectors sharing the same central angle of $\theta$ radians. The outer sector has radius $R\text{ cm}$ , and the inner sector has radius $r\text{ cm}$ , where the difference between the two radii is $R - r = 4\text{ cm}$ . If the area of the emblem is $15\pi\text{ cm}^2$ and the length of the outer arc is $5\pi\text{ cm}$ , what is the value of $\theta$ , in radians?

\frac{3\pi}{8}

\frac{5\pi}{16}

\frac{5\pi}{8}

\frac{3\pi}{4}

\frac{5\pi}{4}

Show answer & explanation

Answer: $\frac{5\pi}{8}$

Answer

The central angle

\theta

is

\frac{5\pi}{8}

radians.

Using the formulas for arc length (

s = R\theta

) and sector area (

A = \frac{1}{2}r^2\theta

), the area of the emblem is

A = \frac{1}{2}\theta(R^2 - r^2) = \frac{1}{2}\theta(R - r)(R + r)

. Substituting

R - r = 4\text{ cm}

gives

15\pi = 2\theta(R + r)

. Since

R\theta = 5\pi

, we have

R + r = \frac{10\pi}{\theta} - 4

. Substituting this into the area equation yields

2\theta\left(\frac{10\pi}{\theta} - 4\right) = 15\pi

, which simplifies to

20\pi - 8\theta = 15\pi

, yielding

\theta = \frac{5\pi}{8}

.

Step-by-Step Solution

1

Express the outer arc length using the radian arc length formula.

s_{\text{outer}} = R\theta = 5\pi

, which gives

R = \frac{5\pi}{\theta}

.

Arc length in radians is given by

s = r\theta

.

2

Express the area of the emblem as the difference between the outer and inner sector areas.

A = \frac{1}{2}R^2\theta - \frac{1}{2}r^2\theta = \frac{1}{2}\theta(R^2 - r^2) = 15\pi

.

The emblem is formed by removing the inner sector from the outer sector.

3

Factor

R^2 - r^2

as

(R - r)(R + r)

and substitute

R - r = 4

.

15\pi = \frac{1}{2}\theta(4)(R + r) = 2\theta(R + r)

.

The difference of squares allows substituting the known difference between radii.

4

Substitute

R = \frac{5\pi}{\theta}

and

r = \frac{5\pi}{\theta} - 4

into the sum

(R + r)

.

R + r = \frac{10\pi}{\theta} - 4

.

Expreing

R + r

solely in terms of

\theta

allows solving a single variable equation.

5

Solve the resulting equation for

\theta

.

2\theta \left(\frac{10\pi}{\theta} - 4\right) = 15\pi \implies 20\pi - 8\theta = 15\pi \implies 8\theta = 5\pi \implies \theta = \frac{5\pi}{8}

.

Distributing

2\theta

cancels

\theta

in the first term and leaves a linear equation in

\theta

.

Key Concept

Sector area and arc length formulas in radians applied to concentric regions

Estimated Time:2m 0s

Question 154Question →

In rhombus $ABCD$ , the measure of interior angle $\angle DAB$ is $120^\circ$ , and the length of diagonal $AC$ is $12$ inches. What is the perimeter, in inches, of rhombus $ABCD$ ?

24

48

48\sqrt{3}

72

96

Show answer & explanation

Answer: 48

Answer

The perimeter of rhombus

ABCD

is 48 inches.

Since consecutive angles in a rhombus are supplementary,

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

. Because all sides of a rhombus are equal in length,

AB = BC

, making

\triangle ABC

an isosceles triangle with a

60^\circ

vertex angle, which implies

\triangle ABC

is equilateral. Therefore, side length

AB = AC = 12

inches, and the perimeter is

4 \times 12 = 48

inches.

Step-by-Step Solution

1

Determine the consecutive angle measure in the rhombus.

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

Consecutive interior angles in a rhombus (which is a parallelogram) are supplementary.

2

Determine the nature of triangle ABC.

\triangle ABC is equilateral with AB = BC = AC = 12 inches.

A rhombus has four equal sides, so AB = BC. An isosceles triangle with a vertex angle of 60 degrees is an equilateral triangle.

3

Calculate the perimeter of the rhombus.

Perimeter = 4 \times 12 = 48 inches.

The perimeter of a rhombus is four times the length of one side.

Key Concept

Properties of a rhombus: all four sides are congruent, consecutive interior angles are supplementary, and a diagonal splitting a 60-degree angle forms an equilateral triangle with adjacent sides.

Question 155Question →

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 156Question →

A circular stained-glass window with a radius of $12\text{ inches}$ contains a sector, $OAB$ , with a central angle measuring $120^\circ$ . A straight piece of lead wire is placed along chord $\overline{AB}$ , dividing sector $OAB$ into a triangular region $\triangle OAB$ and a circular segment bounded by chord $\overline{AB}$ and arc $\widehat{AB}$ . What is the area, in square inches, of the circular segment?

48\pi - 36\sqrt{3}

144\pi - 36\sqrt{3}

24\pi - 36\sqrt{3}

48\pi - 72\sqrt{3}

48\pi - 36

Show answer & explanation

Answer: $48\pi - 36\sqrt{3}$

Answer

48\pi - 36\sqrt{3}

square inches

The area of a circular segment is found by subtracting the area of the central triangle from the area of the sector. The sector area with a

120^\circ

central angle and a radius of

12

is

\frac{120^\circ}{360^\circ} \cdot \pi (12^2) = 48\pi

. The area of the isosceles triangle formed by the two radii and the chord is

\frac{1}{2} (12)(12) \sin(120^\circ) = 36\sqrt{3}

. Thus, the segment area is

48\pi - 36\sqrt{3}

square inches.

Step-by-Step Solution

1

Calculate the area of sector

OAB

Sector Area =

\frac{120^\circ}{360^\circ} \cdot \pi \cdot (12)^2 = \frac{1}{3} \cdot 144\pi = 48\pi\text{ sq in}

The area of a sector with central angle

\theta

in degrees is given by

\frac{\theta}{360^\circ} \pi r^2

.

2

Calculate the area of triangle

\triangle OAB

Triangle Area =

\frac{1}{2} \cdot 12 \cdot 12 \cdot \sin(120^\circ) = 72 \cdot \frac{\sqrt{3}}{2} = 36\sqrt{3}\text{ sq in}

The area of a triangle with two sides

a

and

b

and included angle

\theta

is

\frac{1}{2} a b \sin(\theta)

.

3

Subtract the triangle area from the sector area to find the circular segment area

Segment Area =

48\pi - 36\sqrt{3}\text{ sq in}

A circular segment area is equal to the area of the containing sector minus the area of the central triangle.

Key Concept

Area of a Circular Segment

Estimated Time:2m 0s

Question 157Question →

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 158Question →

Determine whether the following statement is true or false: In any trapezoid, the line segment connecting the midpoints of the non-parallel sides (the midsegment) is parallel to the bases and has a length equal to half the difference of the lengths of the bases.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False.

The statement is False because according to the Trapezoid Midsegment Theorem, the midsegment is parallel to the bases and its length is equal to half the sum (the arithmetic mean) of the base lengths,

m = \frac{b_1 + b_2}{2}

.

Step-by-Step Solution

1

Identify the geometric shape and segment described in the statement.

The shape is a trapezoid with parallel bases of lengths

b_1

and

b_2

, and the segment is the midsegment connecting the midpoints of the non-parallel legs.

Understanding the definition of a trapezoid's midsegment is necessary to evaluate its properties.

2

State the Trapezoid Midsegment Theorem.

The Trapezoid Midsegment Theorem states that the midsegment is parallel to both bases and its length

m

is equal to the average of the two base lengths:

m = \frac{b_1 + b_2}{2}

.

This theorem directly provides the exact mathematical relationship for the length of the midsegment.

3

Compare the theorem with the statement provided in the stem.

The statement claims the length is half the difference,

\frac{|b_1 - b_2|}{2}

, which contradicts the actual formula involving the sum.

Since the statement uses subtraction instead of addition, the statement is false.

Key Concept

Trapezoid Midsegment Theorem

Question 159Question →

In the standard $(x, y)$ coordinate plane, three consecutive vertices of parallelogram $ABCD$ are $A(-2, 1)$ , $B(3, 4)$ , and $C(5, -1)$ . If $(x, y)$ represents the coordinates of vertex $D$ , what is the value of $x + y$ ?

-4

-2

0

6

12

Show answer & explanation

Answer: $-4$

Answer

The sum of the coordinates

x + y

is

-4

.

The diagonals of a parallelogram bisect each other, meaning the midpoint of diagonal

AC

is identical to the midpoint of diagonal

BD

. Finding the midpoint of

AC

gives

\left(\frac{3}{2}, 0\right)

. Setting the midpoint of

BD

to this point gives

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

and

\frac{4+y}{2} = 0

, yielding

x = 0

and

y = -4

. The sum

x + y

is

0 + (-4) = -4

.

Step-by-Step Solution

1

Apply the diagonal midpoint property of parallelograms.

In parallelogram

ABCD

, the diagonals

AC

and

BD

bisect each other at their common midpoint

M

.

Diagonals of any parallelogram share a common midpoint.

2

Calculate the midpoint of diagonal

AC

.

M = \left(\frac{-2 + 5}{2}, \frac{1 + (-1)}{2}\right) = \left(\frac{3}{2}, 0\right)

.

The midpoint formula is

\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

.

3

Set the midpoint of

BD

equal to

M

and solve for

x

and

y

.

\frac{3 + x}{2} = \frac{3}{2} \implies x = 0

and

\frac{4 + y}{2} = 0 \implies y = -4

. So vertex

D

is

(0, -4)

.

Equating coordinates of the common midpoint.

4

Find the sum

x + y

.

x + y = 0 + (-4) = -4

.

Adding the

x

- and

y

-coordinates of vertex

D

.

Key Concept

Properties of Parallelogram Diagonals in Coordinate Geometry

Question 160Question →

An automated lawn sprinkler sweeps through a central angle of $\theta$ radians and waters a sector-shaped region of radius $r$ feet. Due to a mechanical adjustment, the central angle $\theta$ is increased by $20\%$ , while the water pressure is reduced such that the radius $r$ is decreased by $10\%$ . What is the net percentage change in the area of the region watered by the sprinkler?

Decreases by

2.8\%

Increases by

2.8\%

Increases by

8.0\%

Increases by

10.0\%

Decreases by

19.0\%

Show answer & explanation

Answer: Decreases by $2.8\%$

Answer

Decreases by

2.8\%

The area of a circular sector is proportional to the square of its radius and linearly proportional to its central angle (

A = \frac{1}{2} r^2 \theta

). Decreasing the radius by

10\%

scales the radius by

0.90

, which scales

r^2

by

(0.90)^2 = 0.81

. Increasing the central angle by

20\%

scales

\theta

by

1.20

. The overall scaling factor for the sector area is

0.81 \times 1.20 = 0.972

. This corresponds to

97.2\%

of the original area, which is a net decrease of

2.8\%

.

Step-by-Step Solution

1

Express the original sector area in terms of

r

and

\theta

.

A_1 = \frac{1}{2} r^2 \theta

The area of a circular sector with radius

r

and central angle

\theta

in radians is given by

A = \frac{1}{2} r^2 \theta

.

2

Express the new radius and central angle after the percentage adjustments.

r_{new} = 0.90 r

and

\theta_{new} = 1.20 \theta

A

10\%

decrease in radius leaves

90\%

of

r

, and a

20\%

increase in angle yields

120\%

of

\theta

.

3

Substitute the new variables into the sector area formula and simplify.

A_2 = \frac{1}{2} (0.90 r)^2 (1.20 \theta) = \frac{1}{2} (0.81 r^2) (1.20 \theta) = 0.972 \left(\frac{1}{2} r^2 \theta\right) = 0.972 A_1

Squaring

0.90

gives

0.81

, and multiplying

0.81 \times 1.20

yields

0.972

.

4

Calculate the net percentage change from

A_1

to

A_2

.

Percentage Change

= (0.972 - 1) \times 100\% = -2.8\%

A factor of

0.972

means the new area is

97.2\%

of the original area, which is a decrease of

2.8\%

.

Key Concept

Circular Sector Area under proportional changes of parameters

Estimated Time:2m 0s

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