Geometry Practice Questions — GRE General Test — Page 3

Question 41Question →

In a geometric plane, straight lines $L_1$ and $L_2$ intersect at point $P$ . The measure of the obtuse angle formed by the intersection of $L_1$ and $L_2$ is $(5x - 10)^\circ$ , and the measure of an adjacent acute angle is $(2x + 15)^\circ$ . Ray $PQ$ originates from point $P$ , is perpendicular to line $L_1$ , and lies entirely within the interior of the $(5x - 10)^\circ$ obtuse angle. What is the measure, in degrees, of the angle formed between ray $PQ$ and line $L_2$ ?

25^\circ

35^\circ

50^\circ

65^\circ

155^\circ

Show answer & explanation

Answer: $25^\circ$

Answer

25^\circ

Two intersecting lines form adjacent angles that sum to

180^\circ

. Solving

(5x - 10) + (2x + 15) = 180

gives

7x + 5 = 180

, so

x = 25

. The obtuse angle measure is

5(25) - 10 = 115^\circ

. Since ray

PQ

is perpendicular to line

L_1

, it forms a

90^\circ

angle with line

L_1

. The remaining angle between ray

PQ

and line

L_2

within the obtuse angle region is

115^\circ - 90^\circ = 25^\circ

.

Step-by-Step Solution

1

Set up an algebraic equation using the supplementary angle relationship.

(5x - 10) + (2x + 15) = 180

Adjacent angles formed by two intersecting straight lines lie on a straight line and are supplementary, summing to

180^\circ

.

2

Solve for the variable

x

.

7x + 5 = 180 \implies 7x = 175 \implies x = 25

Combining like terms simplifies the linear equation.

3

Calculate the degree measure of the obtuse angle.

Obtuse angle =

5(25) - 10 = 125 - 10 = 115^\circ

Substitute

x = 25

back into the expression

(5x - 10)^\circ

.

4

Calculate the angle between ray

PQ

and line

L_2

.

Angle =

115^\circ - 90^\circ = 25^\circ

Ray

PQ

is perpendicular to line

L_1

(

90^\circ

) and lies inside the

115^\circ

angle, dividing the obtuse angle into a

90^\circ

portion and the remaining angle adjacent to line

L_2

.

Key Concept

Supplementary angles on a straight line and angle subtraction with perpendicular rays

Estimated Time:1m 30s

Question 42Question →

A sector of a circle has an area of $15\pi$ square units and a perimeter of $10 + 6\pi$ units. What is the measure of the central angle of the sector, in degrees?

108^\circ

135^\circ

144^\circ

216^\circ

270^\circ

Show answer & explanation

Answer: $216^\circ$

Answer

The correct answer is

216^\circ

, corresponding to the option stating

216^\circ

.

The perimeter of a sector is defined as

2r + s = 10 + 6\pi

, which yields a radius of

r = 5

and an arc length

s = 6\pi

. Checking with the sector area formula

\frac{1}{2}rs = \frac{1}{2}(5)(6\pi) = 15\pi

confirms these measurements. The total area of the circle is

\pi (5)^2 = 25\pi

. The sector thus constitutes

\frac{15\pi}{25\pi} = \frac{3}{5}

of the circle. Multiplying this fraction by

360^\circ

gives a central angle of

216^\circ

.

Step-by-Step Solution

1

Express the sector perimeter formula and solve for radius and arc length.

Radius

r = 5

and arc length

s = 6\pi

.

The perimeter of a sector equals two radii plus its arc length:

\text{Perimeter} = 2r + s = 10 + 6\pi

. Equating standard and

\pi

-termed components gives

2r = 10 \implies r = 5

and arc length

s = 6\pi

.

2

Verify consistency using the sector area formula.

\text{Area} = 15\pi

, matching the given information.

The area of a sector can also be calculated as

\frac{1}{2} r s = \frac{1}{2}(5)(6\pi) = 15\pi

.

3

Find the total area of the circle and the fraction of the circle occupied by the sector.

Total area

= 25\pi

, area fraction

= \frac{3}{5}

.

The full circle area is

\pi r^2 = \pi (5^2) = 25\pi

. The sector represents

\frac{15\pi}{25\pi} = \frac{3}{5}

of the entire circle.

4

Calculate the central angle

\theta

in degrees.

\theta = 216^\circ

.

Multiply the fraction by

360^\circ

:

\theta = \frac{3}{5} \times 360^\circ = 216^\circ

.

Key Concept

Perimeter, Arc Length, and Area of a Circle Sector

Question 43Question →

In a circle centered at point $O$ with radius $R$ , radii $OA$ and $OB$ bound a sector $AOB$ with a central angle of $120^\circ$ . Point $P$ lies on minor arc $AB$ such that the ratio of the length of arc $AP$ to the length of arc $PB$ is $1:3$ . Segment $PQ$ is drawn perpendicular to radius $OA$ , intersecting $OA$ at point $Q$ . What is the ratio of the area of the region bounded by line segment $PQ$ , line segment $AQ$ , and minor arc $AP$ to the area of sector $AOB$ ?

\frac{4\pi - 3\sqrt{3}}{16\pi}

\frac{4\pi - 3\sqrt{3}}{4\pi}

\frac{4\pi - 3\sqrt{3}}{8\pi}

\frac{\pi - 3}{4\pi}

\frac{1}{4}

Show answer & explanation

Answer: $\frac{4\pi - 3\sqrt{3}}{16\pi}$

Answer

\frac{4\pi - 3\sqrt{3}}{16\pi}

The correct answer

\frac{4\pi - 3\sqrt{3}}{16\pi}

is obtained by finding the central angle of sector

AOP

(

30^\circ

), subtracting the area of right triangle

OQP

(

\frac{\sqrt{3}}{16}R^2

) from the area of sector

AOP

(

\frac{\pi}{12}R^2

), and dividing this bounded area by the total sector

AOB

area (

\frac{\pi}{3}R^2

).

Step-by-Step Solution

1

Determine the central angle

\angle AOP

of sector

AOP

\angle AOP = \frac{1}{1+3} \times 120^\circ = 30^\circ

Arc length is directly proportional to central angle. A

1:3

ratio means sector

AOP

represents

\frac{1}{4}

of the

120^\circ

central angle.

2

Calculate the area of sector

AOP

and the total area of sector

AOB

\text{Area(Sector } AOP) = \frac{30^\circ}{360^\circ} \pi R^2 = \frac{\pi}{12} R^2

, and

\text{Area(Sector } AOB) = \frac{120^\circ}{360^\circ} \pi R^2 = \frac{\pi}{3} R^2

The area of a sector with central angle

\theta

is

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

.

3

Calculate the area of right triangle

OQP

\text{Area}(\triangle OQP) = \frac{\sqrt{3}}{16} R^2

In

\triangle OQP

,

\angle QOP = 30^\circ

and hypotenuse

OP = R

. Thus

OQ = R \cos(30^\circ) = \frac{R\sqrt{3}}{2}

and

PQ = R \sin(30^\circ) = \frac{R}{2}

.

\text{Area} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \left(\frac{R\sqrt{3}}{2}\right)\left(\frac{R}{2}\right) = \frac{\sqrt{3}}{16} R^2

.

4

Calculate the area of the bounded region and take the ratio to the area of sector

AOB

\text{Ratio} = \frac{\frac{\pi}{12}R^2 - \frac{\sqrt{3}}{16}R^2}{\frac{\pi}{3}R^2} = \frac{4\pi - 3\sqrt{3}}{16\pi}

Subtract the triangle area from sector

AOP

area to get the bounded region area, then divide by sector

AOB

area.

Key Concept

Sector Area, Arc Length Proportions, and Geometric Region Subdivision

Estimated Time:3m 0s

Question 44Question →

Two concentric circles centered at point $O$ have radii $r$ and $R$ , where $r < R$ . A sector bounded by radii $OA$ and $OB$ of the outer circle has central angle $\theta^\circ$ . Region $S$ is the region lying inside sector $AOB$ but outside the inner circle. The area of region $S$ is equal to $3$ times the area of the sector of the inner circle bounded by central angle $\theta^\circ$ . If the perimeter of region $S$ is equal to $\frac{11}{6}$ times the length of arc $AB$ , what is the value of $\theta$ ?

\frac{60}{\pi}

\frac{270}{\pi}

\frac{480}{\pi}

\frac{540}{\pi}

\frac{1080}{\pi}

Show answer & explanation

Answer: $\frac{540}{\pi}$

Answer

\frac{540}{\pi}

The area of region

S

is

\frac{\theta}{360}\pi(R^2 - r^2)

, which is equal to

3

times the area of the inner sector

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

. This simplifies to

R^2 - r^2 = 3r^2 \implies R = 2r

. The perimeter of region

S

consists of outer arc length

L_{\text{outer}}

, inner arc length

L_{\text{inner}} = \frac{1}{2} L_{\text{outer}}

, and two straight line segments each of length

R - r = r

. Setting the total perimeter

\frac{3}{2} L_{\text{outer}} + 2r

equal to

\frac{11}{6} L_{\text{outer}}

yields

2r = \frac{1}{3} L_{\text{outer}}

, or

L_{\text{outer}} = 6r

. Substituting

L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{4\pi r \theta}{360}

gives

\frac{4\pi r \theta}{360} = 6r

, which solves to

\theta = \frac{540}{\pi}

.

Step-by-Step Solution

1

Relate outer radius

R

to inner radius

r

using sector areas.

R = 2r

The area of region

S

is

\frac{\theta}{360}\pi(R^2 - r^2)

and the inner sector area is

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

. Setting

\frac{\theta}{360}\pi(R^2 - r^2) = 3 \cdot \frac{\theta}{360}\pi r^2

gives

R^2 - r^2 = 3r^2

, so

R^2 = 4r^2

and

R = 2r

.

2

Express the perimeter of region

S

in terms of outer arc length

L_{\text{outer}}

and radius

r

.

\text{Perimeter}(S) = \frac{3}{2} L_{\text{outer}} + 2r

Region

S

is bounded by outer arc

AB

(

L_{\text{outer}}

), inner arc

CD

(

L_{\text{inner}}

), and two straight segments

AC

and

BD

of length

R - r = 2r - r = r

. Since

R = 2r

,

L_{\text{inner}} = \frac{1}{2} L_{\text{outer}}

, making total perimeter

L_{\text{outer}} + \frac{1}{2} L_{\text{outer}} + 2(r) = \frac{3}{2} L_{\text{outer}} + 2r

.

3

Use the given perimeter relationship to express

L_{\text{outer}}

in terms of

r

.

L_{\text{outer}} = 6r

Setting

\frac{3}{2} L_{\text{outer}} + 2r = \frac{11}{6} L_{\text{outer}}

yields

2r = \left(\frac{11}{6} - \frac{9}{6}\right) L_{\text{outer}} = \frac{1}{3} L_{\text{outer}}

, so

L_{\text{outer}} = 6r

.

4

Solve for

\theta

using the definition of outer arc length.

\theta = \frac{540}{\pi}

L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{\theta}{360}(4\pi r)

. Setting

\frac{4\pi r \theta}{360} = 6r

simplifies to

\frac{\pi \theta}{90} = 6

, giving

\theta = \frac{540}{\pi}

.

Key Concept

Annular sector area and perimeter relations combining arc length formulas and concentric circle geometry.

Estimated Time:2m 30s

Question 45Question →

A circle has a radius of $9$ inches. What is the area, in square inches, of a sector of this circle formed by a central angle of $40^\circ$ ?

2\pi

9\pi

18\pi

27\pi

81\pi

Show answer & explanation

Answer: $9\pi$

Answer

The area of the sector is

9\pi

square inches.

The area of a sector with radius

r

and central angle

\theta

is calculated as

\text{Area} = \pi r^2 \left(\frac{\theta}{360^\circ}\right)

. Substituting

r = 9

and

\theta = 40^\circ

yields

\pi (9^2) \left(\frac{40^\circ}{360^\circ}\right) = 81\pi \left(\frac{1}{9}\right) = 9\pi

square inches.

Step-by-Step Solution

1

Calculate the area of the entire circle using the formula

A_{\text{circle}} = \pi r^2

.

A_{\text{circle}} = \pi (9^2) = 81\pi

square inches.

The full circle area provides the total measure from which the sector fraction is calculated.

2

Determine the fraction of the circle represented by the central angle of

40^\circ

.

\frac{40^\circ}{360^\circ} = \frac{1}{9}

.

A circle contains

360^\circ

in total, so the central angle over

360^\circ

gives the proportion of the circle covered by the sector.

3

Multiply the total area of the circle by the fraction of the sector.

\text{Sector Area} = 81\pi \times \frac{1}{9} = 9\pi

square inches.

Applying the fraction to the total area yields the specific sector area.

Key Concept

Sector Area of a Circle

Question 46Question →

Circle $C_1$ has radius $r_1$ and Circle $C_2$ has radius $r_2$ . An arc on Circle $C_1$ subtended by a central angle of $\theta_1^\circ$ has the exact same length as an arc on Circle $C_2$ subtended by a central angle of $\theta_2^\circ$ . The sector formed by this arc in Circle $C_1$ has an area of $54\pi$ , and the sector formed by this arc in Circle $C_2$ has an area of $36\pi$ . If $\theta_1 + \theta_2 = 150$ , what is the value of $r_1$ ?

Show answer & explanation

Answer: 18

Answer

The radius

r_1

of Circle

C_1

is 18.

Using the relationship

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

, the ratio of the two sector areas gives

\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = 1.5

, so

r_2 = \frac{2}{3} r_1

. Expressing arc length as

L = \frac{108\pi}{r_1}

, the central angles are

\theta_1 = \frac{19440}{r_1^2}

and

\theta_2 = \frac{29160}{r_1^2}

. Setting their sum equal to

150

leads to

\frac{48600}{r_1^2} = 150

, giving

r_1^2 = 324

and

r_1 = 18

.

Step-by-Step Solution

1

Relate sector area to arc length and radius

A_1 = \frac{1}{2} L r_1 = 54\pi

and

A_2 = \frac{1}{2} L r_2 = 36\pi

The area of a sector with arc length

L

and radius

r

is given by

A = \frac{\theta}{360}\pi r^2 = \frac{1}{2} L r

.

2

Find the ratio of

r_1

to

r_2

\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = \frac{3}{2} \implies r_2 = \frac{2}{3}r_1

Dividing the first area equation by the second cancels out

\frac{1}{2}L

.

3

Express central angles in terms of

r_1

\theta_1 = \frac{19440}{r_1^2}

and

\theta_2 = \frac{29160}{r_1^2}

Since

L = \frac{108\pi}{r_1}

, substituting into

\theta_1 = \frac{L}{2\pi r_1} \times 360

yields

\theta_1 = \frac{19440}{r_1^2}

, and substituting into

\theta_2 = \frac{L}{2\pi (\frac{2}{3}r_1)} \times 360

yields

\theta_2 = \frac{29160}{r_1^2}

.

4

Solve for

r_1

using the angle sum equation

r_1 = 18

Summing the angles gives

\frac{19440 + 29160}{r_1^2} = \frac{48600}{r_1^2} = 150 \implies r_1^2 = 324 \implies r_1 = 18

.

Key Concept

Relationship between arc length, radius, central angle, and sector area

Estimated Time:2m 30s

Question 47Question →

In triangle $ABC$ , the measure of angle $A$ is $40^\circ$ and the measure of angle $B$ is $70^\circ$ . If the perimeter of triangle $ABC$ is $22$ centimeters and the length of side $AB$ is $8$ centimeters, what is the length, in centimeters, of side $BC$ ?

4

6

7

8

14

Show answer & explanation

Answer: $6$

Answer

6 centimeters

First, find the measure of angle

C

:

180^\circ - 40^\circ - 70^\circ = 70^\circ

. Because angle

B

and angle

C

both measure

70^\circ

, triangle

ABC

is isosceles with sides

AC

and

AB

of equal length (

AC = AB = 8

cm). Since the perimeter is the sum of all three sides (

AB + AC + BC = 22

), substituting the known side lengths gives

8 + 8 + BC = 22

, which yields

BC = 6

cm.

Step-by-Step Solution

1

Calculate the measure of the third angle, angle

C

.

Angle

C = 180^\circ - (40^\circ + 70^\circ) = 70^\circ

.

The sum of interior angles in any triangle is

180^\circ

.

2

Determine the relationship between the side lengths using the angle measures.

Side

AC = AB = 8

centimeters.

Since angle

B = 70^\circ

and angle

C = 70^\circ

, triangle

ABC

is isosceles. Sides opposite equal angles are equal in length, so the side opposite angle

C

(which is

AB

) equals the side opposite angle

B

(which is

AC

).

3

Use the perimeter formula to find the length of side

BC

.

Length of side

BC = 22 - (8 + 8) = 6

centimeters.

Perimeter is the total boundary distance (

AB + AC + BC = 22

). Substituting known values gives

8 + 8 + BC = 22

, so

BC = 6

.

Key Concept

Isosceles Triangle Side-Angle Properties and Perimeter

Estimated Time:45s

Question 48Question →

Lines $L_1$ and $L_2$ intersect at point $P$ to form an acute angle of $54^\circ$ . Line $M$ passes through point $P$ and is perpendicular to line $L_1$ . Ray $R$ originates at point $P$ and lies in the interior of one of the obtuse angles formed by $L_1$ and $L_2$ . If Ray $R$ bisects the angle formed between line $M$ and line $L_2$ , what is the measure, in degrees, of the acute angle formed by Ray $R$ and line $L_1$ ?

Show answer & explanation

Answer: 72

Answer

The measure of the acute angle formed by Ray

R

and line

L_1

is

72^\circ

.

Lines

L_1

and

L_2

form an obtuse angle of

180^\circ - 54^\circ = 126^\circ

. Perpendicular line

M

consumes

90^\circ

of this angle relative to

L_1

, leaving

36^\circ

between line

M

and line

L_2

. Bisecting this remaining angle gives

18^\circ

. Ray

R

therefore lies

18^\circ

away from perpendicular line

M

, making an acute angle of

90^\circ - 18^\circ = 72^\circ

with line

L_1

.

Step-by-Step Solution

1

Determine the measure of the obtuse angle between lines

L_1

and

L_2

.

The obtuse angle measures

180^\circ - 54^\circ = 126^\circ

.

Adjacent angles along intersecting straight lines are supplementary and sum to

180^\circ

.

2

Find the angle between perpendicular line

M

and line

L_2

.

The angle between line

M

and line

L_2

within the obtuse region is

126^\circ - 90^\circ = 36^\circ

.

Line

M

is perpendicular to line

L_1

, taking up

90^\circ

of the

126^\circ

obtuse angle.

3

Determine the angle formed by Ray

R

after bisecting the

36^\circ

angle.

The angle between Ray

R

and line

M

is

36^\circ / 2 = 18^\circ

.

An angle bisector divides an angle into two equal parts.

4

Calculate the acute angle between Ray

R

and line

L_1

.

The acute angle formed between Ray

R

and line

L_1

is

90^\circ - 18^\circ = 72^\circ

.

Line

M

forms a

90^\circ

angle with line

L_1

. Subtracting the

18^\circ

offset created by Ray

R

yields the acute angle of

72^\circ

.

Key Concept

Supplementary angles, perpendicular lines, and angle bisectors

Estimated Time:2m 0s

Question 49Question →

A right circular cylinder has a base radius of $3$ units and a height of $4$ units. Which of the following statements regarding this cylinder are true? Select all such statements.

Select all that apply

The volume of the cylinder is

36\pi

cubic units.

The total surface area of the cylinder is

42\pi

square units.

The volume of the cylinder is

24\pi

cubic units.

The total surface area of the cylinder is

36\pi

square units.

The ratio of the area of one circular base to the lateral surface area is

1

to

2

.

Show answer & explanation

Answer: The volume of the cylinder is $36\pi$ cubic units.; The total surface area of the cylinder is $42\pi$ square units.

Answer

The correct statements are that the volume of the cylinder is

36\pi

cubic units and the total surface area of the cylinder is

42\pi

square units.

The volume of a cylinder with radius

3

and height

4

is

V = \pi r^2 h = \pi(3)^2(4) = 36\pi

. The total surface area is

A = 2\pi r h + 2\pi r^2 = 2\pi(3)(4) + 2\pi(3)^2 = 24\pi + 18\pi = 42\pi

. Therefore, both statements asserting these exact values are correct.

Step-by-Step Solution

1

Calculate the volume of the right circular cylinder

Volume

V = \pi r^2 h = \pi (3^2)(4) = 36\pi

The formula for the volume of a right circular cylinder is

V = \pi r^2 h

.

2

Calculate the lateral surface area and total surface area of the cylinder

Lateral Area =

2\pi r h = 24\pi

; Total Area =

24\pi + 2(\pi \cdot 3^2) = 42\pi

Total surface area is the sum of the lateral surface area (

2\pi r h

) and the areas of the two circular bases (

2\pi r^2

).

3

Evaluate the base-to-lateral surface area ratio

Ratio =

\frac{9\pi}{24\pi} = \frac{3}{8}

Comparing the base area (

9\pi

) to the lateral area (

24\pi

) simplifies to

3:8

.

Key Concept

Volume and surface area formulas for right circular cylinders

Question 50Question →

In the geometric plane, line $k$ is parallel to line $m$ ( $k \parallel m$ ). Transversal line $t$ intersects line $k$ at point $A$ and line $m$ at point $B$ . Point $C$ lies on line $k$ to the left of $A$ , such that interior acute angle $\angle CAB = (3x - 10)^\circ$ . Ray $AD$ bisects $\angle CAB$ . Ray $BF$ is drawn into the region between lines $k$ and $m$ making an angle $\angle ABF = (x + 35)^\circ$ with transversal segment $AB$ . Ray $AD$ and ray $BF$ intersect at point $P$ inside the parallel region. Line $PB$ is extended past $P$ to intersect line $k$ at point $Q$ . If ray $AD$ is perpendicular to ray $BF$ , what is the measure of the obtuse angle formed at the intersection of line $QB$ and line $k$ ?

121^\circ

59^\circ

118^\circ

128^\circ

149^\circ

Show answer & explanation

Answer: $121^\circ$

Answer

The measure of the obtuse angle formed at the intersection of line

QB

and line

k

is

121^\circ

.

The measure

121^\circ

is correct because ray

AD

bisects

\angle CAB

, giving

\angle PAB = (1.5x - 5)^\circ

. Since

AD \perp BF

,

\triangle APB

is a right triangle where

(1.5x - 5) + (x + 35) = 90

, yielding

x = 24

. Consequently,

\angle QAB = 62^\circ

and

\angle ABQ = 59^\circ

. In

\triangle ABQ

, the acute angle

\angle AQB = 180^\circ - (62^\circ + 59^\circ) = 59^\circ

. The supplementary obtuse angle along line

k

is

180^\circ - 59^\circ = 121^\circ

.

Step-by-Step Solution

1

Express the angle

\angle PAB

in terms of

x

.

Since ray

AD

bisects

\angle CAB = (3x - 10)^\circ

,

\angle PAB = \frac{3x - 10}{2} = (1.5x - 5)^\circ

.

An angle bisector divides an angle into two equal halves.

2

Set up an equation using triangle

\triangle APB

.

In

\triangle APB

,

\angle APB = 90^\circ

because ray

AD \perp

ray

BF

. Therefore,

\angle PAB + \angle ABP = 90^\circ \implies (1.5x - 5) + (x + 35) = 90 \implies 2.5x + 30 = 90 \implies 2.5x = 60 \implies x = 24

.

The acute angles in a right triangle sum to

90^\circ

.

3

Calculate the measures of

\angle CAB

and

\angle ABQ

.

\angle CAB = 3(24) - 10 = 62^\circ

, and

\angle ABQ = \angle ABF = 24 + 35 = 59^\circ

.

Substitute

x = 24

back into the original angle expressions.

4

Determine the acute angle

\angle AQB

in triangle

\triangle ABQ

.

In

\triangle ABQ

, points

Q, C, A

lie on line

k

, so

\angle QAB = \angle CAB = 62^\circ

. Sum of angles in

\triangle ABQ

:

\angle AQB = 180^\circ - (62^\circ + 59^\circ) = 180^\circ - 121^\circ = 59^\circ

.

The interior angles of any triangle sum to

180^\circ

.

5

Find the supplementary obtuse angle at intersection point

Q

.

Obtuse angle

= 180^\circ - 59^\circ = 121^\circ

.

Angles forming a linear pair on a straight line are supplementary.

Key Concept

Parallel Lines, Transversals, Angle Bisectors, and Triangle Angle Sum Theorem

Question 51Question →

A square is inscribed in a circle with center $O$ . The perimeter of the square is $16\sqrt{2}$ units. A sector of this circle has an area equal to the total area of the region inside the circle that lies outside the square. What is the arc length of this sector?

4\pi - 8

8\pi - 16

16\pi - 32

16\pi - 8

8\pi - 32

Show answer & explanation

Answer: $8\pi - 16$

Answer

8\pi - 16

The correct answer is

8\pi - 16

. The inscribed square has side length

4\sqrt{2}

and diagonal

8

. Since the diagonal of an inscribed square is the circle's diameter, the circle has radius

r = 4

and total area

16\pi

. The square has area

32

, so the region outside the square has area

16\pi - 32

. For any sector of radius

r

, the arc length

L

and sector area

A

satisfy

L = \frac{2A}{r}

. Substituting

A = 16\pi - 32

and

r = 4

yields

L = \frac{2(16\pi - 32)}{4} = 8\pi - 16

.

Step-by-Step Solution

1

Find the side length and diagonal of the inscribed square.

Side length

s = \frac{16\sqrt{2}}{4} = 4\sqrt{2}

. Diagonal

d = s\sqrt{2} = (4\sqrt{2})\sqrt{2} = 8

.

The perimeter of a square is

4s

, and the diagonal of a square with side

s

is

s\sqrt{2}

.

2

Determine the radius and total area of the circle.

Diameter equals diagonal

d = 8

, so radius

r = 4

. Area of circle

A_{\text{circle}} = \pi r^2 = 16\pi

.

A square inscribed in a circle has its diagonal aligned with the diameter of the circle.

3

Calculate the area of the region inside the circle but outside the square.

Area of square

A_{\text{square}} = (4\sqrt{2})^2 = 32

. Area outside square

A_{\text{outside}} = 16\pi - 32

.

Subtract the area of the inscribed square from the total area of the circle.

4

Relate sector area to arc length using

L = \frac{2 A_{\text{sector}}}{r}

.

L = \frac{2(16\pi - 32)}{4} = \frac{32\pi - 64}{4} = 8\pi - 16

.

Since sector area

A_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2

and arc length

L = \frac{\theta}{360^\circ}(2\pi r)

, we have

L = \frac{2A_{\text{sector}}}{r}

.

Key Concept

Relationship between circle area, inscribed figures, sector area, and arc length

Estimated Time:1m 30s

Question 52Question →

Four rays, $\vec{OA}$ , $\vec{OB}$ , $\vec{OC}$ , and $\vec{OD}$ , radiate from a common point $O$ in consecutive clockwise order such that $\vec{OA} \perp \vec{OC}$ and $\vec{OB} \perp \vec{OD}$ . If the measure of angle $\angle AOD$ is $3.5$ times the measure of angle $\angle BOC$ , what is the measure, in degrees, of angle $\angle AOB$ ?

36^\circ

40^\circ

50^\circ

65^\circ

140^\circ

Show answer & explanation

Answer: $50^\circ$

Answer

50^\circ

Because

\vec{OA} \perp \vec{OC}

and

\vec{OB} \perp \vec{OD}

, we know

\angle AOB + \angle BOC = 90^\circ

and

\angle BOC + \angle COD = 90^\circ

, which implies

\angle AOB = \angle COD

. The total angle

\angle AOD = \angle AOB + \angle BOD = \angle AOB + 90^\circ

. Using the given condition

\angle AOD = 3.5 \times \angle BOC

, we substitute

\angle BOC = 90^\circ - \angle AOB

to get

\angle AOB + 90^\circ = 3.5(90^\circ - \angle AOB)

, which solves to

\angle AOB = 50^\circ

.

Step-by-Step Solution

1

Set up angle variable definitions and perpendicular relationships.

Let

\angle AOB = x

,

\angle BOC = y

, and

\angle COD = z

. Since

\vec{OA} \perp \vec{OC}

, we have

x + y = 90^\circ

. Since

\vec{OB} \perp \vec{OD}

, we have

y + z = 90^\circ

.

Perpendicular rays form right angles measuring

90^\circ

.

2

Deduce the relationship between

x

,

y

, and

z

, and express

\angle AOD

in terms of

x

.

Subtracting

y

from both equations gives

x = 90^\circ - y

and

z = 90^\circ - y

, so

x = z

. Thus,

\angle AOD = x + y + z = x + 90^\circ

.

Adjacent angles sharing a vertex add up to form the overall combined angle.

3

Formulate and solve the equation based on the given ratio.

We are given

\angle AOD = 3.5 \times \angle BOC

, so

x + 90^\circ = 3.5y

. Substituting

y = 90^\circ - x

yields

x + 90^\circ = 3.5(90^\circ - x) \implies x + 90^\circ = 315^\circ - 3.5x \implies 4.5x = 225^\circ \implies x = 50^\circ

.

Substitution creates a single linear equation in terms of

x = \angle AOB

.

Key Concept

Perpendicular Ray Systems and Angle Addition

Estimated Time:1m 30s

Question 53Question →

In circle $O$ , minor arc $AB$ is subtended by a central angle of $120^\circ$ , and the radius of circle $O$ is $6$ . A smaller circle, circle $C$ , is constructed such that it is tangent to chord $AB$ at its midpoint and tangent to minor arc $AB$ at its midpoint. What is the area of a sector of circle $C$ subtended by a central angle of $90^\circ$ ?

\frac{9\pi}{16}

\frac{9\pi}{8}

\frac{9\pi}{4}

\frac{27\pi}{16}

9\pi

Show answer & explanation

Answer: $\frac{9\pi}{16}$

Answer

\frac{9\pi}{16}

The distance from the center of circle

O

to chord

AB

is

6 \cos(60^\circ) = 3

, leaving a height of

6 - 3 = 3

for the minor segment. Circle

C

is tangent to the chord and arc midpoints, making its diameter equal to

3

and its radius

\frac{3}{2}

. The area of a

90^\circ

sector of circle

C

is

\frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{9\pi}{16}

.

Step-by-Step Solution

1

Find the perpendicular distance from center

O

to chord

AB

.

In isosceles triangle

OAB

with

OA = OB = 6

and central angle

\angle AOB = 120^\circ

, the perpendicular bisector from

O

to

AB

meets

AB

at midpoint

M

.

\angle AOM = 60^\circ

, so

OM = OA \cos(60^\circ) = 6 \times \frac{1}{2} = 3

.

Determining

OM

allows us to find the sagitta (height) of the minor segment bounded by chord

AB

and minor arc

AB

.

2

Calculate the height of the minor segment (diameter of circle

C

).

The height of the minor segment along the radius passing through

M

to arc midpoint

N

is

MN = ON - OM = 6 - 3 = 3

. Since circle

C

is tangent to

AB

at

M

and to the arc at

N

, segment

MN

is a diameter of circle

C

, giving a diameter of

3

.

The space between the chord midpoint and arc midpoint bounds circle

C

, defining its diameter.

3

Find the radius of circle

C

.

Radius

r_C = \frac{\text{diameter}}{2} = \frac{3}{2}

.

The radius of circle

C

is needed to calculate its sector area.

4

Compute the area of the

90^\circ

sector of circle

C

.

\text{Area} = \frac{\theta}{360^\circ} \pi r_C^2 = \frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{1}{4} \pi \left(\frac{9}{4}\right) = \frac{9\pi}{16}.

Multiply the fraction of the full circle represented by the central angle by the total area of circle

C

.

Key Concept

Calculating sector area of an inscribed circle within a circle segment

Estimated Time:2m 0s

Question 54Question →

In a circle centered at point $O$ with a radius of $6$ units, points $P$ and $Q$ lie on the circle such that central angle $\angle POQ$ measures $120^\circ$ . Which of the following statements are true? Select all such statements.

Select all that apply

The length of minor arc

PQ

is

4\pi

.

The area of minor sector

POQ

is

12\pi

.

The perimeter of sector

POQ

is

4\pi

.

The area of sector

POQ

is

36\pi

.

The length of chord

PQ

is

6\sqrt{3}

.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ .; The area of minor sector $POQ$ is $12\pi$ .; The length of chord $PQ$ is $6\sqrt{3}$ .

Answer

The true statements are that the length of minor arc

PQ

is

4\pi

, the area of minor sector

POQ

is

12\pi

, and the length of chord

PQ

is

6\sqrt{3}

.

The arc length of a

120^\circ

central angle is

\frac{1}{3}

of the total circumference

12\pi

, which gives

4\pi

. The sector area is

\frac{1}{3}

of the total area

36\pi

, which gives

12\pi

. Furthermore, by geometry of an isosceles triangle with vertex angle

120^\circ

and congruent sides

6

, the base chord length is

2 \times 6 \sin(60^\circ) = 6\sqrt{3}

.

Step-by-Step Solution

1

Calculate the arc length of minor arc

PQ

Arc length =

4\pi

Arc length equals

\frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi(6) = 4\pi

.

2

Calculate the area of sector

POQ

Sector area =

12\pi

Sector area equals

\frac{\theta}{360^\circ} \times \pi r^2 = \frac{120^\circ}{360^\circ} \times \pi(6^2) = 12\pi

.

3

Determine the perimeter of sector

POQ

Perimeter =

4\pi + 12

The boundary of a sector consists of the curved arc length plus two straight radii:

4\pi + 6 + 6 = 4\pi + 12

.

4

Calculate the length of chord

PQ

Chord

PQ = 6\sqrt{3}

Using triangle

OPQ

with sides

OP = OQ = 6

and

\angle POQ = 120^\circ

, law of cosines or bisecting

\angle POQ

into two

30^\circ

-

60^\circ

-

90^\circ

triangles yields chord length

2 \times (6 \sin 60^\circ) = 2 \times 3\sqrt{3} = 6\sqrt{3}

.

Key Concept

Arc Length, Sector Area, and Chord Relationships in Circles

Estimated Time:1m 30s

Question 55Question →

A solid rectangular box has a length of $6$ centimeters, a width of $4$ centimeters, and a height of $5$ centimeters. What is the total surface area, in square centimeters, of the box?

74

120

148

240

288

Show answer & explanation

Answer: $148$

Answer

The total surface area of the rectangular box is

148

square centimeters.

The total surface area of a rectangular solid with length

l

, width

w

, and height

h

is given by

2(lw + lh + wh)

. Substituting

l = 6

,

w = 4

, and

h = 5

yields

2(6 \cdot 4 + 6 \cdot 5 + 4 \cdot 5) = 2(24 + 30 + 20) = 2(74) = 148

square centimeters.

Step-by-Step Solution

1

Identify the given dimensions of the rectangular box.

Length

l = 6

cm, width

w = 4

cm, and height

h = 5

cm.

These dimensions are required to compute the area of each face.

2

Calculate the surface area of the three distinct face pairs.

Top and bottom faces:

6 \times 4 = 24

; Front and back faces:

6 \times 5 = 30

; Left and right side faces:

4 \times 5 = 20

.

A rectangular solid has six rectangular faces grouped into three identical pairs.

3

Apply the surface area formula

A = 2(lw + lh + wh)

.

A = 2(24 + 30 + 20) = 2(74) = 148

square centimeters.

Summing the area of all six faces gives the total surface area.

Key Concept

Surface Area of a Rectangular Solid

Question 56Question →

In a circle with center $O$ and radius $12$ , radii $OA$ and $OB$ are perpendicular. Point $C$ lies on segment $OA$ such that $C$ is the midpoint of $OA$ . A line segment perpendicular to $OA$ is drawn from point $C$ to intersect minor arc $AB$ at point $D$ . What is the area of the region bounded by line segment $CD$ , line segment $CA$ , and minor arc $AD$ ?

12\pi

24\pi - 18

24\pi - 18\sqrt{3}

36\pi - 18\sqrt{3}

24\pi - 36\sqrt{3}

Show answer & explanation

Answer: $24\pi - 18\sqrt{3}$

Answer

24\pi - 18\sqrt{3}

The area of the region bounded by line segment

CD

, segment

CA

, and minor arc

AD

is obtained by subtracting the area of right triangle

OCD

from the area of sector

OAD

. With

OC = 6

and radius

OD = 12

, triangle

OCD

is a

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

right triangle, giving

CD = 6\sqrt{3}

and central angle

\angle AOD = 60^\circ

. The sector area is

\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = 24\pi

, and the triangle area is

\frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}

. Subtracting the triangle area from the sector area yields

24\pi - 18\sqrt{3}

.

Step-by-Step Solution

1

Determine the length of

OC

and the height

CD

OC = 6

and

CD = 6\sqrt{3}

Since

C

is the midpoint of radius

OA = 12

,

OC = 6

. Triangle

OCD

is a right triangle at

C

with hypotenuse

OD = 12

(radius of circle). By the Pythagorean theorem,

CD = \sqrt{12^2 - 6^2} = \sqrt{108} = 6\sqrt{3}

.

2

Find the central angle

\angle AOD

\angle AOD = 60^\circ

In right triangle

OCD

,

\cos(\angle AOD) = \frac{OC}{OD} = \frac{6}{12} = \frac{1}{2}

, which implies

\angle AOD = 60^\circ

.

3

Calculate the area of sector

OAD

\text{Area(Sector } OAD) = 24\pi

$

The area of a sector with central angle

60^\circ

and radius

12

is

\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = \frac{1}{6} \cdot 144\pi = 24\pi

.

4

Calculate the area of right triangle

OCD

\text{Area(Triangle } OCD) = 18\sqrt{3}

The area of right triangle

OCD

is

\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}

.

5

Subtract the area of triangle

OCD

from the area of sector

OAD

\text{Area of shaded region} = 24\pi - 18\sqrt{3}

The bounded region is formed by removing triangle

OCD

from sector

OAD

.

Key Concept

Calculating the area of a region bounded by a circle arc and line segments by subtracting a right triangle area from a sector area.

Question 57Question →

In a plane, lines $l_1$ and $l_2$ are parallel ( $l_1 \parallel l_2$ ). A transversal line $t$ intersects $l_1$ at point $P$ and $l_2$ at point $Q$ . The measures of two consecutive interior angles on the same side of transversal $t$ are $(4x - 10)^\circ$ on line $l_1$ and $(3x + 50)^\circ$ on line $l_2$ . A third line $l_3$ passes through point $P$ and is perpendicular to line $l_2$ .

Which of the following statements MUST be true? Select all that apply.

Select all that apply

x = 20

The acute angle formed between line

t

and line

l_3

at point

P

measures

20^\circ

.

The obtuse angle formed at point

Q

between line

t

and line

l_2

measures

120^\circ

.

Line

l_3

is parallel to transversal line

t

.

The sum of all four interior angles located between lines

l_1

and

l_2

along transversal line

t

is

360^\circ

.

Show answer & explanation

Answer: $x = 20$ ; The acute angle formed between line $t$ and line $l_3$ at point $P$ measures $20^\circ$ .; The sum of all four interior angles located between lines $l_1$ and $l_2$ along transversal line $t$ is $360^\circ$ .

Answer

The correct statements are that

x = 20

, the acute angle formed between line

t

and line

l_3

at point

P

measures

20^\circ

, and the sum of all four interior angles located between lines

l_1

and

l_2

along transversal line

t

is

360^\circ

.

The statements confirming

x = 20

, the

20^\circ

acute angle between lines

t

and

l_3

, and the

360^\circ

interior angle sum are all correct because consecutive interior angles sum to

180^\circ

, perpendicular lines create

90^\circ

angles, and the four interior angles sum to

360^\circ

.

Step-by-Step Solution

1

Set up an equation for consecutive interior angles.

(4x - 10) + (3x + 50) = 180^\circ

Consecutive interior angles on the same side of a transversal intersecting parallel lines are supplementary.

2

Solve the algebraic equation for

x

.

7x + 40 = 180 \implies 7x = 140 \implies x = 20

Combine like terms and isolate

x

.

3

Calculate the interior angle measures at points

P

and

Q

.

Angle at

P = 4(20) - 10 = 70^\circ

; Angle at

Q = 3(20) + 50 = 110^\circ

Substitute

x = 20

back into the original expressions.

4

Determine perpendicular relationships and angle measure between line

t

and line

l_3

.

Angle between

t

and

l_3

at

P

is

90^\circ - 70^\circ = 20^\circ

Line

l_3 \perp l_2

implies

l_3 \perp l_1

because

l_1 \parallel l_2

. Thus

l_3

forms a

90^\circ

angle with

l_1

at

P

.

5

Sum the four interior angles between lines

l_1

and

l_2

.

70^\circ + 110^\circ + 70^\circ + 110^\circ = 360^\circ

The four interior angles comprise two pairs of interior angles measuring

70^\circ

and

110^\circ

each.

Key Concept

Parallel line angle theorems (consecutive interior angles, alternate interior angles) and perpendicular line relationships.

Question 58Question →

An isosceles triangle has a base of length $10$ inches and a perimeter of $36$ inches. What is the area of the triangle, in square inches?

30

60

65

120

130

Show answer & explanation

Answer: $60$

Answer

60

square inches

Subtracting the base length of

10

inches from the total perimeter of

36

inches leaves

26

inches for the remaining two congruent sides, giving

13

inches each. The perpendicular height divides the isosceles triangle into two right triangles with base

5

inches and hypotenuse

13

inches. By the Pythagorean theorem, the height is

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

inches. Substituting base

10

and height

12

into the area formula

\frac{1}{2} \times \text{base} \times \text{height}

gives

\frac{1}{2} \times 10 \times 12 = 60

square inches.

Step-by-Step Solution

1

Determine the length of the two congruent sides of the isosceles triangle.

Each side has a length of

13

inches.

Since the perimeter is

36

inches and the base is

10

inches, the sum of the two equal sides is

36 - 10 = 26

inches. Dividing by

2

yields

13

inches per side.

2

Calculate the perpendicular height of the triangle using the Pythagorean theorem.

The height of the triangle is

12

inches.

Dropping an altitude from the top vertex to the base bisects the base into two segments of

5

inches. This creates two right triangles with a base of

5

inches and hypotenuse of

13

inches. Thus,

h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12

inches.

3

Calculate the area of the isosceles triangle.

The area is

60

square inches.

Using the triangle area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, we compute

\frac{1}{2} \times 10 \times 12 = 60

.

Key Concept

Properties of Isosceles Triangles and Area Formula

Question 59Question →

A solid right circular cone has a base radius of $r$ and a height of $h = \frac{4}{3}r$ . A solid sphere has a radius of $R$ . If the total surface area of the cone is equal to the total surface area of the sphere, which of the following statements must be true? Select all that apply.

Select all that apply

The base radius of the cone,

r

, is strictly greater than the radius of the sphere,

R

.

The volume of the cone is strictly less than the volume of the sphere.

The ratio of the volume of the cone to the volume of the sphere is

\frac{\sqrt{6}}{4}

.

The volume of the cone is strictly greater than the volume of the sphere.

The ratio of the base radius of the cone to the radius of the sphere,

\frac{r}{R}

, is equal to

\frac{3}{2}

.

Show answer & explanation

Answer: The base radius of the cone, $r$ , is strictly greater than the radius of the sphere, $R$ .; The volume of the cone is strictly less than the volume of the sphere.; The ratio of the volume of the cone to the volume of the sphere is $\frac{\sqrt{6}}{4}$ .

Answer

The statements confirming that the base radius of the cone is strictly greater than the radius of the sphere, that the volume of the cone is strictly less than the volume of the sphere, and that the ratio of the volume of the cone to the volume of the sphere is

\frac{\sqrt{6}}{4}

are all correct.

Equating the total surface area of the cone

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

with the surface area of the sphere

4\pi R^2

yields

\frac{r^2}{R^2} = \frac{3}{2}

, which simplifies to

\frac{r}{R} = \sqrt{\frac{3}{2}}

. Because

\sqrt{1.5} > 1

, the base radius of the cone is strictly greater than the radius of the sphere. Furthermore, evaluating the ratio of their volumes gives

\frac{V_{\text{cone}}}{V_{\text{sphere}}} = \frac{\frac{4}{9}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{1}{3}\left(\frac{r}{R}\right)^3 = \frac{\sqrt{6}}{4} \approx 0.612

. Because this ratio is strictly less than

1

, the volume of the cone is strictly less than the volume of the sphere. Thus, the three true statements are those stating

r > R

, that the volume of the cone is strictly less than the sphere's volume, and that their volume ratio is

\frac{\sqrt{6}}{4}

.

Step-by-Step Solution

1

Calculate the slant height and total surface area of the cone in terms of

r

.

Slant height

l = \sqrt{r^2 + \left(\frac{4}{3}r\right)^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r

. Total surface area

A_{\text{cone}} = \pi r^2 + \pi r l = \pi r^2 + \pi r\left(\frac{5}{3}r\right) = \frac{8}{3}\pi r^2

.

The total surface area of a right circular cone is the sum of its base area

\pi r^2

and lateral area

\pi r l

.

2

Equate the total surface area of the cone to the total surface area of the sphere to find the ratio

\frac{r}{R}

.

\frac{8}{3}\pi r^2 = 4\pi R^2 \implies 2 r^2 = 3 R^2 \implies \frac{r^2}{R^2} = \frac{3}{2} \implies \frac{r}{R} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}.

The surface area of a sphere of radius

R

is

4\pi R^2

.

3

Compare the linear dimensions

r

and

R

.

Since

\frac{r}{R} = \sqrt{1.5} \approx 1.225 > 1

, it follows that

r > R

.

A ratio greater than

1

implies the numerator is larger than the denominator.

4

Express the volumes of both solids and compute their ratio

\frac{V_{\text{cone}}}{V_{\text{sphere}}}

.

V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \left(\frac{4}{3}r\right) = \frac{4}{9}\pi r^3

.

V_{\text{sphere}} = \frac{4}{3}\pi R^3

. Therefore,

\frac{V_{\text{cone}}}{V_{\text{sphere}}} = \frac{\frac{4}{9}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{1}{3}\left(\frac{r}{R}\right)^3 = \frac{1}{3}\left(\frac{3}{2}\right)^{3/2} = \frac{1}{3} \cdot \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{4}

.

Using standard volume formulas for cones and spheres and substituting the known linear dimension ratio.

5

Determine whether the cone's volume is greater than or less than the sphere's volume.

Since

\frac{\sqrt{6}}{4} \approx 0.612 < 1

,

V_{\text{cone}} < V_{\text{sphere}}

.

A volume ratio less than

1

proves the cone has a smaller volume than the sphere.

Key Concept

Analyzing geometric scaling, volume, and total surface area relations between cones and spheres using variable constraints.

Question 60Question →

In the $xy$ -plane, line $l$ passes through the points $(2, 5)$ and $(6, 13)$ . What is the slope of line $l$ ?

Show answer & explanation

Answer: 2

Answer

The slope of line

l

is 2.

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting

(2, 5)

and

(6, 13)

into the formula gives

m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2

.

Step-by-Step Solution

1

Identify the coordinates of the two given points on line

l

.

The points are

(x_1, y_1) = (2, 5)

and

(x_2, y_2) = (6, 13)

.

Two points are required to calculate the slope of a straight line.

2

Apply the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

.

m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2

.

The slope measures the vertical change (rise) divided by the horizontal change (run).

Key Concept

Slope of a line through two points

Geometry

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