Geometry Practice Questions — GRE General Test — Page 8

Question 141Question →

In right triangle $ABC$ , the right angle is at vertex $B$ , $AB = 6$ , and the measure of $\angle ACB$ is $30^\circ$ . Point $P$ lies on segment $BC$ such that the measure of $\angle APB$ is $60^\circ$ . Which of the following statements must be true? Select all that apply.

Select all that apply

The length of segment

AP

is

4\sqrt{3}

.

The length of segment

PC

is

4\sqrt{3}

.

The area of triangle

APC

is

18\sqrt{3}

.

The perimeter of triangle

ABC

is

18 + 6\sqrt{3}

.

The length of segment

BP

is

3\sqrt{3}

.

Show answer & explanation

Answer: The length of segment $AP$ is $4\sqrt{3}$ .; The length of segment $PC$ is $4\sqrt{3}$ .; The perimeter of triangle $ABC$ is $18 + 6\sqrt{3}$ .

Answer

The correct statements are that the length of segment

AP

is

4\sqrt{3}

, the length of segment

PC

is

4\sqrt{3}

, and the perimeter of triangle

ABC

is

18 + 6\sqrt{3}

.

The 30-60-90 triangle ratio (

1 : \sqrt{3} : 2

) establishes

AC = 12

,

BC = 6\sqrt{3}

,

AP = 4\sqrt{3}

, and

BP = 2\sqrt{3}

. Consequently,

PC = 4\sqrt{3}

and the perimeter of triangle

ABC

equals

18 + 6\sqrt{3}

. Thus, the statements asserting that

AP = 4\sqrt{3}

,

PC = 4\sqrt{3}

, and the perimeter of triangle

ABC

is

18 + 6\sqrt{3}

are all correct.

Step-by-Step Solution

1

Analyze main triangle

ABC

Hypotenuse

AC = 12

and base

BC = 6\sqrt{3}

Triangle

ABC

is a 30-60-90 triangle with side opposite

30^\circ

equal to

AB = 6

. Therefore, hypotenuse

AC = 2(6) = 12

and leg

BC = 6\sqrt{3}

.

2

Analyze sub-triangle

ABP

Leg

BP = 2\sqrt{3}

and hypotenuse

AP = 4\sqrt{3}

Triangle

ABP

is a 30-60-90 triangle with side opposite

60^\circ

equal to

AB = 6

. The shorter leg is

BP = \frac{6}{\sqrt{3}} = 2\sqrt{3}

and the hypotenuse is

AP = 2(2\sqrt{3}) = 4\sqrt{3}

.

3

Calculate segment

PC

and area of triangle

APC

PC = 4\sqrt{3}

and

\text{Area}(\triangle APC) = 12\sqrt{3}

PC = BC - BP = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}

. The area of

\triangle APC

is

\frac{1}{2} \times PC \times AB = \frac{1}{2}(4\sqrt{3})(6) = 12\sqrt{3}

.

4

Calculate perimeter of triangle

ABC

Perimeter

= 18 + 6\sqrt{3}

Sum of sides

AB + BC + AC = 6 + 6\sqrt{3} + 12 = 18 + 6\sqrt{3}

.

Key Concept

Properties of 30-60-90 Special Right Triangles and Side Length Ratios (

1 : \sqrt{3} : 2

)

Question 142Question →

In convex quadrilateral $ABCD$ , the diagonals $AC$ and $BD$ intersect at point $P$ at right angles ( $AC \perp BD$ ). The length of diagonal $AC$ is $16$ and the length of diagonal $BD$ is $12$ . Points $E$ , $F$ , $G$ , and $H$ are the midpoints of sides $AB$ , $BC$ , $CD$ , and $DA$ , respectively. Which of the following statements MUST be true? Select all such statements.

Select all that apply

The perimeter of quadrilateral

EFGH

is

28

.

Quadrilateral

EFGH

is a rectangle.

The area of quadrilateral

ABCD

is

96

.

The area of quadrilateral

EFGH

is

96

.

Quadrilateral

EFGH

is a rhombus.

Show answer & explanation

Answer: The perimeter of quadrilateral $EFGH$ is $28$ .; Quadrilateral $EFGH$ is a rectangle.; The area of quadrilateral $ABCD$ is $96$ .

Answer

The true statements are that the perimeter of quadrilateral

EFGH

is

28

, quadrilateral

EFGH

is a rectangle, and the area of quadrilateral

ABCD

is

96

.

Applying the Midpoint Theorem shows that midsegments

EF

and

GH

are parallel to

AC

with length

8

, while

FG

and

HE

are parallel to

BD

with length

6

. This gives a perimeter of

8 + 6 + 8 + 6 = 28

. Because

AC \perp BD

, the adjacent midsegments meet at

90^\circ

, confirming that quadrilateral

EFGH

is a rectangle. Additionally, for any orthodiagonal quadrilateral, the area is

\frac{1}{2} d_1 d_2 = \frac{1}{2} \times 16 \times 12 = 96

.

Step-by-Step Solution

1

Determine the side lengths of midpoint quadrilateral

EFGH

using the Triangle Midpoint Theorem.

EF = GH = \frac{1}{2}AC = 8

and

FG = HE = \frac{1}{2}BD = 6

.

In any triangle, the segment connecting the midpoints of two sides is parallel to the third side and half its length.

2

Calculate the perimeter of quadrilateral

EFGH

.

Perimeter =

EF + FG + GH + HE = 8 + 6 + 8 + 6 = 28

.

The perimeter is the sum of all four side lengths of the quadrilateral.

3

Determine the shape classification of quadrilateral

EFGH

.

EFGH

is a rectangle.

Since

EF \parallel AC

and

FG \parallel BD

, the angle between

EF

and

FG

equals the angle between diagonals

AC

and

BD

. Given

AC \perp BD

, the angle

\angle EFG = 90^\circ

. Opposite sides are equal and parallel with right angles, making

EFGH

a rectangle (and not a rhombus, as adjacent sides

8 \neq 6

).

4

Calculate the area of quadrilateral

ABCD

and quadrilateral

EFGH

.

Area(

ABCD

) =

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

; Area(

EFGH

) =

8 \times 6 = 48

.

The area of a quadrilateral with perpendicular diagonals is half the product of its diagonal lengths. The midpoint quadrilateral has half the area of the outer quadrilateral.

Key Concept

Midpoint Theorem (Varignon's Theorem) and Area of Orthodiagonal Quadrilaterals

Estimated Time:2m 0s

Question 143Question →

A right rectangular prism has a square base. The diagonal of the base has a length of $6\sqrt{2}$ units. A space diagonal of the prism makes a $30^\circ$ angle with the diagonal of the base. What is the height of the prism?

2\sqrt{6}

6\sqrt{6}

6\sqrt{2}

12\sqrt{2}

4\sqrt{6}

Show answer & explanation

Answer: $2\sqrt{6}$

Answer

2\sqrt{6}

The height of a right rectangular prism is perpendicular to the base, forming a right triangle with the base diagonal as one leg and the space diagonal as the hypotenuse. Given that the angle between the space diagonal and the base diagonal is

30^\circ

, this right triangle is a

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

triangle. The base diagonal of length

6\sqrt{2}

is adjacent to the

30^\circ

angle, making it the longer leg (

x\sqrt{3}

). Dividing

6\sqrt{2}

by

\sqrt{3}

gives

h = 2\sqrt{6}

, which correctly matches the length of the leg opposite the

30^\circ

angle.

Step-by-Step Solution

1

Identify the right triangle inside the prism.

The vertical right triangle has legs

d

(base diagonal) and

h

(height), with hypotenuse

D

(space diagonal).

The height of a right prism is perpendicular to its base, forming a right angle with any line segment lying in the base, including the base diagonal.

2

Apply special right triangle ratios for a

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

triangle.

The base diagonal

d = 6\sqrt{2}

is adjacent to the

30^\circ

angle, so

d = x\sqrt{3}

, where

x = h

is the height opposite the

30^\circ

angle.

In a

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

triangle, side lengths follow the ratio

1 : \sqrt{3} : 2

.

3

Solve for height

h

.

h = \frac{6\sqrt{2}}{\sqrt{3}} = \frac{6\sqrt{6}}{3} = 2\sqrt{6}

.

Rationalize the denominator by multiplying top and bottom by

\sqrt{3}

.

Key Concept

Special Right Triangles (

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

) in 3D Space Diagonals

Estimated Time:1m 30s

Question 144Question →

In rhombus $ABCD$ , the side length is $10$ and the length of diagonal $BD$ is $12$ . Line segment $AP$ is drawn perpendicular to side $BC$ , with point $P$ lying on segment $BC$ . What is the length of segment $AP$ ?

Show answer & explanation

Answer: 9.6

Answer

9.6

The diagonals of rhombus

ABCD

intersect perpendicularly at

O

and bisect each other. Given

BD = 12

, half of the diagonal is

BO = 6

. Right triangle

AOB

has hypotenuse

AB = 10

and leg

BO = 6

, so by the Pythagorean theorem, leg

AO = \sqrt{10^2 - 6^2} = 8

. Thus, diagonal

AC = 16

. The area of rhombus

ABCD

is

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

. The area is also equal to

\text{base} \times \text{height} = BC \times AP = 10 \times AP

. Setting

10 \times AP = 96

gives

AP = 9.6

.

Step-by-Step Solution

1

Find half the length of diagonal BD.

Segment BO = 6.

The diagonals of a rhombus bisect each other at right angles.

2

Apply the Pythagorean theorem to right triangle AOB to determine half of diagonal AC.

AO = 8, so diagonal AC = 16.

Triangle AOB has hypotenuse 10 and leg 6, forming a 6-8-10 Pythagorean triple.

3

Calculate the total area of rhombus ABCD from its diagonal lengths.

Area = 96.

The area of a rhombus equals half the product of its two diagonals.

4

Use the alternative area formula (base × height) to solve for altitude AP.

AP = 9.6.

Base BC = 10 and height AP give Area = 10 × AP = 96.

Key Concept

Properties of rhombus diagonals, Pythagorean theorem, and dual area formulas for quadrilaterals

Question 145Question →

In triangle $ABC$ , the measure of angle $A$ is $45^\circ$ and the measure of angle $B$ is $105^\circ$ . The length of side $AB$ is $6$ units. Which of the following statements must be true? Select all that apply.

Select all that apply

The length of the altitude from vertex

B

to side

AC

is

3\sqrt{2}

units.

The length of side

BC

is

6\sqrt{2}

units.

The area of triangle

ABC

is

9 + 9\sqrt{3}

square units.

The length of side

AC

is

6\sqrt{2} + 6\sqrt{6}

units.

The area of triangle

ABC

is

18 + 18\sqrt{3}

square units.

Show answer & explanation

Answer: The length of the altitude from vertex $B$ to side $AC$ is $3\sqrt{2}$ units.; The length of side $BC$ is $6\sqrt{2}$ units.; The area of triangle $ABC$ is $9 + 9\sqrt{3}$ square units.

Answer

The correct statements are: the length of the altitude from vertex

B

to side

AC

is

3\sqrt{2}

units, the length of side

BC

is

6\sqrt{2}

units, and the area of triangle

ABC

is

9 + 9\sqrt{3}

square units.

By drawing altitude

BD

perpendicular to

AC

, triangle

ABC

decomposes into two special right triangles. In the

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

triangle

ABD

, the hypotenuse is

6

, yielding leg lengths

BD = AD = 3\sqrt{2}

. In the

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

triangle

BCD

, side

BD = 3\sqrt{2}

is opposite the

30^\circ

angle, making hypotenuse

BC = 2 \times 3\sqrt{2} = 6\sqrt{2}

and long leg

CD = 3\sqrt{6}

. Combining

AD

and

CD

gives base

AC = 3\sqrt{2} + 3\sqrt{6}

, leading to an area of

\frac{1}{2}(3\sqrt{2} + 3\sqrt{6})(3\sqrt{2}) = 9 + 9\sqrt{3}

.

Step-by-Step Solution

1

Determine the third angle of triangle

ABC

\angle C = 180^\circ - (45^\circ + 105^\circ) = 30^\circ

.

The sum of interior angles in any triangle is

180^\circ

.

2

Drop an altitude

BD

perpendicular to side

AC

Altitude

BD

splits triangle

ABC

into two right triangles:

\triangle ABD

(

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

) and

\triangle BCD

(

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

).

In

\triangle ABD

,

\angle A = 45^\circ

and

\angle ADB = 90^\circ

, leaving

\angle ABD = 45^\circ

. In

\triangle BCD

,

\angle CBD = 105^\circ - 45^\circ = 60^\circ

and

\angle C = 30^\circ

.

3

Calculate side lengths in

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

triangle

ABD

AD = BD = \frac{AB}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}

units.

The ratio of sides in a

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

triangle is

1:1:\sqrt{2}

.

4

Calculate side lengths in

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

triangle

BCD

Hypotenuse

BC = 2 \times BD = 6\sqrt{2}

units, and long leg

CD = BD \times \sqrt{3} = 3\sqrt{2} \times \sqrt{3} = 3\sqrt{6}

units.

The ratio of sides opposite

30^\circ:60^\circ:90^\circ

is

1:\sqrt{3}:2

.

5

Calculate total base

AC

and area of triangle

ABC

AC = AD + CD = 3\sqrt{2} + 3\sqrt{6}

units. Area =

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (3\sqrt{2} + 3\sqrt{6})(3\sqrt{2}) = \frac{1}{2} (18 + 18\sqrt{3}) = 9 + 9\sqrt{3}

square units.

Standard formula for triangle area is

\frac{1}{2} b h

.

Key Concept

Decomposing an oblique triangle with

45^\circ

and

30^\circ

angles into

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

and

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

special right triangles.

Estimated Time:2m 0s

Question 146Question →

A robot starting at point $P$ travels due east for $4$ meters to point $Q$ . At point $Q$ , the robot turns $60^\circ$ counterclockwise from its original direction and travels in a straight line for $8$ meters, stopping at point $R$ . What is the straight-line distance, in meters, between point $P$ and point $R$ ?

4\sqrt{3}

4\sqrt{5}

4\sqrt{7}

8 + 4\sqrt{3}

12

Show answer & explanation

Answer: $4\sqrt{7}$

Answer

The straight-line distance between point

P

and point

R

is

4\sqrt{7}

meters.

Dropping a vertical perpendicular from

R

to the extended line

PQ

forms a

30^\circ

-

60^\circ

-

90^\circ

right triangle

QSR

with hypotenuse

8

. The base extension

QS

equals

4

and the altitude

RS

equals

4\sqrt{3}

. Applying the Pythagorean theorem to the right triangle

PSR

with legs

PS = 4 + 4 = 8

and

RS = 4\sqrt{3}

gives

PR = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{64 + 48} = \sqrt{112} = 4\sqrt{7}

meters.

Step-by-Step Solution

1

Construct a right triangle by extending segment

PQ

past

Q

to a point

S

directly below

R

, dropping altitude

RS \perp PS

.

Triangle

QSR

is formed with

\angle RQS = 60^\circ

and

\angle QSR = 90^\circ

.

Decomposing the angled path into perpendicular horizontal and vertical components allows the use of special right triangles and the Pythagorean theorem.

2

Use

30^\circ

-

60^\circ

-

90^\circ

side ratios (

1 : \sqrt{3} : 2

) in triangle

QSR

where hypotenuse

QR = 8

.

Horizontal segment

QS = 8 \times \frac{1}{2} = 4

meters, and vertical altitude

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

meters.

The leg opposite the

30^\circ

angle is half the hypotenuse, and the leg opposite the

60^\circ

angle is

\sqrt{3}

times the short leg.

3

Calculate the total horizontal distance

PS

and vertical distance

RS

for the large right triangle

PSR

.

Total base

PS = PQ + QS = 4 + 4 = 8

meters, and height

RS = 4\sqrt{3}

meters.

Point

P

, point

Q

, and point

S

are collinear on the horizontal line.

4

Apply the Pythagorean theorem to right triangle

PSR

to find distance

PR

.

PR = \sqrt{PS^2 + RS^2} = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{64 + 48} = \sqrt{112} = 4\sqrt{7}

meters.

The straight-line distance

PR

is the hypotenuse of right triangle

PSR

.

Key Concept

Combining

30^\circ

-

60^\circ

-

90^\circ

special right triangles with the Pythagorean theorem to solve multi-step 2D path/distance problems.

Estimated Time:1m 45s

Question 147Question →

In the geometric configuration formed by adjacent triangles $ABC$ and $ACD$ sharing segment $AC$ , $\angle ABC = 90^\circ$ and $\angle ACD = 90^\circ$ . The lengths of the sides of triangle $ABC$ are $AB = 9$ and $BC = 12$ . If $\angle ADC = 30^\circ$ , what is the length of segment $AD$ ?

Show answer & explanation

Answer: 30

Answer

30

Applying the Pythagorean theorem to right triangle ABC gives AC = sqrt(9^2 + 12^2) = 15. In right triangle ACD with angle ADC = 30 degrees, side AC is opposite the 30-degree angle. By the 30-60-90 triangle side ratio (1 : sqrt(3) : 2), the hypotenuse AD is equal to twice the short leg AC, giving AD = 2 * 15 = 30.

Step-by-Step Solution

1

Calculate hypotenuse AC of right triangle ABC using the Pythagorean theorem.

AC = 15

In right triangle ABC with legs 9 and 12, AC^2 = 9^2 + 12^2 = 225, so AC = 15.

2

Identify the side relationships in special right triangle ACD.

AC is the shorter leg opposite the 30-degree angle ADC, and AD is the hypotenuse.

Since angle ACD is 90 degrees and angle ADC is 30 degrees, triangle ACD is a 30-60-90 right triangle.

3

Compute the hypotenuse AD from short leg AC.

AD = 30

In any 30-60-90 right triangle, the hypotenuse is twice the length of the leg opposite the 30-degree angle.

Key Concept

Pythagorean Theorem and Special Right Triangles

Estimated Time:1m 30s

Question 148Question →

In convex quadrilateral $ABCD$ , $\angle ABC = 90^\circ$ and $\angle ADC = 90^\circ$ . If $AB = BC$ , $AD = 6$ , and $CD = 8$ , what is the area of quadrilateral $ABCD$ ?

48

49

55

73

74

Show answer & explanation

Answer: 49

Answer

49

The correct answer is 49. Dividing quadrilateral

ABCD

along diagonal

AC

creates two right triangles:

\triangle ADC

with legs 6 and 8, and

\triangle ABC

with hypotenuse

AC

and equal legs. Using the Pythagorean theorem on

\triangle ADC

gives hypotenuse

AC = \sqrt{6^2 + 8^2} = 10

, and its area is

\frac{1}{2}(6)(8) = 24

. For isosceles right triangle

\triangle ABC

,

AB^2 + BC^2 = 10^2 \implies 2(AB^2) = 100 \implies AB^2 = 50

, so its area is

\frac{1}{2}(50) = 25

. Adding both triangle areas yields

24 + 25 = 49

.

Step-by-Step Solution

1

Divide the quadrilateral into two right triangles using diagonal

AC

.

Quadrilateral

ABCD

is split into

\triangle ADC

and

\triangle ABC

, both of which are right-angled triangles sharing hypotenuse

AC

.

Diagonal

AC

connects the vertices opposite the

90^\circ

angles.

2

Calculate the length of diagonal

AC

using

\triangle ADC

.

AC^2 = AD^2 + CD^2 = 6^2 + 8^2 = 36 + 64 = 100 \implies AC = 10

.

\triangle ADC

is a right triangle with legs of length 6 and 8.

3

Find the area of

\triangle ADC

.

\text{Area}(\triangle ADC) = \frac{1}{2} \times AD \times CD = \frac{1}{2} \times 6 \times 8 = 24

.

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

4

Determine the area of isosceles right triangle

\triangle ABC

.

Let

AB = BC = s

. Since

s^2 + s^2 = AC^2

, we have

2s^2 = 100 \implies s^2 = 50

. Thus,

\text{Area}(\triangle ABC) = \frac{1}{2} s^2 = \frac{1}{2} \times 50 = 25

.

\triangle ABC

is a right triangle with equal legs

s

and hypotenuse

AC = 10

.

5

Sum the areas of the two triangles to get the total area.

\text{Total Area} = 24 + 25 = 49

.

The total area of the quadrilateral is the sum of the areas of its non-overlapping constituent triangles.

Key Concept

Polygon area decomposition using diagonal partitioning and the Pythagorean theorem

Estimated Time:1m 30s

Question 149Question →

In convex pentagon $ABCDE$ , the measure of interior angle $\angle A$ is $100^\circ$ . The measures of the remaining four interior angles, $\angle B$ , $\angle C$ , $\angle D$ , and $\angle E$ , are in the ratio $2 : 3 : 3 : 3$ . What is the measure, in degrees, of $\angle B$ ?

40^\circ

80^\circ

108^\circ

120^\circ

160^\circ

Show answer & explanation

Answer: $80^\circ$

Answer

The correct answer is

80^\circ

, which corresponds to the measure of

\angle B

.

The sum of interior angles of a 5-sided polygon (pentagon) is

(5-2) \times 180^\circ = 540^\circ

. Subtracting

\angle A = 100^\circ

leaves

440^\circ

for the remaining four angles. The total ratio parts for these four angles is

2 + 3 + 3 + 3 = 11

. Dividing

440^\circ

by 11 gives

40^\circ

per ratio unit. Since

\angle B

corresponds to 2 ratio parts, its measure is

2 \times 40^\circ = 80^\circ

.

Step-by-Step Solution

1

Calculate the sum of all interior angles of the pentagon.

The sum of interior angles for an

n

-sided polygon is given by

(n - 2) \times 180^\circ

. For a pentagon (

n = 5

), the sum is

(5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ

.

Determining the total interior angle sum is necessary to find the sum of the unknown angles.

2

Subtract the known angle measure

\angle A

from the total interior angle sum.

The combined sum of angles

\angle B + \angle C + \angle D + \angle E = 540^\circ - 100^\circ = 440^\circ

.

Isolating the sum of the remaining four angles allows distribution according to the given ratio.

3

Determine the value of one ratio unit.

The sum of ratio parts is

2 + 3 + 3 + 3 = 11

parts. One part is equal to

440^\circ / 11 = 40^\circ

.

Finding the magnitude of a single ratio unit enables calculation of any individual angle.

4

Multiply the single unit value by the ratio coefficient for

\angle B

.

Since

\angle B

corresponds to 2 parts,

\angle B = 2 \times 40^\circ = 80^\circ

.

This yields the requested measure of angle

\angle B

.

Key Concept

Sum of interior angles of an n-sided polygon:

(n - 2) \times 180^\circ

, combined with proportional partitioning of angle sums.

Alternative Method

Express the angles in terms of a variable

x

. Let the remaining angles be

2x, 3x, 3x, 3x

. Set up the equation

100^\circ + 2x + 3x + 3x + 3x = 540^\circ

, simplify to

100 + 11x = 540

, solve for

x = 40^\circ

, and then evaluate

\angle B = 2x = 80^\circ

.

Estimated Time:1m 15s

Question 150Question →

In trapezoid $ABCD$ , side $AB$ is parallel to side $CD$ , and $\angle DAB = 90^\circ$ . If $AB = 14$ , $CD = 8$ , and $BC = 10$ , what is the area of trapezoid $ABCD$ ?

40

66

88

110

176

Show answer & explanation

Answer: 88

Answer

88

By drawing an altitude from point

C

perpendicular to base

AB

at point

H

, the trapezoid is partitioned into a rectangle

ADCH

and a right triangle

CHB

. The base of triangle

CHB

is

HB = 14 - 8 = 6

, and its hypotenuse is

BC = 10

. Applying the Pythagorean theorem yields

CH = \sqrt{10^2 - 6^2} = 8

. Using the trapezoid area formula

\frac{b_1 + b_2}{2} \times h

gives

\frac{14 + 8}{2} \times 8 = 88

.

Step-by-Step Solution

1

Decompose the trapezoid into a rectangle and a right triangle.

Draw a line segment from vertex

C

perpendicular to base

AB

, intersecting

AB

at point

H

. Quadrilateral

ADCH

is a rectangle, and triangle

CHB

is a right triangle with right angle at

H

.

Splitting the trapezoid allows the perpendicular height to be determined using right triangle properties.

2

Determine the horizontal base of right triangle

CHB

.

Since

ADCH

is a rectangle,

AH = CD = 8

. Therefore,

HB = AB - AH = 14 - 8 = 6

.

Opposite sides of a rectangle are equal in length.

3

Calculate height

CH

using the Pythagorean theorem.

In right triangle

CHB

,

CH^2 + HB^2 = BC^2 \implies CH^2 + 6^2 = 10^2 \implies CH^2 + 36 = 100 \implies CH = \sqrt{64} = 8

.

The height of the trapezoid corresponds to leg

CH

of right triangle

CHB

.

4

Calculate the area of trapezoid

ABCD

.

\text{Area} = \frac{AB + CD}{2} \times CH = \frac{14 + 8}{2} \times 8 = 11 \times 8 = 88.

The area formula for a trapezoid is half the sum of the parallel bases multiplied by the perpendicular height.

Key Concept

Decomposing a trapezoid into a rectangle and a right triangle to calculate height via the Pythagorean theorem

Estimated Time:1m 15s

Question 151Question →

An isosceles trapezoid has parallel base lengths of $10$ and $26$ , and an altitude of $15$ . A line segment connects the midpoints of the two non-parallel legs, dividing the figure into two smaller trapezoids. What is the area of the larger of these two smaller trapezoids?

Show answer & explanation

Answer: 165

Answer

165

The midsegment of a trapezoid connects the midpoints of the non-parallel legs, and its length is the average of the two parallel bases:

\frac{10 + 26}{2} = 18

. Because the line connects midpoints, it also bisects the altitude, making the height of each smaller trapezoid equal to

\frac{15}{2} = 7.5

. The larger of the two resulting trapezoids has bases of lengths

18

and

26

. Using the trapezoid area formula

\text{Area} = \frac{b_1 + b_2}{2} \times h

, we obtain

\frac{18 + 26}{2} \times 7.5 = 22 \times 7.5 = 165

.

Step-by-Step Solution

1

Calculate the length of the midsegment connecting the midpoints of the non-parallel legs.

The midsegment length is

18

.

The midsegment of a trapezoid is parallel to the bases and its length equals the average of the two base lengths:

\frac{10 + 26}{2} = 18

.

2

Determine the height of the smaller subtrapezoid.

The height of the subtrapezoid is

7.5

.

The segment connecting the midpoints of the legs bisects the overall altitude of

15

, giving a height of

\frac{15}{2} = 7.5

for each subtrapezoid.

3

Compute the area of the larger subtrapezoid.

The area is

165

.

The larger subtrapezoid is bounded by the midsegment (length

18

) and the bottom base (length

26

). Applying the trapezoid area formula yields

\text{Area} = \frac{18 + 26}{2} \times 7.5 = 22 \times 7.5 = 165

.

Key Concept

Trapezoid Midsegment Theorem and Subdivided Area Calculation

Question 152Question →

In a convex polygon, the sum of the measures of all interior angles except one is $1,200^\circ$ . If the measure of the remaining interior angle is an integer degree, what is the number of sides of the polygon?

7

8

9

10

11

Show answer & explanation

Answer: 9

Answer

The number of sides of the polygon is 9.

The interior angle sum of an

n

-sided polygon is

(n - 2) \times 180^\circ

. Because the polygon is convex, the missing angle

x

must be between

0^\circ

and

180^\circ

. Adding

x

to

1,200^\circ

gives a total sum between

1,200^\circ

and

1,380^\circ

. The only multiple of

180^\circ

in this range is

1,260^\circ

, which corresponds to

(n - 2) = 7

, or

n = 9

sides.

Step-by-Step Solution

1

Set up the inequality for the interior angle sum of a convex polygon

The sum of the interior angles of an

n

-sided convex polygon is

S = (n - 2) \times 180^\circ

. Let

x

be the measure of the remaining interior angle, where

0^\circ < x < 180^\circ

.

Every interior angle of a convex polygon must be strictly greater than

0^\circ

and strictly less than

180^\circ

.

2

Formulate the bounds for

(n - 2) \times 180^\circ

Since

S = 1,200^\circ + x

, we have

1,200^\circ < (n - 2) \times 180^\circ < 1,200^\circ + 180^\circ = 1,380^\circ

.

Adding the bounds of

x

(

0^\circ

to

180^\circ

) to the given sum of

1,200^\circ

establishes the valid range for the total angle sum.

3

Solve for the integer value of

n

Dividing the inequality by

180^\circ

gives

6.67 < n - 2 < 7.67

. Since

n - 2

must be an integer,

n - 2 = 7

, which means

n = 9

.

The only integer in the range

(6.67, 7.67)

is 7, corresponding to a 9-sided polygon with a remaining angle of

x = 1,260^\circ - 1,200^\circ = 60^\circ

.

Key Concept

Polygon Interior Angle Sum Theorem

Alternative Method

Divide

1,200^\circ

by

180^\circ

to get

6.67

. Since

(n - 2)

must be an integer and the remaining angle is positive and less than

180^\circ

, round

6.67

up to the next integer, 7. Thus,

n - 2 = 7

, so

n = 9

.

Estimated Time:1m 15s

Question 153Question →

In the $xy$ -plane, quadrilateral $ABCD$ has vertices $A(0, 0)$ , $B(6, 0)$ , $C(8, 4)$ , and $D(2, 4)$ . Which of the following statements must be true? Select all such statements.

Select all that apply

Quadrilateral

ABCD

is a parallelogram.

The area of quadrilateral

ABCD

is

24

.

The diagonals

AC

and

BD

bisect each other at the point

(4, 2)

.

The length of diagonal

AC

is equal to the length of diagonal

BD

.

The diagonals

AC

and

BD

are perpendicular to each other.

Show answer & explanation

Answer: Quadrilateral $ABCD$ is a parallelogram.; The area of quadrilateral $ABCD$ is $24$ .; The diagonals $AC$ and $BD$ bisect each other at the point $(4, 2)$ .

Answer

The correct statements are that quadrilateral

ABCD

is a parallelogram, its area is

24

, and its diagonals bisect each other at

(4, 2)

.

Quadrilateral

ABCD

is a parallelogram because both pairs of opposite sides are congruent and parallel (

AB = DC = 6

along the horizontal line, and

AD = BC = \sqrt{20}

). The area is base times height, which is

6 \times 4 = 24

. The diagonals bisect each other at their common midpoint

(4, 2)

.

Step-by-Step Solution

1

Determine side lengths and slopes to classify the quadrilateral

Side

AB

is horizontal with length

6

; side

DC

is horizontal with length

6

. Side

AD

has components

(2,4)

and length

\sqrt{20}

; side

BC

has components

(2,4)

and length

\sqrt{20}

. Since opposite sides are parallel and congruent,

ABCD

is a parallelogram.

To verify if the quadrilateral is a parallelogram.

2

Calculate the area of the quadrilateral

Area =

\text{base} \times \text{height} = 6 \times 4 = 24

.

To evaluate the area statement.

3

Find the midpoints and lengths of diagonals

AC

and

BD

Midpoint of

AC = (4, 2)

and midpoint of

BD = (4, 2)

, so they bisect each other. Length

AC = \sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}

, and length

BD = \sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2}

.

To check diagonal bisection and length equality.

4

Determine the slopes of the diagonals to check for perpendicularity

Slope of

AC = \frac{4}{8} = \frac{1}{2}

; slope of

BD = \frac{4}{-4} = -1

. Product of slopes

= \frac{1}{2} \times (-1) = -\frac{1}{2} \neq -1

, so they are not perpendicular.

To verify whether the diagonals intersect at right angles.

Key Concept

Properties of quadrilaterals in the coordinate plane, including parallelogram identification, area calculation, midpoint theorem for diagonals, and perpendicular slope test.

Estimated Time:1m 30s

Question 154Question →

In rhombus $ABCD$ , the perimeter is $52$ and the length of diagonal $AC$ is $10$ . What is the area of rhombus $ABCD$ ?

60

96

120

169

240

Show answer & explanation

Answer: 120

Answer

The area of rhombus

ABCD

is 120.

Because all four sides of a rhombus are equal in length, a perimeter of 52 implies each side measures 13. The diagonals of a rhombus intersect at right angles and bisect each other. Given diagonal AC has a length of 10, half of AC is 5. Using the Pythagorean theorem on one of the right triangles formed by the intersecting diagonals gives a half-diagonal length of sqrt(13^2 - 5^2) = 12 for BD. Thus, the total length of diagonal BD is 24. The area of the rhombus is (1/2) * 10 * 24 = 120.

Step-by-Step Solution

1

Find the side length of rhombus

ABCD

.

Side length

s = 52 / 4 = 13

.

A rhombus has four equal sides, so its perimeter divided by 4 gives the length of one side.

2

Use the properties of rhombus diagonals to find half of diagonal

BD

.

Half of

AC = 5

. Half of

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

.

The diagonals of a rhombus bisect each other at right angles, forming four congruent right triangles whose hypotenuse is the side length (

13

) and legs are the half-diagonals.

3

Calculate the full length of diagonal

BD

.

Length of diagonal

BD = 2 \times 12 = 24

.

The full diagonal length is twice the length of its half.

4

Calculate the area of rhombus

ABCD

.

Area

= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 24 = 120

.

The area of any rhombus is equal to half the product of its diagonal lengths.

Key Concept

Properties of a Rhombus: Perpendicular Bisecting Diagonals and Area Formula

Estimated Time:1m 30s

Question 155Question →

In a regular polygon, the ratio of the measure of an interior angle to the measure of an exterior angle is $7:2$ . What is the total number of diagonals of this polygon?

Show answer & explanation

Answer: 27

Answer

The total number of diagonals of the regular polygon is

27

.

The correct answer is

27

. An interior angle and an exterior angle of a polygon are supplementary (

180^\circ

). Given the ratio

7:2

, the exterior angle is

\frac{2}{9} \times 180^\circ = 40^\circ

. Since the sum of exterior angles of any convex polygon is

360^\circ

, the number of sides is

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

. Using the formula for the number of diagonals,

\frac{n(n-3)}{2}

, we obtain

\frac{9(9-3)}{2} = 27

.

Step-by-Step Solution

1

Determine the measure of the exterior angle using the given interior-to-exterior ratio.

Exterior angle measure =

40^\circ

At any vertex of a polygon, the interior angle and exterior angle sum to

180^\circ

. With a ratio of

7:2

, the exterior angle represents

\frac{2}{7+2} = \frac{2}{9}

of the total

180^\circ

.

2

Calculate the number of sides (

n

) of the regular polygon.

n = 9

The sum of the exterior angles of any convex polygon is

360^\circ

. Since all exterior angles in a regular polygon are equal,

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

.

3

Calculate the total number of diagonals using the formula

\frac{n(n-3)}{2}

.

Number of diagonals =

27

Substituting

n = 9

into

\frac{n(n-3)}{2}

yields

\frac{9 \times 6}{2} = 27

.

Key Concept

Interior and exterior angle properties of regular polygons, and the diagonal counting formula for convex polygons.

Question 156Question →

A convex hexagon has five interior angles measuring $115^\circ$ , $125^\circ$ , $130^\circ$ , $140^\circ$ , and $150^\circ$ . What is the measure, in degrees, of the exterior angle adjacent to the sixth interior angle?

60^\circ

80^\circ

120^\circ

140^\circ

150^\circ

Show answer & explanation

Answer: $120^\circ$

Answer

The measure of the exterior angle adjacent to the sixth interior angle is

120^\circ

.

The sum of the interior angles of a 6-sided polygon (hexagon) is given by

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

. The sum of the five given interior angles is

115^\circ + 125^\circ + 130^\circ + 140^\circ + 150^\circ = 660^\circ

, which leaves

720^\circ - 660^\circ = 60^\circ

for the sixth interior angle. Since an interior angle and its adjacent exterior angle are supplementary, the exterior angle is

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

.

Step-by-Step Solution

1

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

Using

(n - 2) \times 180^\circ

with

n = 6

, the total interior angle sum is

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

.

The sum of interior angles of any convex polygon with

n

sides is

(n - 2) \times 180^\circ

.

2

Find the sum of the five given interior angles.

115^\circ + 125^\circ + 130^\circ + 140^\circ + 150^\circ = 660^\circ

.

Summing the five known angle values is necessary to find the remaining sixth angle.

3

Determine the measure of the sixth interior angle.

720^\circ - 660^\circ = 60^\circ

.

Subtracting the sum of the five interior angles from the total interior angle sum yields the sixth interior angle.

4

Calculate the supplementary exterior angle.

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

.

An interior angle and its adjacent exterior angle form a straight line and are supplementary (

180^\circ

).

Key Concept

Interior and Exterior Angles of Convex Polygons

Geometry

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes