Geometry Alıştırma Soruları — GRE General Test

Soru 1Soru →

Sector $S_1$ belongs to a circle with radius $r$ and has a central angle of measure $\theta^\circ$ , where $0 < \theta < 360$ . Sector $S_2$ belongs to a circle with radius $2r$ and has a central angle of measure $\left(\frac{\theta}{2}\right)^\circ$ . Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

The area of sector

S_2

is twice the area of sector

S_1

.

The arc length of sector

S_2

is equal to the arc length of sector

S_1

.

The perimeter of sector

S_2

is twice the perimeter of sector

S_1

.

The ratio of the area of sector

S_1

to its arc length is half the ratio of the area of sector

S_2

to its arc length.

The perimeter of sector

S_2

is equal to the perimeter of sector

S_1

.

Cevabı ve açıklamayı göster

Cevap: The area of sector $S_2$ is twice the area of sector $S_1$ .; The arc length of sector $S_2$ is equal to the arc length of sector $S_1$ .; The ratio of the area of sector $S_1$ to its arc length is half the ratio of the area of sector $S_2$ to its arc length.

Cevap

The true statements are that the area of sector

S_2

is twice the area of sector

S_1

, the arc length of sector

S_2

is equal to the arc length of sector

S_1

, and the ratio of area to arc length for sector

S_1

is half that of sector

S_2

.

The area of sector

S_2

is twice that of

S_1

because quadrupling

r^2

combined with halving the central angle results in a factor of 2. The arc length of sector

S_2

equals that of

S_1

because doubling

r

and halving the angle cancel each other out. The area-to-arc-length ratio of any sector reduces to

\frac{R}{2}

, so sector

S_1

with radius

r

has ratio

\frac{r}{2}

, which is half the ratio

r

of sector

S_2

.

Adım Adım Çözüm

1

Calculate and compare the sector areas.

\text{Area}(S_1) = \frac{\theta}{360}\pi r^2

and

\text{Area}(S_2) = \frac{\theta/2}{360}\pi (2r)^2 = \frac{\theta}{720}\pi (4r^2) = \frac{2\theta}{360}\pi r^2 = 2 \cdot \text{Area}(S_1)

.

The sector area formula is

\text{Area} = \frac{\text{angle}}{360^\circ} \pi R^2

. Doubling the radius quadruples

R^2

, while halving the angle reduces the fraction by half, producing a net doubling of area.

2

Calculate and compare the arc lengths.

\text{Arc}(S_1) = \frac{\theta}{360}(2\pi r)

and

\text{Arc}(S_2) = \frac{\theta/2}{360}(2\pi \cdot 2r) = \frac{\theta}{360}(2\pi r) = \text{Arc}(S_1)

.

The arc length formula is

\text{Arc} = \frac{\text{angle}}{360^\circ} (2\pi R)

. Doubling the radius doubles

R

, while halving the angle halves the fraction, leaving the product unchanged.

3

Calculate and compare the sector perimeters.

\text{Perimeter}(S_1) = 2r + \text{Arc}(S_1)

and

\text{Perimeter}(S_2) = 4r + \text{Arc}(S_2) = 4r + \text{Arc}(S_1)

.

The perimeter of a sector consists of two straight radii and the curved arc length.

2 \cdot \text{Perimeter}(S_1) = 4r + 2\text{Arc}(S_1) \neq \text{Perimeter}(S_2)

.

4

Evaluate the area-to-arc-length ratios for both sectors.

For

S_1

,

\frac{\text{Area}(S_1)}{\text{Arc}(S_1)} = \frac{\frac{\theta}{360}\pi r^2}{\frac{\theta}{360}2\pi r} = \frac{r}{2}

. For

S_2

,

\frac{\text{Area}(S_2)}{\text{Arc}(S_2)} = \frac{2r}{2} = r

.

The ratio of area to arc length simplifies to

\frac{R}{2}

for any sector, so doubling the radius doubles this ratio.

Anahtar Kavram

Geometric properties of circle sectors, including proportional relationships between radius, central angle, arc length, sector area, and total sector perimeter.

Tahmini Süre:2m 0s

Soru 2Soru →

In circle $O$ , sector $AOB$ has a central angle of $60^\circ$ and a radius of $12$ . A smaller circle $C$ is inscribed within sector $AOB$ such that it is tangent to radii $OA$ and $OB$ , as well as to arc $AB$ . What is the area of the region inside sector $AOB$ that lies outside circle $C$ ?

4\pi

8\pi

12\pi

16\pi

18\pi

Cevabı ve açıklamayı göster

Cevap: $8\pi$

Cevap

The area of the region inside sector

AOB

outside circle

C

is

8\pi

.

The correct answer is derived by first finding the area of sector

AOB

using

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

. Then, analyzing the geometry of the inscribed circle reveals that the line from

O

to the center of circle

C

bisects the

60^\circ

angle. In the resulting

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

right triangle, the hypotenuse length is

2r

, making the total radius of sector

AOB

equal to

2r + r = 3r = 12

, which yields

r = 4

. The area of circle

C

is

\pi (4^2) = 16\pi

. Subtracting the circle area from the sector area gives

24\pi - 16\pi = 8\pi

.

Adım Adım Çözüm

1

Calculate the area of sector

AOB

Sector area

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

The area of a sector with central angle

\theta

and radius

R

is

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

.

2

Find the radius

r

of the inscribed circle

C

Distance from

O

to center of circle

C

is

OP = 2r

, so total radius

R = OP + r = 3r = 12 \implies r = 4

The line segment connecting center

O

to center

P

of circle

C

bisects the

60^\circ

angle into two

30^\circ

angles. A perpendicular dropped from

P

to radius

OA

forms a

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

right triangle where

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

, giving

OP = 2r

.

3

Calculate the area of circle

C

Area of circle

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

The area of a circle with radius

r

is

\pi r^2

.

4

Subtract the area of circle

C

from the area of sector

AOB

24\pi - 16\pi = 8\pi

The desired region is the difference between the full sector area and the enclosed circle's area.

Anahtar Kavram

Inscribed circles in sectors and central angle sector area calculations

Tahmini Süre:2m 30s

Soru 3Soru →

An equilateral triangle $ABC$ with side length $6\sqrt{3}$ is inscribed in a circle with center $O$ . What is the area of sector $AOB$ , divided by $\pi$ ?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The area of sector

AOB

divided by

\pi

is 12.

Since triangle

ABC

is equilateral, its three vertices divide the circle into three congruent arcs of

120^\circ

each. Thus, central angle

\angle AOB = 120^\circ

. The relationship between the side length

s

of an inscribed equilateral triangle and the radius

R

of its circumscribed circle is

s = R\sqrt{3}

. Given

s = 6\sqrt{3}

, we solve for

R

to find

R = 6

. The area of sector

AOB

is

\frac{120^\circ}{360^\circ} \pi R^2 = \frac{1}{3} \pi (6^2) = 12\pi

. Dividing this area by

\pi

yields

12

.

Adım Adım Çözüm

1

Find the central angle

\angle AOB

corresponding to side

AB

of the inscribed equilateral triangle.

\angle AOB = 120^\circ

An inscribed equilateral triangle divides the

360^\circ

circle into three equal central angles.

2

Calculate the radius

R

of the circumscribed circle from the given side length

s = 6\sqrt{3}

.

R = 6

In an inscribed equilateral triangle,

s = R\sqrt{3}

. Substituting

6\sqrt{3} = R\sqrt{3}

gives

R = 6

.

3

Compute the area of sector

AOB

and divide by

\pi

.

12

\text{Sector Area} = \frac{120^\circ}{360^\circ} \times \pi \times 6^2 = 12\pi

. Dividing by

\pi

leaves

12

.

Anahtar Kavram

Relationship between inscribed shapes, circle radii, and sector area

Soru 4Soru →

A right circular cylinder has base radius $r$ and height $h$ . The height of the cylinder is increased by $50\%$ , and its base radius is decreased by $20\%$ . Which of the following statements about the modified cylinder compared to the original cylinder must be true? Select all that apply.

Geçerli olan tümünü seçin

The volume of the cylinder decreases by

4\%

.

The lateral surface area of the cylinder increases by

20\%

.

The combined area of the two circular bases decreases by

20\%

.

If

h = r

, the total surface area of the cylinder decreases.

If

h = 4r

, the total surface area of the cylinder decreases.

Cevabı ve açıklamayı göster

Cevap: The volume of the cylinder decreases by $4\%$ .; The lateral surface area of the cylinder increases by $20\%$ .; If $h = r$ , the total surface area of the cylinder decreases.

Cevap

The statements confirming a

4\%

volume decrease, a

20\%

lateral surface area increase, and a total surface area decrease when

h = r

are all true.

The volume of the modified cylinder decreases by

4\%

because

0.8^2 \times 1.5 = 0.96

. The lateral surface area increases by

20\%

because

0.8 \times 1.5 = 1.20

. When

h = r

, the original total surface area

4\pi r^2

reduces to

3.68\pi r^2

, confirming a decrease.

Adım Adım Çözüm

1

Express original and modified dimensions algebraically.

Original: radius

= r

, height

= h

. Modified: radius

= 0.8r

, height

= 1.5h

.

Decreasing radius by

20\%

multiplies it by

1 - 0.20 = 0.80

, while increasing height by

50\%

multiplies it by

1 + 0.50 = 1.50

.

2

Calculate the ratio of modified volume to original volume.

V_{new} = \pi (0.8r)^2 (1.5h) = 0.64 \times 1.5 \pi r^2 h = 0.96 V_{old}

, indicating a

4\%

volume decrease.

Volume of a cylinder is given by

V = \pi r^2 h

.

3

Calculate the ratio of modified lateral surface area to original lateral surface area.

L_{new} = 2\pi (0.8r)(1.5h) = 2.4\pi r h = 1.20 L_{old}

, indicating a

20\%

lateral surface area increase.

Lateral surface area of a cylinder is given by

L = 2\pi r h

.

4

Evaluate the base area change and total surface area under specific height-to-radius ratios.

Base area becomes

(0.8)^2 = 0.64

of original (

36\%

decrease). For

h = r

, total surface area changes from

4\pi r^2

to

3.68\pi r^2

(decrease). For

h = 4r

, total surface area changes from

10\pi r^2

to

10.88\pi r^2

(increase).

Total surface area is the sum of lateral surface area and base area,

A = 2\pi r^2 + 2\pi r h

.

Anahtar Kavram

Impact of dimensional scaling on volume, lateral surface area, and total surface area of cylinders

Soru 5Soru →

A vertical flagpole stands on flat, horizontal ground. An observer at point $S$ on the ground measures the angle of elevation to the top of the flagpole to be $60^\circ$ . A second observer at point $P$ on the ground, located $30$ feet further from the base of the flagpole along the same line extending from the base through $S$ , measures the angle of elevation to the top of the flagpole to be $30^\circ$ . What is the height of the flagpole, in feet?

15\sqrt{3}

30\sqrt{3}

45

30

15

Cevabı ve açıklamayı göster

Cevap: $15\sqrt{3}$

Cevap

The height of the flagpole is

15\sqrt{3}

feet.

Let

B

be the base and

T

be the top of the flagpole. The right triangle

TBS

has angles

30^\circ

,

60^\circ

, and

90^\circ

at

B

. If

BS = x

, then the height

TB = x\sqrt{3}

. In right triangle

TBP

, the angle at

P

is

30^\circ

, so the base

BP = TB\sqrt{3} = (x\sqrt{3})\sqrt{3} = 3x

. Since

BP = BS + 30

, we set up

3x = x + 30

, yielding

x = 15

. Therefore, the height

TB = 15\sqrt{3}

feet.

Adım Adım Çözüm

1

Model the scenario using right triangles.

Let

B

be the base of the flagpole and

T

be the top. Triangle

TBS

is a

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

right triangle at

B

, and triangle

TBP

is also a

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

right triangle at

B

.

The flagpole is perpendicular to the horizontal ground.

2

Express side lengths in terms of base distance

x = BS

.

In

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

triangle

TBS

,

TB = x\sqrt{3}

and

ST = 2x

.

The side opposite the

60^\circ

angle is

\sqrt{3}

times the side opposite the

30^\circ

angle.

3

Set up the equation for triangle

TBP

.

In

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

triangle

TBP

, the side opposite the

30^\circ

angle is

TB = x\sqrt{3}

, so the adjacent side

BP = (x\sqrt{3})\sqrt{3} = 3x

.

The side opposite the

60^\circ

angle (

BP

) is

\sqrt{3}

times the side opposite the

30^\circ

angle (

TB

).

4

Solve for

x

using the given distance

SP = 30

.

Since

BP = BS + SP

, we have

3x = x + 30

, which simplifies to

2x = 30

, so

x = 15

.

The total horizontal distance

BP

is the sum of segments

BS

and

SP

.

5

Calculate the height

TB

.

TB = 15\sqrt{3}

.

Substitute

x = 15

into

TB = x\sqrt{3}

.

Anahtar Kavram

Special Right Triangle Ratios (

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

)

Tahmini Süre:1m 30s

Soru 6Soru →

In convex quadrilateral $ABCD$ , diagonal $AC$ is drawn. It is given that $AB = 9$ , $BC = 12$ , $CD = 8$ , and $DA = 15$ , with $\angle ABC = 90^\circ$ . Which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

The length of diagonal

AC

is 15.

The area of quadrilateral

ABCD

is 114.

The perimeter of quadrilateral

ABCD

is 44.

A triangle can be formed using side lengths of 8, 15, and 32.

The sum of the interior angles of quadrilateral

ABCD

is

360^\circ

.

Cevabı ve açıklamayı göster

Cevap: The length of diagonal $AC$ is 15.; The perimeter of quadrilateral $ABCD$ is 44.; The sum of the interior angles of quadrilateral $ABCD$ is $360^\circ$ .

Cevap

The correct statements are that the length of diagonal

AC

is 15, the perimeter of quadrilateral

ABCD

is 44, and the sum of the interior angles of quadrilateral

ABCD

is

360^\circ

.

The statement specifying that the length of diagonal

AC

is 15 is correct because right triangle

ABC

has leg lengths 9 and 12, giving hypotenuse

\sqrt{81 + 144} = 15

. The statement specifying that the perimeter is 44 is correct because summing the outer side lengths yields

9 + 12 + 8 + 15 = 44

. The statement specifying that the interior angle sum is

360^\circ

is correct because every convex quadrilateral has an interior angle sum of

(4 - 2) \times 180^\circ = 360^\circ

.

Adım Adım Çözüm

1

Calculate the length of diagonal

AC

using triangle

ABC

.

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

.

Since

\angle ABC = 90^\circ

, triangle

ABC

is a right triangle, allowing the application of the Pythagorean theorem.

2

Calculate the perimeter of quadrilateral

ABCD

.

Perimeter

= 9 + 12 + 8 + 15 = 44

.

The perimeter of a polygon is the sum of all its outer side lengths.

3

Verify the sum of the interior angles for quadrilateral

ABCD

.

Interior angle sum

= (4 - 2) \times 180^\circ = 360^\circ

.

The formula

(n - 2) \times 180^\circ

applies to all convex polygons.

4

Evaluate the incorrect claims regarding area and triangle formation.

Triangle

ADC

is not a right triangle (

8^2 + 15^2 \neq 15^2

), so its area is not 60; and side lengths 8, 15, and 32 violate the triangle inequality theorem (

8 + 15 = 23 < 32

).

Right triangle formulas require a right angle, and valid triangle side lengths must satisfy the triangle inequality theorem.

Anahtar Kavram

Properties of convex quadrilaterals, right triangle side relationships, and triangle inequality bounds

Tahmini Süre:1m 30s

Soru 7Soru →

In the plane, line $L_1$ is parallel to line $L_2$ . A transversal line $T$ intersects $L_1$ and $L_2$ . One of the acute angles formed at the intersection of $L_1$ and $T$ measures $(4x - 10)^\circ$ , and an alternate interior angle on $L_2$ measures $(2x + 30)^\circ$ . What is the degree measure of one of the obtuse angles formed at the intersection of $L_1$ and $T$ ?

20^\circ

50^\circ

70^\circ

110^\circ

140^\circ

Cevabı ve açıklamayı göster

Cevap: $110^\circ$

Cevap

The degree measure of the obtuse angle is

110^\circ

.

Since lines

L_1

and

L_2

are parallel, alternate interior angles are equal in measure. Setting

(4x - 10)^\circ = (2x + 30)^\circ

gives

2x = 40

, so

x = 20

. Substituting

x = 20

into

4x - 10

gives an acute angle of

70^\circ

. Because angles on a straight line are supplementary, the obtuse angle measures

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

, which corresponds to the correct choice.

Adım Adım Çözüm

1

Set alternate interior angles equal to solve for

x

.

4x - 10 = 2x + 30 \implies 2x = 40 \implies x = 20

When two parallel lines are cut by a transversal, alternate interior angles are congruent.

2

Substitute

x = 20

into the expression for the acute angle.

Acute angle

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

This yields the degree measure of the acute angle formed by the intersection.

3

Calculate the measure of the supplementary obtuse angle.

Obtuse angle

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

Adjacent angles along a straight line are supplementary and sum to

180^\circ

.

Anahtar Kavram

Alternate Interior Angles and Supplementary Angles

Tahmini Süre:50s

Soru 8Soru →

In triangle $ABC$ , the length of side $AB$ is $15$ and the length of side $BC$ is $25$ . The area of triangle $ABC$ is $150$ . If angle $ABC$ is obtuse, what is the length of side $AC$ ?

20

5\sqrt{34}

10\sqrt{13}

34

40

Cevabı ve açıklamayı göster

Cevap: $10\sqrt{13}$

Cevap

10\sqrt{13}

The correct answer

10\sqrt{13}

is obtained by drawing altitude

AH = 12

to line

BC

. In right triangle

ABH

, the base projection is

BH = \sqrt{15^2 - 12^2} = 9

. Because angle

ABC

is obtuse, point

H

lies outside segment

BC

, so

CH = 25 + 9 = 34

. Applying the Pythagorean theorem to right triangle

AHC

gives

AC = \sqrt{12^2 + 34^2} = \sqrt{1300} = 10\sqrt{13}

.

Adım Adım Çözüm

1

Calculate the height (altitude)

h

perpendicular to line

BC

.

Since

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

, we have

150 = \frac{1}{2} \times 25 \times h

, which simplifies to

h = 12

.

The area formula for any triangle connects the base length and its corresponding perpendicular altitude.

2

Determine the location of the altitude foot

H

and calculate segment

BH

.

Draw altitude

AH

to the line containing

BC

. In right triangle

ABH

,

AB = 15

and

AH = 12

, so

BH = \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9

.

The Pythagorean theorem applies to right triangle

ABH

formed by the altitude.

3

Account for the obtuse angle condition to find segment

CH

.

Because angle

ABC

is obtuse, the altitude foot

H

lies on the extension of segment

CB

beyond vertex

B

. Therefore,

CH = CB + BH = 25 + 9 = 34

.

For an obtuse triangle, the altitude to one of the adjacent sides falls outside the triangle.

4

Calculate side length

AC

using right triangle

AHC

.

AC = \sqrt{AH^2 + CH^2} = \sqrt{12^2 + 34^2} = \sqrt{144 + 1156} = \sqrt{1300} = 10\sqrt{13}

.

Applying the Pythagorean theorem to right triangle

AHC

yields the hypotenuse

AC

.

Anahtar Kavram

Triangle Area, Altitude Projection, and Extended Pythagorean Theorem

Soru 9Soru →

A triangle has a base of length $10$ and an area of $30$ . Which of the following statements could be true about this triangle? Select all that apply.

Geçerli olan tümünü seçin

The altitude corresponding to the base of length

10

is

6

.

The triangle is a right triangle.

The perimeter of the triangle is

22

.

The perimeter of the triangle is

30

.

The perimeter of the triangle is

24

.

Cevabı ve açıklamayı göster

Cevap: The altitude corresponding to the base of length $10$ is $6$ .; The triangle is a right triangle.; The perimeter of the triangle is $30$ .

Cevap

The altitude perpendicular to the given base must be 6, the triangle can be a right triangle, and the perimeter can be equal to 30.

The area formula directly forces the altitude to be 6. Setting the altitude at one endpoint of the base constructs a valid right triangle. Furthermore, the minimum perimeter of any triangle with base 10 and height 6 is

10 + 2\sqrt{61} \approx 25.62

, so any perimeter value greater than or equal to

25.62

(such as 30) is attainable.

Adım Adım Çözüm

1

Calculate the required altitude of the triangle.

Using

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

,

30 = \frac{1}{2}(10)h \implies h = 6

.

The area and base are fixed, determining a unique height.

2

Evaluate whether the triangle can be a right triangle.

A right triangle with legs

10

and

6

has area

\frac{1}{2} \times 10 \times 6 = 30

.

Choosing the altitude to meet the base at an endpoint creates a right angle.

3

Determine the lower bound for the perimeter of the triangle.

The third vertex lies on a line parallel to the base at a distance of

6

. By symmetry, the minimum sum of the remaining two sides occurs when the triangle is isosceles with base

10

split into two segments of length

5

. Each equal side is

\sqrt{5^2 + 6^2} = \sqrt{61} \approx 7.8102

. The minimum perimeter is

10 + 2\sqrt{61} \approx 25.62

.

The shortest path from two fixed base endpoints to a parallel line is formed when the reflection creates equal angles, making the triangle isosceles.

4

Assess the possible perimeter values based on the lower bound.

Perimeters

22

and

24

are strictly below the minimum boundary of

\approx 25.62

and are impossible. Since the perimeter can take any value in

[10 + 2\sqrt{61}, \infty)

, a perimeter of

30

is possible.

Continuous movement of the top vertex increases the side lengths smoothly without bound.

Anahtar Kavram

Triangle area formula

\text{Area} = \frac{1}{2}bh

and geometric optimization of perimeter for a given base and height.

Soru 10Soru →

In the $xy$ -plane, line $k$ passes through the points $(1, -2)$ and $(5, 6)$ . Line $m$ is the perpendicular bisector of the line segment connecting $(1, -2)$ and $(5, 6)$ . Which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

Line

m

passes through the point

(7, 0)

.

Line

m

has a

y

-intercept of

(0, 3.5)

.

The distance from the origin

(0, 0)

to the midpoint of the line segment is

\sqrt{13}

.

The slope of line

m

is

2

.

Line

m

is parallel to the line with equation

2x - y = 4

.

Cevabı ve açıklamayı göster

Cevap: Line $m$ passes through the point $(7, 0)$ .; Line $m$ has a $y$ -intercept of $(0, 3.5)$ .; The distance from the origin $(0, 0)$ to the midpoint of the line segment is $\sqrt{13}$ .

Cevap

The statements asserting that line

m

passes through

(7, 0)

, that line

m

has a

y

-intercept of

(0, 3.5)

, and that the distance from the origin to the midpoint of the segment is

\sqrt{13}

are all true.

Line

k

has slope

\frac{6 - (-2)}{5 - 1} = 2

and segment midpoint

(3, 2)

. Line

m

, as the perpendicular bisector, has slope

-\frac{1}{2}

and equation

y = -\frac{1}{2}x + 3.5

. Substituting

x = 7

gives

y = 0

, so

(7, 0)

lies on line

m

. Substituting

x = 0

gives

y = 3.5

, confirming the

y

-intercept. Finally, the distance from

(0,0)

to the midpoint

(3,2)

is

\sqrt{3^2 + 2^2} = \sqrt{13}

. Thus, all three of these statements are correct.

Adım Adım Çözüm

1

Find the midpoint and slope of the segment connecting

(1, -2)

and

(5, 6)

.

Midpoint

M = \left(\frac{1+5}{2}, \frac{-2+6}{2}\right) = (3, 2)

. Slope of segment

m_k = \frac{6 - (-2)}{5 - 1} = \frac{8}{4} = 2

.

Line

m

is the perpendicular bisector, so it passes through midpoint

M

and has a slope perpendicular to

m_k

.

2

Determine the slope and equation of line

m

.

Perpendicular slope

m_m = -\frac{1}{m_k} = -\frac{1}{2}

. Point-slope equation through

(3, 2)

:

y - 2 = -\frac{1}{2}(x - 3) \implies y = -\frac{1}{2}x + \frac{7}{2}

.

Perpendicular lines have negative reciprocal slopes.

3

Evaluate each given statement against the calculated values.

Line

m

contains

(7, 0)

because

0 = -\frac{1}{2}(7) + \frac{7}{2}

. The

y

-intercept is

(0, 3.5)

. Distance from

(0,0)

to

(3,2)

is

\sqrt{3^2 + 2^2} = \sqrt{13}

. Statements claiming slope is

2

or that line

m

is parallel to

2x - y = 4

are incorrect.

Determines which statements satisfy all mathematical conditions.

Anahtar Kavram

Perpendicular Bisectors, Slopes of Perpendicular Lines, and Distance Formula

Soru 11Soru →

In regular octagon $ABCDEFGH$ with side length $2$ , diagonals $AD$ and $BE$ intersect at point $P$ . What is the area of quadrilateral $ABCP$ ?

2 + \sqrt{2}

2\sqrt{2}

4

4 + \sqrt{2}

1 + 2\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $2 + \sqrt{2}$

Cevap

2 + \sqrt{2}

In regular octagon

ABCDEFGH

, each interior angle measures

135^\circ

. Segment

ABCD

forms an isosceles trapezoid where

AD

is parallel to

BC

. The perpendicular distance from

AD

to

BC

is

2 \sin(45^\circ) = \sqrt{2}

, and

AD = 2 + 2(2 \cos(45^\circ)) = 2 + 2\sqrt{2}

. Diagonal

BE

is parallel to side

CD

, so

\angle CBE = 45^\circ

, which implies

\angle ABE = 135^\circ - 45^\circ = 90^\circ

. Triangle

\triangle ABP

is therefore a right isosceles triangle with legs

AB = 2

and

BP = 2

, yielding hypotenuse

AP = 2\sqrt{2}

. Quadrilateral

ABCP

is a trapezoid with parallel sides

BC = 2

and

AP = 2\sqrt{2}

and height

h = \sqrt{2}

. Its area is

\frac{2 + 2\sqrt{2}}{2} \times \sqrt{2} = 2 + \sqrt{2}

.

Adım Adım Çözüm

1

Calculate the interior angle of a regular octagon

Interior angle =

\frac{(8-2) \times 180^\circ}{8} = 135^\circ

Regular polygon interior angle formula.

2

Determine the orientation and length of diagonal

AD

AD \parallel BC

, height between them is

h = 2 \sin(45^\circ) = \sqrt{2}

, and base

AD = 2 + 2(2 \cos(45^\circ)) = 2 + 2\sqrt{2}

Quadrilateral

ABCD

forms an isosceles trapezoid with side length

2

and interior angles

135^\circ

.

3

Determine the direction of diagonal

BE

and find intersection point

P

\angle CBE = 180^\circ - 135^\circ = 45^\circ

, making

\angle ABE = 135^\circ - 45^\circ = 90^\circ

Quadrilateral

BCDE

is an isosceles trapezoid with

BE \parallel CD

.

4

Calculate the dimensions of right triangle

\triangle ABP

and segment

AP

\triangle ABP

is a right triangle at

B

with

\angle BAP = 45^\circ

,

AB = 2

,

BP = 2

, and hypotenuse

AP = 2\sqrt{2}

Since

AP \parallel BC

, alternate interior angle relationships yield

\angle BAP = 45^\circ

.

5

Calculate the area of quadrilateral

ABCP

\text{Area}(ABCP) = \frac{BC + AP}{2} \times h = \frac{2 + 2\sqrt{2}}{2} \times \sqrt{2} = (1 + \sqrt{2})\sqrt{2} = 2 + \sqrt{2}

Quadrilateral

ABCP

is a trapezoid with parallel bases

BC = 2

and

AP = 2\sqrt{2}

and height

\sqrt{2}

.

Anahtar Kavram

Properties of regular polygons, isosceles trapezoids, diagonal angles, and area decomposition.

Tahmini Süre:2m 30s

Soru 12Soru →

In right trapezoid $ABCD$ , side $AB$ is parallel to side $DC$ , and $\angle DAB = 90^\circ$ . The lengths of the sides are $AD = 12$ , $AB = 9$ , and $BC = 13$ , with $DC > AB$ . Point $E$ lies on the line containing segment $DC$ such that segment $BE$ is perpendicular to segment $BC$ . What is the length of segment $CE$ ?

16.9

25.0

28.8

33.8

36.0

Cevabı ve açıklamayı göster

Cevap: $33.8$

Cevap

The length of segment

CE

is

33.8

.

Dropping altitude

BP

from

B

to side

DC

creates a rectangle

ABPD

with height

BP = 12

and a right triangle

\triangle BPC

with legs

BP = 12

and hypotenuse

BC = 13

. Applying the Pythagorean theorem yields

PC = 5

. Since segment

BE

is perpendicular to

BC

, triangle

\triangle BCE

is a right triangle with right angle at

B

and altitude

BP

to hypotenuse

CE

. Using the leg rule

BC^2 = PC \times CE

, we find

13^2 = 5 \times CE

, which simplifies to

CE = 169/5 = 33.8

.

Adım Adım Çözüm

1

Find the length of segment

PC

by dropping a perpendicular from

B

to

DC

.

PC = 5

Let

P

be the foot of the perpendicular from

B

to line

DC

. Since

ABCD

is a right trapezoid with

AD \perp DC

,

ABPD

forms a rectangle. Thus

BP = AD = 12

and

DP = AB = 9

. In right triangle

\triangle BPC

,

BP^2 + PC^2 = BC^2 \implies 12^2 + PC^2 = 13^2 \implies PC = \sqrt{169 - 144} = 5

.

2

Set up a similar triangle relationship for right triangle

\triangle BCE

.

\triangle BPC \sim \triangle EBC

In

\triangle BCE

,

\angle CBE = 90^\circ

and

BP \perp CE

. Thus, right triangle

\triangle BPC

is similar to right triangle

\triangle EBC

because they share

\angle C

.

3

Solve for the length of

CE

.

CE = 33.8

By the geometric mean theorem (or ratio of corresponding sides in similar triangles),

\frac{BC}{CE} = \frac{PC}{BC} \implies BC^2 = PC \times CE

. Substituting

BC = 13

and

PC = 5

yields

13^2 = 5 \times CE \implies 169 = 5 \times CE \implies CE = \frac{169}{5} = 33.8

.

Anahtar Kavram

Pythagorean Theorem and Geometric Mean Theorem (Similar Right Triangles)

Tahmini Süre:1m 40s

Soru 13Soru →

In circle $O$ , a sector with a central angle of $60^\circ$ has an area of $6\pi$ . In circle $P$ , a sector with a central angle of $120^\circ$ has an arc length equal to the arc length of the sector in circle $O$ . What is the area of the sector in circle $P$ ?

12\pi

6\pi

3\pi

9\pi

\frac{3\pi}{2}

Cevabı ve açıklamayı göster

Cevap: $3\pi$

Cevap

3\pi

The sector area of

3\pi

is correctly calculated by first determining the radius of circle

O

(

r_1 = 6

), using it to find the arc length (

2\pi

), setting that as the arc length for the sector in circle

P

to find its radius (

r_2 = 3

), and finally computing its sector area

\frac{120^\circ}{360^\circ} \cdot \pi (3^2) = 3\pi

.

Adım Adım Çözüm

1

Find the radius of circle

O

Radius

r_1 = 6

The sector area formula is

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

. Given

A = 6\pi

and

\theta = 60^\circ

, we have

\frac{60^\circ}{360^\circ} \cdot \pi r_1^2 = 6\pi \Rightarrow \frac{1}{6} r_1^2 = 6 \Rightarrow r_1^2 = 36 \Rightarrow r_1 = 6

.

2

Calculate the arc length of the sector in circle

O

Arc length

L = 2\pi

The arc length formula is

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

. Substituting

\theta = 60^\circ

and

r_1 = 6

, we get

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

.

3

Find the radius of circle

P

Radius

r_2 = 3

The sector in circle

P

has an arc length equal to

2\pi

and a central angle of

120^\circ

. Using

L = \frac{120^\circ}{360^\circ} \cdot 2\pi r_2

, we get

2\pi = \frac{1}{3} \cdot 2\pi r_2 \Rightarrow r_2 = 3

.

4

Calculate the area of the sector in circle

P

Sector area

= 3\pi

Using the sector area formula for circle

P

:

A_2 = \frac{120^\circ}{360^\circ} \cdot \pi r_2^2 = \frac{1}{3} \cdot \pi (3^2) = 3\pi

.

Anahtar Kavram

Relationship between central angles, radii, arc lengths, and sector areas in circles

Soru 14Soru →

In the figure, line $l_1$ is parallel to line $l_2$ . Point $A$ lies on line $l_1$ , while points $B$ and $C$ lie on line $l_2$ such that $B$ is to the left of $C$ . Line segments $AB$ and $AC$ extend from line $l_1$ to line $l_2$ to form $\triangle ABC$ . The measure of interior angle $\angle BAC$ is $(2x + 10)^\circ$ , the measure of interior angle $\angle ABC$ is $(3x - 15)^\circ$ , and the measure of the exterior angle at vertex $C$ along line $l_2$ is $(6x - 35)^\circ$ . What is the degree measure of angle $\angle BAC$ ?

30^\circ

35^\circ

50^\circ

70^\circ

75^\circ

Cevabı ve açıklamayı göster

Cevap: $70^\circ$

Cevap

The degree measure of angle

\angle BAC

is

70^\circ

.

According to the Exterior Angle Theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two non-adjacent (remote) interior angles. Setting up the equation

(2x + 10) + (3x - 15) = 6x - 35

gives

5x - 5 = 6x - 35

, which simplifies to

x = 30

. Substituting

x = 30

into the expression for

\angle BAC

,

(2x + 10)^\circ

, yields

2(30) + 10 = 70^\circ

.

Adım Adım Çözüm

1

Apply the Exterior Angle Theorem to set up an algebraic equation.

(2x + 10) + (3x - 15) = 6x - 35

The measure of an exterior angle of a triangle equals the sum of the measures of its two remote interior angles.

2

Simplify and solve the linear equation for

x

.

5x - 5 = 6x - 35 \implies x = 30

Combining like terms yields

5x - 5 = 6x - 35

. Subtracting

5x

from both sides gives

-5 = x - 35

, so

x = 30

.

3

Substitute

x = 30

into the expression for angle

\angle BAC

.

\text{m}\angle BAC = 2(30) + 10 = 70^\circ

The question asks for the degree measure of angle

\angle BAC

, which is given by

(2x + 10)^\circ

.

Anahtar Kavram

Exterior Angle Theorem and Angle Relationships in Parallel Lines

Soru 15Soru →

In a circle centered at point $O$ , sector $AOB$ has a radius of length $r$ , a central angle of $\theta$ degrees, an arc length of $L$ , an area of $A_s$ , and a perimeter of $P$ . If the ratio of the sector area to the arc length is $\frac{A_s}{L} = 4$ , and the ratio of the sector area to the sector perimeter is $\frac{A_s}{P} = \frac{4}{3}$ , which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

The radius

r

of the circle is

8

.

The area

A_s

of sector

AOB

is

32

.

The central angle

\theta

is greater than

60^\circ

.

The perimeter

P

of sector

AOB

is

24

.

The arc length

L

of sector

AOB

is

8\pi

.

Cevabı ve açıklamayı göster

Cevap: The radius $r$ of the circle is $8$ .; The area $A_s$ of sector $AOB$ is $32$ .; The perimeter $P$ of sector $AOB$ is $24$ .

Cevap

The true statements are that the radius of the circle is 8, the area of the sector is 32, and the perimeter of the sector is 24.

The ratio of sector area to arc length simplifies directly to half the radius,

\frac{r}{2} = 4

, giving a radius of

8

. Expressing the sector perimeter as the sum of the arc length and two radii (

L + 16

) and using

\frac{A_s}{P} = \frac{4L}{L + 16} = \frac{4}{3}

yields

L = 8

. From this, the sector area is

A_s = 32

and the sector perimeter is

P = 24

.

Adım Adım Çözüm

1

Relate sector area

A_s

and arc length

L

to find the radius

r

.

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

and

L = \frac{\theta}{360}2\pi r

, so

\frac{A_s}{L} = \frac{r}{2} = 4 \implies r = 8

.

Dividing the sector area formula by the arc length formula cancels the central angle fraction and

\pi

.

2

Set up the ratio equation for

\frac{A_s}{P}

using

P = L + 2r

.

Since

r = 8

,

P = L + 16

. Given

A_s = 4L

, we have

\frac{4L}{L + 16} = \frac{4}{3} \implies 3L = L + 16 \implies 2L = 16 \implies L = 8

.

Expressing both sector area and perimeter in terms of arc length allows direct solution for

L

.

3

Calculate sector area

A_s

and perimeter

P

.

A_s = 4(8) = 32

and

P = 8 + 2(8) = 24

.

Substituting

L = 8

and

r = 8

gives the exact values for area and perimeter.

4

Determine the central angle

\theta

.

8 = \frac{\theta}{360} \cdot 2\pi(8) \implies 8 = \frac{16\pi \theta}{360} \implies \theta = \left(\frac{90}{\pi}\right)^\circ \approx 28.65^\circ

.

Plugging

L = 8

and

r = 8

into the arc length formula gives

\theta \approx 28.65^\circ

, which is less than

60^\circ

.

Anahtar Kavram

Relating sector area, arc length, and perimeter using proportional ratios and fundamental circle formulas.

Soru 16Soru →

A solid right circular cylinder has a base radius of $r$ and a height of $4r$ . A solid sphere has a radius of $R$ . If the total surface area of the sphere is equal to the total surface area of the cylinder, what is the ratio of the volume of the sphere to the volume of the cylinder?

\frac{5\sqrt{10}}{12}

\frac{\sqrt{10}}{2}

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

\frac{5\sqrt{10}}{6}

\frac{5}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{5\sqrt{10}}{12}$

Cevap

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

\frac{5\sqrt{10}}{12}

.

The total surface area of the cylinder is the sum of its lateral area and two circular bases:

2\pi r(4r) + 2\pi r^2 = 10\pi r^2

. Setting this equal to the sphere's surface area

4\pi R^2

yields

R/r = \sqrt{5/2} = \sqrt{10}/2

. The ratio of the sphere's volume

\frac{4}{3}\pi R^3

to the cylinder's volume

\pi r^2 (4r) = 4\pi r^3

is

\frac{1}{3}(R/r)^3 = \frac{1}{3} \left(\frac{\sqrt{10}}{2}\right)^3 = \frac{5\sqrt{10}}{12}

.

Adım Adım Çözüm

1

Calculate the total surface area of the cylinder.

\text{TSA}_{\text{cyl}} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r(4r) = 10\pi r^2

A cylinder's total surface area consists of two circular bases (

2\times\pi r^2

) and the lateral surface area (

2\pi r h

).

2

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

R

to

r

.

4\pi R^2 = 10\pi r^2 \implies R^2 = \frac{5}{2}r^2 \implies \frac{R}{r} = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}

The total surface area of a sphere of radius

R

is

4\pi R^2

.

3

Express the volumes of both figures in terms of

r

and

R

.

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

and

V_{\text{cyl}} = \pi r^2 h = \pi r^2 (4r) = 4\pi r^3

The volume of a sphere is

\frac{4}{3}\pi R^3

and the volume of a cylinder is

\pi r^2 h

.

4

Compute the ratio of the volume of the sphere to the volume of the cylinder.

\frac{V_{\text{sph}}}{V_{\text{cyl}}} = \frac{\frac{4}{3}\pi R^3}{4\pi r^3} = \frac{1}{3}\left(\frac{R}{r}\right)^3 = \frac{1}{3}\left(\frac{\sqrt{10}}{2}\right)^3 = \frac{1}{3} \cdot \frac{10\sqrt{10}}{8} = \frac{5\sqrt{10}}{12}

Substitute

\frac{R}{r} = \frac{\sqrt{10}}{2}

into the ratio expression.

Anahtar Kavram

Relating 3D surface area formulas to volume formulas for cylinders and spheres

Tahmini Süre:2m 0s

Soru 17Soru →

Two straight lines, $L_1$ and $L_2$ , intersect at point $O$ . One of the angles formed by their intersection measures $(3x - 15)^\circ$ , and the vertically opposite angle measures $(x + 25)^\circ$ . What is the degree measure of an angle adjacent to $(3x - 15)^\circ$ ?

20^\circ

45^\circ

135^\circ

67.5^\circ

112.5^\circ

Cevabı ve açıklamayı göster

Cevap: $135^\circ$

Cevap

135^\circ

Since vertically opposite angles are equal, setting

3x - 15 = x + 25

yields

x = 20

. Substituting

x = 20

into

3x - 15

gives an angle of

45^\circ

. Because adjacent angles on intersecting lines form a straight line (a linear pair), they are supplementary. Therefore, the adjacent angle measures

180^\circ - 45^\circ = 135^\circ

.

Adım Adım Çözüm

1

Set the vertically opposite angle expressions equal to each other.

3x - 15 = x + 25

Vertically opposite angles formed by two intersecting lines are equal in measure.

2

Solve for the variable

x

.

2x = 40 \implies x = 20

Subtract

x

and add

15

to both sides of the equation.

3

Calculate the degree measure of the angle

(3x - 15)^\circ

.

3(20) - 15 = 60 - 15 = 45^\circ

Substitute

x = 20

back into the angle expression.

4

Calculate the measure of an adjacent angle.

180^\circ - 45^\circ = 135^\circ

Adjacent angles along a straight line are supplementary and sum to

180^\circ

.

Anahtar Kavram

Vertically opposite angles are equal, and adjacent angles forming a linear pair are supplementary.

Soru 18Soru →

A sector of a circle with a radius of $12$ units has an area of $24\pi$ square units. What is the measure, in degrees, of the central angle of the sector?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

The measure of the central angle of the sector is

60^\circ

.

The area of the full circle is

\pi r^2 = \pi (12)^2 = 144\pi

. The sector area represents a fraction of the total area, specifically

\frac{24\pi}{144\pi} = \frac{1}{6}

. Multiplying this fraction by the full circle's central angle of

360^\circ

gives

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

.

Adım Adım Çözüm

1

Find the total area of the circle

The total area of the circle is

\pi \times 12^2 = 144\pi

.

The area of a full circle with radius

r

is given by

A = \pi r^2

.

2

Relate the sector area to the total circle area

The fraction of the circle represented by the sector is

\frac{24\pi}{144\pi} = \frac{1}{6}

.

The area of a sector is proportional to the fraction of the total central angle (

360^\circ

) it subtends.

3

Calculate the central angle in degrees

\theta = \frac{1}{6} \times 360^\circ = 60^\circ.

Multiply the fraction of the circle by

360^\circ

to get the central angle measure.

Anahtar Kavram

Relationship between central angle measure, total circle area, and sector area

Soru 19Soru →

A triangle has a base of length $14$ centimeters and an area of $42$ square centimeters. What is the height, in centimeters, of the triangle perpendicular to this base?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The height of the triangle perpendicular to the base is 6 centimeters.

The area of a triangle is given by

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

. Substituting

42

for the area and

14

for the base yields

42 = \frac{1}{2} \times 14 \times h

, which simplifies to

42 = 7h

. Dividing

42

by

7

gives

h = 6

.

Adım Adım Çözüm

1

Write down the triangle area formula.

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

The area and base are known values, and the goal is to solve for the perpendicular height.

2

Substitute the given values into the formula.

42 = \frac{1}{2} \times 14 \times \text{height}

Plugging in

\text{Area} = 42

and

\text{base} = 14

creates a single-variable algebraic equation.

3

Solve for the height.

42 = 7 \times \text{height} \implies \text{height} = 6

Simplifying

\frac{1}{2} \times 14

to

7

and dividing both sides by

7

yields the height.

Anahtar Kavram

The area of a triangle is calculated using the formula Area = (1/2) * base * height.

Soru 20Soru →

Lines $l_1$ and $l_2$ intersect at point $P$ to form an acute angle measuring $44^\circ$ . Line $b_1$ is the angle bisector of this acute angle. A third line, $l_3$ , is drawn perpendicular to $b_1$ and intersects line $l_1$ at point $Q$ (where $Q \neq P$ ). What is the measure, in degrees, of the acute angle formed by the intersection of line $l_2$ and line $l_3$ ?

22^\circ

46^\circ

68^\circ

78^\circ

112^\circ

Cevabı ve açıklamayı göster

Cevap: $68^\circ$

Cevap

The measure of the acute angle formed by the intersection of line

l_2

and line

l_3

is

68^\circ

.

The correct answer is

68^\circ

. Line

b_1

divides the

44^\circ

angle between

l_1

and

l_2

into two

22^\circ

angles. Because line

l_3

is perpendicular to

b_1

, it forms a right triangle with

l_1

and

b_1

, making the angle between

l_1

and

l_3

equal to

90^\circ - 22^\circ = 68^\circ

. Considering the large triangle formed by lines

l_1

,

l_2

, and

l_3

, two of its angles are

44^\circ

and

68^\circ

. Thus, the third interior angle at the intersection of

l_2

and

l_3

is

180^\circ - (44^\circ + 68^\circ) = 68^\circ

, which is acute.

Adım Adım Çözüm

1

Determine the angle formed by the angle bisector

b_1

with line

l_1

and line

l_2

.

Since line

b_1

bisects the

44^\circ

acute angle between

l_1

and

l_2

, the angle between

l_1

and

b_1

is

\frac{44^\circ}{2} = 22^\circ

, and the angle between

l_2

and

b_1

is also

22^\circ

.

An angle bisector divides an angle into two equal congruent parts.

2

Find the measure of the interior angle between line

l_1

and line

l_3

.

Line

l_3

is perpendicular to

b_1

, forming a right triangle with

l_1

and

b_1

. The interior angle between

l_1

and

l_3

is

180^\circ - 90^\circ - 22^\circ = 68^\circ

.

The sum of interior angles in any triangle is

180^\circ

.

3

Calculate the interior angle at the intersection of line

l_2

and line

l_3

in the main triangle formed by

l_1

,

l_2

, and

l_3

.

The interior angle at the intersection of

l_2

and

l_3

is

180^\circ - 44^\circ - 68^\circ = 68^\circ

.

The interior angles of the triangle formed by lines

l_1

,

l_2

, and

l_3

must sum to

180^\circ

.

4

Verify that the calculated angle is acute.

Since

68^\circ < 90^\circ

, the acute angle formed by lines

l_2

and

l_3

is

68^\circ

.

An angle measuring strictly less than

90^\circ

is defined as an acute angle.

Anahtar Kavram

Angle bisector properties, perpendicular lines, and interior angle sum theorem for triangles.

Geometry

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