Geometry Practice Questions — GRE General Test — Page 2

Question 21Question →

In a circle, an arc of length $4\pi$ corresponds to a central angle of $40^\circ$ . If the area of the sector formed by this central angle is $k\pi$ , what is the value of $k$ ?

Show answer & explanation

Answer: 36

Answer

The value of

k

is 36.

Using the arc length equation

4\pi = \frac{40}{360} \cdot 2\pi r

, we solve for the radius

r = 18

. Substituting

r = 18

into the sector area formula

A = \frac{40}{360} \cdot \pi (18)^2

results in

A = 36\pi

. Thus,

k = 36

.

Step-by-Step Solution

1

Calculate the radius of the circle using the arc length formula

Radius

r = 18

Arc length is related to central angle and radius by

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

. Substituting

L = 4\pi

and

\theta = 40^\circ

gives

4\pi = \frac{1}{9} \cdot 2\pi r \implies r = 18

.

2

Calculate the area of the sector using the radius and central angle

Sector Area

A = 36\pi

Sector area is calculated using

A = \frac{\theta}{360^\circ} \cdot \pi r^2

. Substituting

\theta = 40^\circ

and

r = 18

gives

A = \frac{1}{9} \cdot \pi (18^2) = 36\pi

.

3

Extract the coefficient

k

from

k\pi

k = 36

Comparing

36\pi

to

k\pi

directly yields

k = 36

.

Key Concept

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

Question 22Question →

In a circle centered at point $O$ , sector $OAB$ has a central angle of $60^\circ$ and a radius of $12$ . A smaller circle is inscribed inside sector $OAB$ such that it is tangent to radius $OA$ , radius $OB$ , and arc $AB$ . If the area of the region inside sector $OAB$ that lies outside the inscribed circle is expressed in the form $k\pi$ , what is the value of $k$ ?

Show answer & explanation

Answer: 8

Answer

The correct value of

k

is 8.

By using the geometry of the

30^\circ

-

60^\circ

-

90^\circ

right triangle formed by the angle bisector and the radius of tangency, the radius of the inscribed circle is found to be

r = 4

. Subtracting its area (

16\pi

) from the sector's area (

24\pi

) gives

8\pi

, so

k = 8

.

Step-by-Step Solution

1

Determine the relationship between the radius of the larger circle

R

and the radius of the inscribed circle

r

.

OP = 2r

and

R = 3r

.

The center

P

of the inscribed circle lies on the angle bisector of

\angle AOB = 60^\circ

, creating a

30^\circ

angle with radius

OA

. The perpendicular distance from

P

to radius

OA

is

r

, so

\sin(30^\circ) = \frac{r}{OP} = \frac{1}{2}

, giving

OP = 2r

. Since the inscribed circle touches arc

AB

,

OP + r = R

, so

3r = R

.

2

Calculate the radius

r

of the inscribed circle.

r = 4

.

Given

R = 12

, solving

3r = 12

yields

r = 4

.

3

Compute the area of sector

OAB

.

\text{Area}_{\text{sector}} = 24\pi

.

The formula for the area of a sector is

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

. Here,

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

.

4

Compute the area of the inscribed circle.

\text{Area}_{\text{circle}} = 16\pi

.

The area of a circle with radius

r = 4

is

\pi r^2 = \pi (4^2) = 16\pi

.

5

Subtract the area of the inscribed circle from the area of sector

OAB

to find

k

.

k = 8

.

\text{Area}_{\text{region}} = 24\pi - 16\pi = 8\pi

, which means

k = 8

.

Key Concept

Inscribed shapes within sectors, arc length, and sector area relations

Question 23Question →

Two sides of a triangle have lengths of $5$ centimeters and $9$ centimeters. Which of the following could be the length, in centimeters, of the third side? Select all such lengths.

Select all that apply

4

6

10

14

16

Show answer & explanation

Answer: $6$ ; $10$

Answer

The possible lengths for the third side are

6

centimeters and

10

centimeters.

According to the Triangle Inequality Theorem, the length of the third side

x

must be strictly greater than the difference of the two given sides (

9 - 5 = 4

) and strictly less than their sum (

9 + 5 = 14

). This establishes the valid range as

4 < x < 14

. The options

6

centimeters and

10

centimeters are the only values provided that fall strictly within this interval.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to determine the allowable range for the third side.

The length of the third side,

x

, must satisfy

|9 - 5| < x < 9 + 5

, which simplifies to

4 < x < 14

.

The sum of the lengths of any two sides of a non-degenerate triangle must be strictly greater than the length of the remaining side.

2

Check each candidate option against the inequality

4 < x < 14

.

The values

6

and

10

satisfy

4 < x < 14

, whereas

4

,

14

, and

16

do not.

Only numbers strictly within the open interval

(4, 14)

can form a valid triangle with side lengths

5

and

9

.

Key Concept

Triangle Inequality Theorem

Question 24Question →

In the geometric plane, line $l_1$ is parallel to line $l_2$ . Point $A$ lies on line $l_1$ and point $B$ lies on line $l_2$ . Point $C$ is located between lines $l_1$ and $l_2$ such that $C$ lies to the right of both $A$ and $B$ .

The acute angle between segment $AC$ and the ray extending to the right from $A$ along line $l_1$ measures $x^\circ$ .

The acute angle between segment $BC$ and the ray extending to the right from $B$ along line $l_2$ measures $y^\circ$ .

The interior angle $\angle ACB = 118^\circ$ , and $y = 2x - 14$ .

Line $k$ passes through point $C$ and is perpendicular to segment $AC$ . Line $k$ intersects line $l_2$ at point $E$ , where point $E$ lies to the right of point $B$ .

What is the measure, in degrees, of the acute angle $\angle CEB$ formed by line $k$ and line $l_2$ ?

16^\circ

44^\circ

46^\circ

62^\circ

74^\circ

Show answer & explanation

Answer: $46^\circ$

Answer

The measure of the acute angle

\angle CEB

is

46^\circ

.

By the parallel lines angle property, drawing a parallel line through vertex

C

shows that

\angle ACB = x + y = 118^\circ

. Substituting

y = 2x - 14

yields

3x - 14 = 118

, giving

x = 44^\circ

. Since line

l_1

is parallel to line

l_2

, segment

AC

intersects line

l_2

at an acute angle of

44^\circ

. Line

k

is constructed perpendicular to segment

AC

, so the acute angle formed by line

k

and line

l_2

is complementary to

44^\circ

, which gives

90^\circ - 44^\circ = 46^\circ

.

Step-by-Step Solution

1

Set up the parallel lines zig-zag relationship to find

x

and

y

.

\angle ACB = x + y = 118^\circ

By drawing an auxiliary line through

C

parallel to

l_1

and

l_2

, the interior angle

\angle ACB

facing left equals the sum of the alternate interior angles

x

and

y

.

2

Substitute the given algebraic relation

y = 2x - 14

into the sum equation.

x + (2x - 14) = 118 \implies 3x - 14 = 118 \implies 3x = 132 \implies x = 44^\circ

Solving the linear system gives the exact value of angle

x

.

3

Determine the angle that line

AC

makes with line

l_2

.

Line

AC

intersects line

l_2

at an acute angle of

44^\circ

.

Since

l_1 \parallel l_2

, alternate interior angles formed by transversal line

AC

are equal (

x = 44^\circ

).

4

Calculate the acute angle

\angle CEB

between line

k

and line

l_2

.

\angle CEB = 90^\circ - 44^\circ = 46^\circ

Line

k

is perpendicular to segment

AC

, so the angle it forms with line

l_2

is the complementary angle to the angle line

AC

forms with line

l_2

.

Key Concept

Parallel Line Angle Relationships and Perpendicular Line Complements

Question 25Question →

Two concentric circles centered at point $O$ have radii of $6$ units and $10$ units. A central angle $\theta$ defines a sector that intersects the region between the concentric circles, creating a region bounded by an outer arc, an inner arc, and two straight line segments. If the area of this bounded region is $16\pi$ square units, what is the perimeter of the region?

8\pi

4 + 8\pi

8 + 8\pi

8 + 16\pi

20 + 8\pi

Show answer & explanation

Answer: $8 + 8\pi$

Answer

The perimeter of the region is

8 + 8\pi

units.

The region between the concentric circles has an area equal to the fraction of the central angle times the area of the ring:

\frac{\theta}{360^\circ} \cdot \pi (10^2 - 6^2) = 64\pi \left(\frac{\theta}{360^\circ}\right)

. Setting this equal to

16\pi

gives

\frac{\theta}{360^\circ} = \frac{1}{4}

. The outer arc length is

\frac{1}{4}(2\pi \cdot 10) = 5\pi

, and the inner arc length is

\frac{1}{4}(2\pi \cdot 6) = 3\pi

. The region is bounded on the sides by two segments of length

10 - 6 = 4

units each. The total perimeter is

5\pi + 3\pi + 4 + 4 = 8 + 8\pi

.

Step-by-Step Solution

1

Express the area of the bounded region in terms of the central angle fraction

f = \frac{\theta}{360^\circ}

.

\text{Area} = f \cdot \pi(R^2 - r^2) = f \cdot \pi(10^2 - 6^2) = 64\pi f

.

The area of the region between two concentric sector arcs is the difference between the outer sector area and the inner sector area.

2

Solve for the fraction

f

using the given area of

16\pi

.

64\pi f = 16\pi \implies f = \frac{16\pi}{64\pi} = \frac{1}{4}

.

Equating the algebraic expression for the region's area to

16\pi

allows finding the proportion of the circle represented by the central angle.

3

Calculate the lengths of the outer arc, inner arc, and straight boundary segments.

\text{Outer Arc} = \frac{1}{4}(2\pi \cdot 10) = 5\pi

,

\text{Inner Arc} = \frac{1}{4}(2\pi \cdot 6) = 3\pi

,

\text{Segment Length} = 10 - 6 = 4

.

Arc lengths use the formula

f \cdot 2\pi r

, and each straight boundary segment is the radial distance between the two circles.

4

Sum all boundary components to find the total perimeter.

\text{Perimeter} = 5\pi + 3\pi + 4 + 4 = 8 + 8\pi

.

The complete perimeter of the bounded region consists of the outer arc, inner arc, and two radial segments.

Key Concept

Annular Sector Area and Perimeter

Estimated Time:1m 30s

Question 26Question →

In triangle $ABC$ , the length of side $AB$ is $7$ and the length of side $BC$ is $15$ . If the area of triangle $ABC$ is $42$ , which of the following could be the length of side $AC$ ? Select all such lengths.

Select all that apply

2\sqrt{37}

16

20

22

\sqrt{274}

Show answer & explanation

Answer: $2\sqrt{37}$ ; $20$

Answer

The possible lengths of side

AC

are

2\sqrt{37}

and

20

.

Using the triangle area formula

\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin B

, we find

\sin B = \frac{4}{5}

. Because

\sin B

is positive in both Quadrant I and Quadrant II, angle

B

can be either acute or obtuse. If angle

B

is acute,

\cos B = \frac{3}{5}

, giving

AC = \sqrt{7^2 + 15^2 - 2(7)(15)(0.6)} = 2\sqrt{37}

. If angle

B

is obtuse,

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

, giving

AC = \sqrt{7^2 + 15^2 - 2(7)(15)(-0.6)} = 20

. Both values represent valid triangle configurations.

Step-by-Step Solution

1

Determine the sine of angle

B

using the area formula.

\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B \implies 42 = \frac{1}{2} \cdot 7 \cdot 15 \cdot \sin B \implies \sin B = \frac{84}{105} = \frac{4}{5}

.

The area of a triangle with two given sides and an included angle is

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

.

2

Find the possible values for

\cos B

.

Since

\sin B = \frac{4}{5}

,

\cos B

can be either

\frac{3}{5}

(if angle

B

is acute) or

-\frac{3}{5}

(if angle

B

is obtuse).

Sine is positive in both the first and second quadrants, permitting both acute and obtuse angles for triangle

ABC

.

3

Calculate the length of side

AC

when angle

B

is acute.

AC^2 = 7^2 + 15^2 - 2(7)(15)\left(\frac{3}{5}\right) = 49 + 225 - 126 = 148 \implies AC = \sqrt{148} = 2\sqrt{37}

.

Apply the Law of Cosines:

AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B

with

\cos B = 0.6

.

4

Calculate the length of side

AC

when angle

B

is obtuse.

AC^2 = 7^2 + 15^2 - 2(7)(15)\left(-\frac{3}{5}\right) = 49 + 225 + 126 = 400 \implies AC = \sqrt{400} = 20

.

Apply the Law of Cosines with

\cos B = -0.6

.

Key Concept

Triangle Area via Included Angle and Dual Solutions in Non-Right Triangles

Question 27Question →

In triangle $PQR$ , the measure of angle $PQR$ is $90^\circ$ . Point $S$ lies on segment $PR$ such that line segment $QS$ is perpendicular to $PR$ . If $PS = 4$ and $SR = 9$ , what is the area of triangle $PQR$ ?

18

39

78

117

234

Show answer & explanation

Answer: $39$

Answer

The area of triangle

PQR

is 39.

By the geometric mean theorem (right triangle altitude theorem), the altitude

QS

satisfies

QS^2 = PS \cdot SR = 4 \cdot 9 = 36

, so

QS = 6

. The hypotenuse

PR = PS + SR = 4 + 9 = 13

. Substituting base

13

and height

6

into the triangle area formula

\frac{1}{2}bh

gives

\frac{1}{2} \times 13 \times 6 = 39

.

Step-by-Step Solution

1

Find the length of hypotenuse

PR

PR = PS + SR = 4 + 9 = 13

Point

S

lies on segment

PR

, so the total length of the hypotenuse is the sum of its two segments.

2

Calculate altitude

QS

using the geometric mean theorem for right triangles

QS = \sqrt{PS \times SR} = \sqrt{4 \times 9} = \sqrt{36} = 6

In a right triangle, the altitude to the hypotenuse is the geometric mean of the two segments into which the hypotenuse is divided.

3

Calculate the area of triangle

PQR

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 13 \times 6 = 39

The area of any triangle is half the product of its base and corresponding height.

Key Concept

Altitude to the hypotenuse in right triangles and triangle area calculation

Alternative Method

Alternatively, use similar triangles

\triangle PQS \sim \triangle QRS

. The ratio of corresponding sides gives

\frac{PQ}{PS} = \frac{PR}{PQ} \implies PQ^2 = PS \cdot PR = 4 \cdot 13 = 52

, and

QR^2 = SR \cdot PR = 9 \cdot 13 = 117

. Since

\triangle PQR

is a right triangle at

Q

, its area is

\frac{1}{2} \cdot PQ \cdot QR = \frac{1}{2} \sqrt{52 \cdot 117} = \frac{1}{2} \sqrt{6084} = \frac{1}{2} \cdot 78 = 39

.

Estimated Time:1m 30s

Question 28Question →

In the $xy$ -plane, line $L_1$ passes through the origin $(0,0)$ and the point $(4, 3)$ . Line $L_2$ is formed by reflecting line $L_1$ across the vertical line $x = 2$ and then translating the resulting line downward by $5$ units. If line $L_2$ intersects the $y$ -axis at the point $(0, k)$ , what is the value of $k$ ?

-5

-2

\frac{1}{3}

3

8

Show answer & explanation

Answer: $-2$

Answer

The correct value of

k

is

-2

.

Line

L_1

has slope

\frac{3}{4}

and equation

y = \frac{3}{4}x

. Reflecting across

x = 2

replaces

x

with

4 - x

, transforming the equation into

y = \frac{3}{4}(4 - x) = 3 - \frac{3}{4}x

. Translating downward by

5

units gives

y = -\frac{3}{4}x - 2

. Setting

x = 0

gives the

y

-intercept

(0, -2)

, making

-2

the correct value.

Step-by-Step Solution

1

Find the equation of line

L_1

.

The slope of line

L_1

passing through

(0,0)

and

(4,3)

is

m = \frac{3 - 0}{4 - 0} = \frac{3}{4}

. Thus, the equation is

y = \frac{3}{4}x

.

Establishing the initial linear equation is necessary before applying coordinate transformations.

2

Apply the reflection across the line

x = 2

.

Reflecting any point

(x, y)

across

x = 2

transforms its x-coordinate to

2(2) - x = 4 - x

. Substituting

4 - x

into the equation gives

y = \frac{3}{4}(4 - x) = 3 - \frac{3}{4}x

.

Reflection across a vertical line

x = a

preserves the y-values while mapping

x \mapsto 2a - x

.

3

Apply the downward vertical translation by

5

units.

Subtracting

5

from the equation yields

y = \left(3 - \frac{3}{4}x\right) - 5 = -\frac{3}{4}x - 2

.

Translating a graph downward by

c

units subtracts

c

from the output

y

.

4

Determine the y-intercept of line

L_2

.

Setting

x = 0

in

y = -\frac{3}{4}x - 2

yields

y = -2

, so

k = -2

.

The y-intercept of a line occurs where

x = 0

.

Key Concept

Coordinate Transformations of Lines: Reflection across

x = a

and Vertical Translations

Estimated Time:2m 0s

Question 29Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $60^\circ$ , and the length of side $AB$ is $6$ . Square $BDEF$ is inscribed in $\triangle ABC$ such that vertex $D$ lies on side $AB$ , vertex $E$ lies on hypotenuse $AC$ , and vertex $F$ lies on side $BC$ . What is the length of segment $BE$ ?

9\sqrt{2} - 3\sqrt{6}

9 - 3\sqrt{3}

3\sqrt{6} - 3\sqrt{2}

9\sqrt{3} - 9

9\sqrt{2} - 3\sqrt{3}

Show answer & explanation

Answer: $9\sqrt{2} - 3\sqrt{6}$

Answer

The length of segment

BE

is

9\sqrt{2} - 3\sqrt{6}

.

The correct answer is

9\sqrt{2} - 3\sqrt{6}

. Since

\triangle ADE

is a

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

triangle with short leg

AD = 6 - x

and long leg

DE = x

, the side length of the square is

x = 9 - 3\sqrt{3}

. Segment

BE

is the diagonal of the square, which equals

x\sqrt{2} = (9 - 3\sqrt{3})\sqrt{2} = 9\sqrt{2} - 3\sqrt{6}

.

Step-by-Step Solution

1

Analyze the properties of right triangle

ABC

and the inscribed square

BDEF

.

In

\triangle ABC

,

\angle B = 90^\circ

and

\angle A = 60^\circ

, so

\angle C = 30^\circ

. The side

AB = 6

. Let

x

be the side length of square

BDEF

.

Establishing the variable

x

allows us to express the dimensions of the smaller triangles formed by the square.

2

Identify the side lengths and angles of the smaller triangle

\triangle ADE

.

Since

BD = x

lies on side

AB

,

AD = AB - BD = 6 - x

. Because

DE \parallel BC

,

\angle ADE = 90^\circ

and

\angle A = 60^\circ

, making

\triangle ADE

a

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

right triangle with

DE = x

.

In a

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

triangle, the side opposite the

60^\circ

angle is

\sqrt{3}

times the side adjacent to the

60^\circ

angle.

3

Set up and solve the equation for the side length

x

of the square.

Since

DE = AD \cdot \sqrt{3}

, we have

x = (6 - x)\sqrt{3} \implies x(1 + \sqrt{3}) = 6\sqrt{3} \implies x = \frac{6\sqrt{3}}{\sqrt{3} + 1} = 3(3 - \sqrt{3}) = 9 - 3\sqrt{3}

.

Rationalizing the denominator

\frac{6\sqrt{3}(\sqrt{3}-1)}{2}

yields the exact side length of the square.

4

Calculate the diagonal length

BE

of square

BDEF

.

Segment

BE

is the diagonal of square

BDEF

. In a

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

right triangle

\triangle BDE

, the hypotenuse is

x\sqrt{2}

. Thus,

BE = (9 - 3\sqrt{3})\sqrt{2} = 9\sqrt{2} - 3\sqrt{6}

.

Applying the special right triangle ratio

1:1:\sqrt{2}

for the square's diagonal gives the required segment length.

Key Concept

Combining

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

side ratios (

1:\sqrt{3}:2

) and

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

hypotenuse ratios (

1:1:\sqrt{2}

) to solve composite geometric figures.

Estimated Time:2m 30s

Question 30Question →

In $\triangle ABC$ , point $D$ lies on side $BC$ such that the ratio of the length of segment $BD$ to the length of segment $DC$ is $3 : 2$ . Point $E$ is the midpoint of line segment $AD$ . If the area of $\triangle ABC$ is $60$ square units, which of the following statements must be true? Select all that apply.

Select all that apply

The area of

\triangle ABD

is

36

square units.

The area of

\triangle ABE

is

18

square units.

The area of

\triangle BEC

is

30

square units.

The area of

\triangle ADC

is

30

square units.

The area of

\triangle ECD

is

15

square units.

Show answer & explanation

Answer: The area of $\triangle ABD$ is $36$ square units.; The area of $\triangle ABE$ is $18$ square units.; The area of $\triangle BEC$ is $30$ square units.

Answer

The statements asserting that the area of triangle ABD is 36 square units, the area of triangle ABE is 18 square units, and the area of triangle BEC is 30 square units are all true.

Triangles sharing a vertex and having bases along the same straight line share the same height. Thus, their areas are in the exact ratio of their bases. Since segment BD is 3/5 of BC, triangle ABD has an area of (3/5) * 60 = 36 square units. Median BE divides triangle ABD into two equal areas of 18 square units each. Median CE divides triangle ADC (area 24) into two equal areas of 12 square units each. Summing triangles EBD (18) and ECD (12) gives an area of 30 square units for triangle BEC.

Step-by-Step Solution

1

Determine the areas of triangles ABD and ADC using the base ratio.

Area of triangle ABD = 36 square units, and Area of triangle ADC = 24 square units.

Triangles ABD and ADC share the same altitude from vertex A to line BC. Therefore, their areas are directly proportional to their base lengths BD and DC. Since BD : DC = 3 : 2, BD is 3/5 of BC and DC is 2/5 of BC.

2

Calculate the area of triangle ABE.

Area of triangle ABE = 18 square units.

Point E is the midpoint of segment AD. In triangle ABD, line segment BE is a median from vertex B to side AD. A median bisects a triangle into two region of equal area, so Area(ABE) = 1/2 * Area(ABD) = 1/2 * 36 = 18 square units.

3

Calculate the area of triangle ECD.

Area of triangle ECD = 12 square units.

Similarly, segment CE is a median in triangle ADC from vertex C to side AD. Thus, Area(ECD) = 1/2 * Area(ADC) = 1/2 * 24 = 12 square units.

4

Calculate the area of triangle BEC.

Area of triangle BEC = 30 square units.

Triangle BEC is formed by combining triangles EBD and ECD. Since Area(EBD) = 18 and Area(ECD) = 12, Area(BEC) = 18 + 12 = 30 square units.

Key Concept

Triangles sharing a vertex and altitude have areas proportional to their bases; a median divides a triangle into two equal areas.

Estimated Time:2m 0s

Question 31Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $30^\circ$ , and hypotenuse $AC = 12$ . Point $D$ lies on hypotenuse $AC$ such that $BD$ is perpendicular to $AC$ , and point $E$ lies on leg $BC$ such that $DE$ is perpendicular to $BC$ . What is the area of quadrilateral $ABDE$ ?

\frac{135\sqrt{3}}{8}

\frac{117\sqrt{3}}{8}

\frac{27\sqrt{3}}{2}

\frac{63\sqrt{3}}{4}

\frac{144\sqrt{3} - 9}{8}

Show answer & explanation

Answer: $\frac{135\sqrt{3}}{8}$

Answer

The area of quadrilateral

ABDE

is

\frac{135\sqrt{3}}{8}

.

The area of quadrilateral

ABDE

is obtained by subtracting the area of the smaller right triangle

\triangle DEC

from the area of the outer right triangle

\triangle ABC

. Since both triangles share the

60^\circ

angle at vertex

C

, repeated application of the

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

ratio (

1:\sqrt{3}:2

) gives

\text{Area}(\triangle ABC) = 18\sqrt{3}

and

\text{Area}(\triangle DEC) = \frac{9\sqrt{3}}{8}

. Subtraction yields

\frac{135\sqrt{3}}{8}

.

Step-by-Step Solution

1

Calculate side lengths and area of

\triangle ABC

.

BC = 6

,

AB = 6\sqrt{3}

, and

\text{Area}(\triangle ABC) = 18\sqrt{3}

.

In

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

triangle

ABC

with hypotenuse

AC = 12

, the side opposite

30^\circ

(

BC

) is half the hypotenuse (

BC = 6

), and the side opposite

60^\circ

(

AB

) is

6\sqrt{3}

. The area is

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

.

2

Find the hypotenuse

DC

of

\triangle BDC

.

DC = 3

.

Since

BD \perp AC

,

\triangle BDC

is a right triangle with

\angle BDC = 90^\circ

and

\angle C = 60^\circ

, making

\angle DBC = 30^\circ

. Its hypotenuse is

BC = 6

. The side opposite

30^\circ

is

DC = \frac{1}{2} BC = 3

.

3

Determine the sides and area of

\triangle DEC

.

EC = \frac{3}{2}

,

DE = \frac{3\sqrt{3}}{2}

, and

\text{Area}(\triangle DEC) = \frac{9\sqrt{3}}{8}

.

Since

DE \perp BC

,

\triangle DEC

is another

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

right triangle with hypotenuse

DC = 3

. Thus

EC = \frac{3}{2}

and

DE = \frac{3\sqrt{3}}{2}

. Its area is

\frac{1}{2} \times \frac{3}{2} \times \frac{3\sqrt{3}}{2} = \frac{9\sqrt{3}}{8}

.

4

Subtract

\text{Area}(\triangle DEC)

from

\text{Area}(\triangle ABC)

to get the area of quadrilateral

ABDE

.

\text{Area}(ABDE) = 18\sqrt{3} - \frac{9\sqrt{3}}{8} = \frac{135\sqrt{3}}{8}

.

Quadrilateral

ABDE

is formed by removing

\triangle DEC

from

\triangle ABC

.

Key Concept

Iterative application of

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

special right triangle side ratios (

1:\sqrt{3}:2

) along dropped altitudes.

Estimated Time:2m 30s

Question 32Question →

An isosceles triangle has two sides of length $10$ units each and a base of length $12$ units. A line segment parallel to the base cuts through the triangle, creating a smaller top triangle with an area of $12$ square units. What is the perimeter of the resulting trapezoid?

Show answer & explanation

Answer: 28

Answer

The perimeter of the trapezoid is 28.

First, find the height of the original isosceles triangle with sides 10, 10, and base 12 by applying the Pythagorean theorem to half of the base:

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

. The area of the original triangle is

\frac{1}{2} \times 12 \times 8 = 48

square units. Because the segment is parallel to the base, the smaller top triangle is similar to the original triangle. The ratio of their areas is

\frac{12}{48} = \frac{1}{4}

, which means the linear scale factor is

\sqrt{\frac{1}{4}} = \frac{1}{2}

. Thus, the top triangle has legs of length

5

and a base of length

6

. The remaining non-parallel sides of the trapezoid each measure

10 - 5 = 5

units, and its bottom base is

12

units. Summing these four side lengths gives

5 + 5 + 6 + 12 = 28

.

Step-by-Step Solution

1

Calculate the height and area of the original isosceles triangle

The altitude to the base bisects the base into two segments of length

6

. The altitude length is

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

. The area of the original triangle is

\frac{1}{2} \times 12 \times 8 = 48

square units.

Splitting the isosceles triangle along its altitude creates two right triangles with hypotenuse 10 and base leg 6.

2

Determine the linear scale factor of the smaller top triangle

The ratio of the area of the smaller triangle to the original triangle is

\frac{12}{48} = \frac{1}{4}

. Taking the square root yields a linear scale factor of

k = \sqrt{\frac{1}{4}} = \frac{1}{2}

.

A line parallel to the base forms a smaller triangle similar to the original triangle, and the ratio of areas of similar triangles is equal to the square of their linear scale factor.

3

Find the side lengths of the smaller triangle and the remaining side segments

The sides of the smaller triangle are

\frac{1}{2} \times 10 = 5

,

\frac{1}{2} \times 10 = 5

, and base

\frac{1}{2} \times 12 = 6

. The non-parallel side segments of the trapezoid are each

10 - 5 = 5

.

Multiplying the dimensions of the original triangle by the linear scale factor gives the side lengths of the top triangle.

4

Calculate the perimeter of the trapezoid

Perimeter =

5 + 5 + 6 + 12 = 28

.

Sum the lengths of the four boundary segments of the trapezoid.

Key Concept

Properties of isosceles triangles, Pythagorean theorem, area calculations, and similar triangle area ratios

Question 33Question →

In the $xy$ -plane, $A(1, 2)$ , $B(7, 4)$ , and $D(3, 8)$ are three vertices of rhombus $ABCD$ . A line $k$ passes through the origin $(0, 0)$ and bisects the area of rhombus $ABCD$ . What is the slope of line $k$ ?

\frac{6}{5}

\frac{5}{6}

1

-1

\frac{3}{4}

Show answer & explanation

Answer: $\frac{6}{5}$

Answer

The slope of line

k

is

\frac{6}{5}

.

Any line that divides a parallelogram or rhombus into two equal areas must pass through its center of symmetry, which is the midpoint of its diagonals. The midpoint of diagonal

BD

with endpoints

(7, 4)

and

(3, 8)

is

\left(\frac{7+3}{2}, \frac{4+8}{2}\right) = (5, 6)

. Since line

k

passes through the origin

(0, 0)

and

(5, 6)

, its slope is

\frac{6 - 0}{5 - 0} = \frac{6}{5}

.

Step-by-Step Solution

1

Identify the key geometric property of area-bisecting lines for parallelograms and rhombuses.

Any line that bisects the area of a rhombus must pass through its center of symmetry (the intersection point of its diagonals).

A rhombus is centrally symmetric about the intersection point of its diagonals, so any line through this point divides the rhombus into two congruent regions.

2

Find the coordinates of the center of symmetry by calculating the midpoint of diagonal

BD

.

Midpoint

M = \left(\frac{7+3}{2}, \frac{4+8}{2}\right) = (5, 6)

.

Opposite vertices

B(7,4)

and

D(3,8)

define one of the diagonals of rhombus

ABCD

.

3

Calculate the slope of line

k

passing through the origin

(0, 0)

and center point

M(5, 6)

.

\text{Slope } m = \frac{6 - 0}{5 - 0} = \frac{6}{5}.

The slope formula between

(x_1, y_1)

and

(x_2, y_2)

is

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

.

Key Concept

Center of Symmetry and Area Bisectors of Quadrilaterals

Estimated Time:2m 0s

Question 34Question →

In right trapezoid $ABCD$ , segment $AB$ is parallel to segment $CD$ , $\angle DAB = 90^\circ$ , $AD = 12$ , $CD = 15$ , and $BC = 13$ . Point $E$ lies on segment $CD$ such that quadrilateral $ABCE$ is a parallelogram. What is the perimeter of triangle $ADE$ ?

Show answer & explanation

Answer: 30

Answer

30

Decomposing right trapezoid

ABCD

by dropping a perpendicular from

B

to

CD

forms a right triangle with height

12

and hypotenuse

13

. The Pythagorean theorem gives the base of this right triangle as

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

. Subtracting this from

CD = 15

yields

AB = 10

. Because

ABCE

is a parallelogram,

CE = AB = 10

, which leaves

DE = CD - CE = 15 - 10 = 5

. Triangle

ADE

is a right triangle with legs

AD = 12

and

DE = 5

, giving hypotenuse

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

. The perimeter of triangle

ADE

is

12 + 5 + 13 = 30

.

Step-by-Step Solution

1

Calculate the horizontal projection of segment

BC

onto base

CD

The length of the horizontal projection is

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

Segment

AD = 12

defines the perpendicular distance between parallel lines

AB

and

CD

2

Determine the length of parallel base

AB

AB = 15 - 5 = 10

The total length of base

CD = 15

is the sum of

AB

and the horizontal projection of slant side

BC

3

Calculate the length of segment

DE

DE = 15 - 10 = 5

Quadrilateral

ABCE

is a parallelogram, which implies

CE = AB = 10

4

Compute the hypotenuse

AE

and the total perimeter of triangle

ADE

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

, so

\text{Perimeter} = 12 + 5 + 13 = 30

Triangle

ADE

is a right-angled triangle with right angle at vertex

D

Key Concept

Trapezoid height decomposition, parallelogram side properties, and Pythagorean theorem application

Question 35Question →

A circle has a radius of $6$ . A central angle of $60^\circ$ intercepts an arc on the circle. What is the perimeter of the sector defined by this central angle?

2\pi

6\pi

2\pi + 6

2\pi + 12

12\pi

Show answer & explanation

Answer: $2\pi + 12$

Answer

2\pi + 12

The sector's perimeter includes the curved arc length and the two straight radii bounding it. The central angle of

60^\circ

represents

\frac{60}{360} = \frac{1}{6}

of the full circle. The arc length is

\frac{1}{6} \times 2\pi(6) = 2\pi

. Adding the two radii of length

6

gives

2\pi + 6 + 6 = 2\pi + 12

.

Step-by-Step Solution

1

Calculate the arc length of the sector

Arc length =

2\pi

The arc length formula is

\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r

. Substituting

\theta = 60^\circ

and

r = 6

yields

\frac{60}{360} \times 12\pi = \frac{1}{6} \times 12\pi = 2\pi

.

2

Calculate the perimeter of the sector by adding the arc length to the two radii

Perimeter =

2\pi + 12

The perimeter of a sector consists of the arc length plus two radii (

2r

). Thus,

\text{Perimeter} = 2\pi + 2(6) = 2\pi + 12

.

Key Concept

Perimeter of a sector equals arc length plus twice the radius (

L + 2r

).

Estimated Time:45s

Question 36Question →

A circle has a radius of $10$ units. A sector within this circle is defined by a central angle of $72^\circ$ . Which of the following statements about this sector are true? Select all such statements.

Select all that apply

The arc length of the sector is

4\pi

units.

The area of the sector is

20\pi

square units.

The area of the sector is

100\pi

square units.

The arc length of the sector is

20\pi

units.

The ratio of the sector's area to the circle's total area is

1

to

5

.

Show answer & explanation

Answer: The arc length of the sector is $4\pi$ units.; The area of the sector is $20\pi$ square units.; The ratio of the sector's area to the circle's total area is $1$ to $5$ .

Answer

The statements confirming an arc length of

4\pi

units, a sector area of

20\pi

square units, and a sector-to-total area ratio of

1

to

5

are correct.

The central angle fraction is

\frac{72^\circ}{360^\circ} = \frac{1}{5}

. Multiplying the full circumference

20\pi

by

\frac{1}{5}

gives an arc length of

4\pi

. Multiplying the total area

100\pi

by

\frac{1}{5}

gives a sector area of

20\pi

. The ratio of sector area to total area is also equal to

\frac{1}{5}

.

Step-by-Step Solution

1

Find the central angle fraction of the circle

The fraction is

\frac{72^\circ}{360^\circ} = \frac{1}{5}

Arc length and sector area are proportional to the central angle relative to a full

360^\circ

turn.

2

Calculate the arc length of the sector

Arc Length

= \frac{1}{5} \times 2\pi(10) = 4\pi

units

The arc length is the circle's circumference multiplied by the central angle fraction.

3

Calculate the area of the sector

Sector Area

= \frac{1}{5} \times \pi(10)^2 = 20\pi

square units

The sector area is the circle's total area multiplied by the central angle fraction.

4

Determine the area ratio

Ratio

= \frac{\text{Sector Area}}{\text{Total Area}} = \frac{20\pi}{100\pi} = \frac{1}{5}

The ratio of the sector area to the total area is identical to the central angle fraction.

Key Concept

Arc Length and Sector Area Formulas

Estimated Time:1m 0s

Question 37Question →

Three straight lines, $R$ , $S$ , and $T$ , all intersect at a single point $P$ . Line $R$ is perpendicular to line $S$ . Line $T$ intersects line $R$ such that one of the acute angles formed between line $R$ and line $T$ measures $35^\circ$ . Which of the following degree measures represent angles formed between any two of the intersecting lines at point $P$ ? Select all that apply.

Select all that apply

55^\circ

90^\circ

145^\circ

115^\circ

70^\circ

Show answer & explanation

Answer: $55^\circ$ ; $90^\circ$ ; $145^\circ$

Answer

The degree measures

55^\circ

,

90^\circ

, and

145^\circ

are all valid angle measures formed by the intersecting lines at point

P

.

The intersecting lines form several distinct angle measures at point

P

. The right angle between lines

R

and

S

measures

90^\circ

. The acute angle between lines

T

and

S

is complementary to the

35^\circ

angle, giving

90^\circ - 35^\circ = 55^\circ

. The obtuse angle between lines

T

and

R

along straight line

R

is supplementary to

35^\circ

, giving

180^\circ - 35^\circ = 145^\circ

. Therefore, the options stating

55^\circ

,

90^\circ

, and

145^\circ

are all correct.

Step-by-Step Solution

1

Identify the angle measure between perpendicular lines R and S.

Since line

R \perp S

, the angle between them is

90^\circ

.

Perpendicular lines intersect at right angles (

90^\circ

).

2

Calculate the acute angle between line T and line S.

Angle between

T

and

S

=

90^\circ - 35^\circ = 55^\circ

.

The given

35^\circ

angle between

R

and

T

and the adjacent angle between

T

and

S

form the

90^\circ

right angle between

R

and

S

.

3

Determine the supplementary obtuse angles formed by line T with line R and line S.

Supplementary angle to

35^\circ

is

180^\circ - 35^\circ = 145^\circ

. Supplementary angle to

55^\circ

is

180^\circ - 55^\circ = 125^\circ

.

Adjacent angles along a straight line sum to

180^\circ

.

4

Compare calculated angle measures with the options.

The valid angle measures are

35^\circ

,

55^\circ

,

90^\circ

,

125^\circ

, and

145^\circ

. Thus

55^\circ

,

90^\circ

, and

145^\circ

are correct.

Matching calculated angle measures with the choices provided.

Key Concept

Perpendicular line relationships and supplementary angle properties of intersecting lines.

Estimated Time:1m 0s

Question 38Question →

A circle has a circumference of $20\pi$ . A sector of this circle is defined by a central angle that intercepts an arc of length $5\pi$ . What is the area of this sector?

10\pi

25\pi

50\pi

100\pi

5\pi

Show answer & explanation

Answer: $25\pi$

Answer

The area of the sector is

25\pi

.

First, the radius is found using

2\pi r = 20\pi

, which gives

r = 10

. The total area of the circle is

\pi (10)^2 = 100\pi

. The fraction of the circle defined by the sector is

\frac{5\pi}{20\pi} = \frac{1}{4}

. Multiplying the total area by this fraction gives

\frac{1}{4} \times 100\pi = 25\pi

.

Step-by-Step Solution

1

Find the radius of the circle from the circumference.

Since

C = 2\pi r = 20\pi

, solving for

r

gives

r = 10

.

The radius is required to calculate the total circle area.

2

Determine the fraction of the circle that the sector represents.

\text{Fraction} = \frac{\text{Arc Length}}{\text{Circumference}} = \frac{5\pi}{20\pi} = \frac{1}{4}.

The ratio of arc length to circumference gives the proportion of the total circle occupied by the sector.

3

Calculate the total area of the circle.

A = \pi r^2 = \pi (10)^2 = 100\pi.

The area of a circle with radius 10 is

100\pi

.

4

Multiply the total area by the sector fraction.

\text{Sector Area} = \frac{1}{4} \times 100\pi = 25\pi.

Applying the proportional fraction yields the sector area.

Key Concept

The ratio of a sector's arc length to the full circumference is equal to the ratio of the sector's area to the full circle's area.

Question 39Question →

In the geometric plane, line $k$ is parallel to line $m$ . A transversal line $t$ intersects line $k$ at point $P$ and line $m$ at point $Q$ . Ray $PR$ extends along line $k$ to the right of $P$ , and ray $QS$ extends along line $m$ to the right of $Q$ . The measure of interior angle $\angle RPQ$ is represented by $(3x + 20)^\circ$ and the measure of interior angle $\angle PQS$ is represented by $(2x + 10)^\circ$ . Which of the following statements must be true? Select all such statements.

Select all that apply

The value of

x

is

30

.

The measure of angle

\angle RPQ

is

110^\circ

.

The acute angle formed by line

t

and line

k

at point

P

measures

70^\circ

.

The value of

x

is

12

.

Angles

\angle RPQ

and

\angle PQS

are complementary angles.

Show answer & explanation

Answer: The value of $x$ is $30$ .; The measure of angle $\angle RPQ$ is $110^\circ$ .; The acute angle formed by line $t$ and line $k$ at point $P$ measures $70^\circ$ .

Answer

The true statements are that

x = 30

, the measure of angle

\angle RPQ

is

110^\circ

, and the acute angle formed by line

t

and line

k

at point

P

measures

70^\circ

.

Lines

k

and

m

are parallel, so interior angles on the same side of transversal

t

(angles

\angle RPQ

and

\angle PQS

) are supplementary. Solving

(3x + 20) + (2x + 10) = 180

gives

x = 30

. Substituting

x = 30

yields

\angle RPQ = 110^\circ

and

\angle PQS = 70^\circ

. The adjacent angle to

\angle RPQ

at point

P

is

180^\circ - 110^\circ = 70^\circ

, which is acute.

Step-by-Step Solution

1

Set up the geometric equation using the parallel line angle relationship.

(3x + 20) + (2x + 10) = 180

When two parallel lines are cut by a transversal, consecutive interior angles on the same side of the transversal are supplementary (their sum is

180^\circ

).

2

Solve the linear equation for

x

.

5x + 30 = 180 \implies 5x = 150 \implies x = 30

Combine like terms and isolate

x

using basic algebra.

3

Calculate the specific angle measures.

m

\angle RPQ = 3(30) + 20 = 110^\circ

and m

\angle PQS = 2(30) + 10 = 70^\circ

Substitute

x = 30

into the given algebraic angle expressions.

4

Determine the supplementary acute angle at point

P

.

180^\circ - 110^\circ = 70^\circ

Angles along a straight line form a linear pair and sum to

180^\circ

.

Key Concept

Parallel Lines and Consecutive Interior Angles

Question 40Question →

Circle $K$ has a radius of $10$ . Points $P$ and $Q$ lie on circle $K$ such that central angle $\angle POQ$ measures $72^\circ$ , where point $O$ is the center of circle $K$ . Which of the following statements are true? Select all that apply.

Select all that apply

The length of minor arc

PQ

is

4\pi

.

The area of sector

POQ

is

20\pi

.

The perimeter of sector

POQ

is

4\pi

.

The area of sector

POQ

is

40\pi

.

The ratio of the area of sector

POQ

to the total area of circle

K

is

1

to

4

.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ .; The area of sector $POQ$ is $20\pi$ .

Answer

The correct statements are that the length of minor arc

PQ

is

4\pi

, and the area of sector

POQ

is

20\pi

.

The central angle of

72^\circ

corresponds to

\frac{72^\circ}{360^\circ} = \frac{1}{5}

of the entire circle. The circle has a circumference of

2\pi(10) = 20\pi

and an area of

\pi(10)^2 = 100\pi

. Taking

\frac{1}{5}

of the circumference gives an arc length of

4\pi

, and taking

\frac{1}{5}

of the total area gives a sector area of

20\pi

. Both of these statements are mathematically accurate.

Step-by-Step Solution

1

Find the central angle fraction relative to the full circle

\text{Fraction} = \frac{72^\circ}{360^\circ} = \frac{1}{5}

A complete circle subtends

360^\circ

, so the arc and sector comprise one-fifth of the circle.

2

Calculate the circumference and length of minor arc

PQ

\text{Circumference} = 2\pi(10) = 20\pi, \quad \text{Arc Length } PQ = \frac{1}{5} \times 20\pi = 4\pi

Arc length is the central angle fraction multiplied by the total circumference.

3

Calculate the total circle area and area of sector

POQ

\text{Total Area} = \pi(10)^2 = 100\pi, \quad \text{Sector Area } POQ = \frac{1}{5} \times 100\pi = 20\pi

Sector area is the central angle fraction multiplied by the total area of the circle.

4

Evaluate sector perimeter and area ratio for remaining options

\text{Perimeter} = 4\pi + 2(10) = 4\pi + 20; \quad \text{Ratio} = \frac{20\pi}{100\pi} = \frac{1}{5}

Sector perimeter requires adding the two straight radii to the arc length, and the ratio of sector area to total area is

1:5

.

Key Concept

Arc Length and Sector Area of a Circle

Estimated Time:1m 30s

Geometry

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