Geometry Alıştırma Soruları — GRE General Test — Sayfa 4

Soru 61Soru →

In triangle $XYZ$ , the length of side $XY$ is $x + 3$ , the length of side $YZ$ is $2x - 1$ , and the length of side $XZ$ is $12$ , where $x$ is an integer. Which of the following could be the perimeter of triangle $XYZ$ ? Select all such perimeters.

Geçerli olan tümünü seçin

26

23

44

62

31

Cevabı ve açıklamayı göster

Cevap: 26; 44

Cevap

The perimeters 26 and 44 are the valid values among the choices provided.

The perimeter formula

P = 3x + 14

must yield an integer value corresponding to an integer

x

in the range

4 \le x \le 15

. The perimeters equal to 26 (for

x = 4

) and 44 (for

x = 10

) fall within this valid range and satisfy the triangle inequality.

Adım Adım Çözüm

1

Express the perimeter in terms of

x

.

\text{Perimeter } P = (x + 3) + (2x - 1) + 12 = 3x + 14

The perimeter of a triangle is the sum of its three side lengths.

2

Apply the Triangle Inequality Theorem to determine valid bounds for

x

.

Condition 1:

(x + 3) + (2x - 1) > 12 \implies 3x + 2 > 12 \implies 3x > 10 \implies x > 3.33

.
Condition 2:

(x + 3) + 12 > 2x - 1 \implies x + 15 > 2x - 1 \implies x < 16

.
Condition 3:

(2x - 1) + 12 > x + 3 \implies 2x + 11 > x + 3 \implies x > -8

(naturally satisfied for positive

x

).
Thus,

4 \le x \le 15

for integer

x

.

For any non-degenerate triangle, the sum of any two side lengths must be strictly greater than the third side length.

3

Evaluate the allowable perimeters for valid integer values of

x

.

The minimum valid perimeter corresponds to

x = 4

, giving

P = 3(4) + 14 = 26

.
The maximum valid perimeter corresponds to

x = 15

, giving

P = 3(15) + 14 = 59

.
Checking options:
- For

26

:

3x + 14 = 26 \implies x = 4

(valid).
- For

44

:

3x + 14 = 44 \implies x = 10

(valid).

Substituting valid integer

x

values identifies which proposed perimeters satisfy all conditions.

Anahtar Kavram

Triangle Inequality Theorem & Algebraic Bounds on Side Lengths

Tahmini Süre:2m 0s

Soru 62Soru →

A sector of a circle with a radius of $10$ units has a total perimeter of $20 + 5\pi$ units. What is the area of this sector?

25\pi

50\pi

5\pi

100\pi

100 + 25\pi

Cevabı ve açıklamayı göster

Cevap: $25\pi$

Cevap

The area of the sector is

25\pi

.

The total perimeter of a sector with radius

r

and arc length

L

is given by

P = 2r + L

. Substituting

r = 10

gives

20 + 5\pi = 20 + L

, so

L = 5\pi

. Using the sector area formula

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

, the area is

\frac{1}{2} \times 10 \times 5\pi = 25\pi

.

Adım Adım Çözüm

1

Set up the formula for the perimeter of a sector.

P = 2r + L

, where

r = 10

is the radius and

L

is the arc length.

A sector's perimeter is bounded by two straight radii and one curved arc.

2

Solve for the arc length

L

.

20 + 5\pi = 2(10) + L \implies L = 5\pi

.

Subtracting the combined length of the two radii (

20

) isolates the arc length.

3

Calculate the area of the sector.

Sector Area =

\frac{1}{2} r L = \frac{1}{2} (10)(5\pi) = 25\pi

.

The area of a sector can be directly evaluated using half the product of its radius and arc length.

Anahtar Kavram

Perimeter, Arc Length, and Area of a Circular Sector

Alternatif Yöntem

Find the central angle

\theta

first: Since

L = 5\pi

and circumference

C = 2\pi(10) = 20\pi

, the fraction of the circle is

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

, which corresponds to

\theta = 90^\circ

. The area is then

\frac{1}{4} \times \pi(10^2) = 25\pi

.

Tahmini Süre:1m 30s

Soru 63Soru →

In a right triangle, the lengths of the two legs are in a ratio of $3:4$ . If the perimeter of the triangle is $36$ centimeters, what is the area of the triangle, in square centimeters?

Cevabı ve açıklamayı göster

Cevap: 54

Cevap

The area of the triangle is 54 square centimeters.

Since the ratio of the legs of the right triangle is 3:4, the triangle forms a classic 3-4-5 right triangle proportion. The perimeter is

3x + 4x + 5x = 12x

. Setting

12x = 36

yields

x = 3

. The legs are therefore

9

cm and

12

cm. Calculating the area using

\frac{1}{2} \times 9 \times 12

gives

54

square centimeters.

Adım Adım Çözüm

1

Express the side lengths in terms of a variable

x

Legs are

3x

and

4x

, and hypotenuse is

5x

By the Pythagorean theorem, a right triangle with legs in ratio 3:4 has hypotenuse ratio

\sqrt{3^2 + 4^2} = 5

.

2

Solve for

x

using the given perimeter

x = 3

The sum of all three sides is

3x + 4x + 5x = 12x = 36

, giving

x = 3

.

3

Calculate the actual leg lengths and area

Legs are

9

cm and

12

cm; Area is

54

\text{cm}^2

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

\frac{1}{2} \times 9 \times 12 = 54

.

Anahtar Kavram

Perimeter and area of right triangles using standard side ratios

Soru 64Soru →

In $\triangle ABC$ , point $D$ lies on segment $BC$ such that segment $AD$ is perpendicular to $BC$ . The ratio of the area of $\triangle ABD$ to the area of $\triangle ADC$ is $5 : 16$ . If $AB = 13$ and the perimeter of $\triangle ABC$ is $54$ , what is the area of $\triangle ABC$ ?

96

104

126

130

252

Cevabı ve açıklamayı göster

Cevap: 126

Cevap

126

The correct answer is 126. Since triangles

\triangle ABD

and

\triangle ADC

share height

AD

, their areas are in proportion to their bases

BD:DC = 5:16

. Setting

BD = 5k

and

DC = 16k

, the Pythagorean theorem yields altitude

AD = \sqrt{169 - 25k^2}

and hypotenuse

AC = \sqrt{169 + 231k^2}

. Substituting these into the perimeter equation

13 + 21k + AC = 54

yields

k = 1

(after rejecting an extraneous root). Thus

BC = 21

and

AD = 12

, making the area

\frac{1}{2} \times 21 \times 12 = 126

.

Adım Adım Çözüm

1

Relate the areas of the sub-triangles to their base lengths.

\frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ADC)} = \frac{\frac{1}{2} \cdot BD \cdot AD}{\frac{1}{2} \cdot DC \cdot AD} = \frac{BD}{DC} = \frac{5}{16}

. Thus,

BD = 5k

and

DC = 16k

for some positive constant

k

, giving

BC = 21k

.

Triangles sharing the same altitude have areas proportional to their bases.

2

Express altitude

AD

and side

AC

in terms of

k

using the Pythagorean theorem.

In right

\triangle ABD

:

AD = \sqrt{AB^2 - BD^2} = \sqrt{13^2 - (5k)^2} = \sqrt{169 - 25k^2}

. In right

\triangle ADC

:

AC = \sqrt{AD^2 + DC^2} = \sqrt{(169 - 25k^2) + (16k)^2} = \sqrt{169 + 231k^2}

.

Since

AD \perp BC

, both

\triangle ABD

and

\triangle ADC

are right triangles.

3

Set up and solve the perimeter equation for

k

.

Perimeter

= AB + BC + AC = 13 + 21k + \sqrt{169 + 231k^2} = 54 \implies \sqrt{169 + 231k^2} = 41 - 21k

. Squaring both sides yields

169 + 231k^2 = 1681 - 1722k + 441k^2 \implies 210k^2 - 1722k + 1512 = 0 \implies 5k^2 - 41k + 36 = 0

. Factoring gives

(5k - 36)(k - 1) = 0

, so

k = 1

or

k = 7.2

.

The given perimeter allows over-constraining the side length expressions to a quadratic in

k

.

4

Test roots for validity and calculate final triangle area.

For

k = 7.2

,

41 - 21(7.2) = -110.2 < 0

, which is extraneous. For

k = 1

,

BD = 5

,

DC = 16

,

BC = 21

,

AD = 12

, and

AC = 20

. Area

(\triangle ABC) = \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot 21 \cdot 12 = 126

.

Extraneous roots introduced by squaring must be discarded, and the valid root gives the exact area.

Anahtar Kavram

Decomposing triangles into adjacent right triangles, leveraging shared altitudes for area ratios, and applying algebraic perimeter constraints with Pythagorean equations.

Soru 65Soru →

In the $xy$ -plane, line $k$ is defined by the equation $2x + 4y = 9$ . Line $m$ is perpendicular to line $k$ . What is the slope of line $m$ ?

2

-\frac{1}{2}

-2

\frac{1}{2}

-\frac{9}{4}

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

2

First, rewrite the line equation

2x + 4y = 9

in slope-intercept form

y = mx + b

: subtracting

2x

from both sides gives

4y = -2x + 9

, and dividing by

4

yields

y = -\frac{1}{2}x + \frac{9}{4}

. Thus, the slope of line

k

is

m_k = -\frac{1}{2}

. Since line

m

is perpendicular to line

k

, its slope is the negative reciprocal of

-\frac{1}{2}

, which is

2

.

Adım Adım Çözüm

1

Convert the equation of line

k

to slope-intercept form (

y = mx + b

).

4y = -2x + 9 \implies y = -\frac{1}{2}x + \frac{9}{4}

.

The coefficient of

x

in slope-intercept form gives the slope of line

k

(

m_k = -\frac{1}{2}

).

2

Calculate the slope of perpendicular line

m

.

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

.

Perpendicular lines have slopes that are negative reciprocals of each other.

Anahtar Kavram

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

Tahmini Süre:45s

Soru 66Soru →

Two concentric circles are centered at point $O$ . The inner circle has a radius of $6$ units, and the outer circle has a radius of $6\sqrt{3}$ units. Radii $OA$ and $OB$ of the outer circle form a central angle $\angle AOB = 60^\circ$ and intersect the inner circle at points $C$ and $D$ , respectively. Which of the following statements must be true regarding the region and boundary lengths defined by these figures? Select all that apply.

Geçerli olan tümünü seçin

The area of the region bounded by arc

AB

, arc

CD

, segment

AC

, and segment

BD

is equal to

12\pi

square units.

The straight-line distance between points

A

and

B

is

6\sqrt{3}

units.

The area of the circular segment bounded by chord

AB

and minor arc

AB

is

18\pi - 27\sqrt{3}

square units.

The ratio of the length of arc

CD

to the length of arc

AB

is

1 : 3

.

The perimeter of the region bounded by arc

AB

, arc

CD

, segment

AC

, and segment

BD

is

2\pi(1 + \sqrt{3}) + 6\sqrt{3}

units.

Cevabı ve açıklamayı göster

Cevap: The area of the region bounded by arc $AB$ , arc $CD$ , segment $AC$ , and segment $BD$ is equal to $12\pi$ square units.; The straight-line distance between points $A$ and $B$ is $6\sqrt{3}$ units.; The area of the circular segment bounded by chord $AB$ and minor arc $AB$ is $18\pi - 27\sqrt{3}$ square units.

Cevap

The true statements are: the area of the region bounded by arc AB, arc CD, segment AC, and segment BD is 12π square units; the straight-line distance between points A and B is 6√3 units; and the area of the circular segment bounded by chord AB and minor arc AB is 18π - 27√3 square units.

The area of the region between the two concentric arcs is obtained by subtracting the inner sector area (6π) from the outer sector area (18π), giving 12π square units. The triangle OAB is equilateral because it has two sides of length 6√3 and an included angle of 60°, making chord AB equal to 6√3. Subtracting the area of this equilateral triangle (27√3) from the outer sector area (18π) yields the area of the circular segment bounded by chord AB and arc AB, which is 18π - 27√3.

Adım Adım Çözüm

1

Calculate the area of sector OAB and sector OCD to evaluate the area of the annular sector region.

Sector OAB area = (60/360) * π * (6√3)^2 = (1/6) * 108π = 18π. Sector OCD area = (60/360) * π * (6^2) = (1/6) * 36π = 6π. Region area = 18π - 6π = 12π.

The area between two concentric sector arcs bounded by the same radii is the difference in sector areas.

2

Determine the length of chord AB using triangle properties.

In triangle OAB, OA = OB = 6√3 and angle AOB = 60°. An isosceles triangle with a 60° vertex angle is equilateral, so AB = 6√3.

All internal angles of an isosceles triangle with a 60° angle must equal 60°.

3

Calculate the area of circular segment AB.

Area of equilateral triangle OAB = (√3 / 4) * (6√3)^2 = 27√3. Segment area = Sector OAB area - Triangle OAB area = 18π - 27√3.

A circular segment's area is found by subtracting the area of the subtended triangle from the area of the corresponding sector.

4

Evaluate arc length ratio and total perimeter of the annular sector.

Arc CD / Arc AB = 6 / (6√3) = 1 / √3 ≠ 1 / 3. Total perimeter = Arc AB + Arc CD + 2*(R - r) = 2√3π + 2π + 2*(6√3 - 6) = 2π(1 + √3) + 12√3 - 12.

Verifies that statements regarding ratio 1:3 and incomplete perimeter calculations are mathematically false.

Anahtar Kavram

Concentric circle geometry, arc length proportions, sector area calculations, and circular segment area formulation.

Soru 67Soru →

In right triangle $ABC$ , the lengths of the two legs perpendicular to each other are $9$ and $12$ . What is the perimeter of triangle $ABC$ ?

21

30

36

42

54

Cevabı ve açıklamayı göster

Cevap: $36$

Cevap

The perimeter of triangle

ABC

is

36

.

Using the Pythagorean theorem, the hypotenuse length is

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

. Summing the two legs (

9

and

12

) with the hypotenuse (

15

) gives a total perimeter of

36

.

Adım Adım Çözüm

1

Calculate the length of the hypotenuse using the Pythagorean theorem.

Hypotenuse

c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

.

For any right triangle with legs

a

and

b

and hypotenuse

c

,

a^2 + b^2 = c^2

holds.

2

Sum the lengths of all three sides to find the perimeter.

Perimeter

= 9 + 12 + 15 = 36

.

The perimeter of a polygon is the total distance around its outer boundary.

Anahtar Kavram

Pythagorean Theorem (

a^2 + b^2 = c^2

) and Perimeter of Triangles

Tahmini Süre:45s

Soru 68Soru →

In triangle $ABC$ , point $D$ lies on side $AC$ such that segment $BD$ is perpendicular to side $AC$ . The length of segment $AD$ is $9$ and the length of segment $DC$ is $16$ . If the ratio of the length of side $AB$ to the length of side $BC$ is $3 : 4$ , what is the area of triangle $ABC$ ?

72

96

150

300

1,800

Cevabı ve açıklamayı göster

Cevap: 150

Cevap

150

Applying the Pythagorean theorem to right triangles

ABD

and

CBD

gives

AB^2 = 81 + BD^2

and

BC^2 = 256 + BD^2

. Squaring the ratio

\frac{AB}{BC} = \frac{3}{4}

yields

\frac{81 + BD^2}{256 + BD^2} = \frac{9}{16}

. Cross-multiplying gives

1296 + 16 BD^2 = 2304 + 9 BD^2

, so

7 BD^2 = 1008 \implies BD = 12

. The base

AC = 9 + 16 = 25

, so the area of triangle

ABC

is

\frac{1}{2} \times 25 \times 12 = 150

.

Adım Adım Çözüm

1

Set up expressions for side lengths

AB

and

BC

using the Pythagorean theorem on right triangles

ABD

and

CBD

.

AB^2 = 9^2 + BD^2 = 81 + BD^2

and

BC^2 = 16^2 + BD^2 = 256 + BD^2

.

Segment

BD

is an altitude perpendicular to

AC

, dividing triangle

ABC

into two right triangles.

2

Use the given side ratio

\frac{AB}{BC} = \frac{3}{4}

to solve for the height

BD

.

\frac{81 + BD^2}{256 + BD^2} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \implies 16(81 + BD^2) = 9(256 + BD^2) \implies 7 BD^2 = 1008 \implies BD^2 = 144 \implies BD = 12

.

Squaring both sides of the ratio allows substitution of the expressions for

AB^2

and

BC^2

.

3

Determine the total length of base

AC

and compute the area of triangle

ABC

.

AC = AD + DC = 9 + 16 = 25

. Area

= \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 25 \times 12 = 150

.

The area of a triangle is given by half the product of its base and corresponding altitude.

Anahtar Kavram

Properties of altitudes in triangles, Pythagorean theorem, and ratio setup for area determination.

Tahmini Süre:2m 0s

Soru 69Soru →

In a circle centered at point $O$ , the ratio of the area of sector $AOB$ to the area of the entire circle is $3:8$ . If the total perimeter of sector $AOB$ is $24 + 9\pi$ , what is the radius of the circle?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The radius of the circle is 12.

Since the ratio of the sector area to the total circle area is

3:8

, the sector accounts for

\frac{3}{8}

of the entire circumference

2\pi r

, making the arc length

L = \frac{3}{8}(2\pi r) = \frac{3}{4}\pi r

. The perimeter of the sector is

2r + L = 2r + \frac{3}{4}\pi r

. Equating this to

24 + 9\pi

gives

2r = 24

, so

r = 12

.

Adım Adım Çözüm

1

Relate sector area ratio to arc length

Arc length

L = \frac{3}{4}\pi r

The fraction of the circle occupied by the sector is

\frac{3}{8}

, so the arc length is

\frac{3}{8}

of the circle's circumference

2\pi r

.

2

Formulate the perimeter expression for sector

AOB

Perimeter =

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

The boundary of a sector includes two straight radii of length

r

plus the curved arc length

L

.

3

Equate to given perimeter and solve for

r

r = 12

Comparing

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

to

24 + 9\pi

, setting

2r = 24

yields

r = 12

, which also satisfies

\frac{3}{4}\pi (12) = 9\pi

.

Anahtar Kavram

Perimeter of a circle sector and fractional relationship between sector area, central angle, and arc length.

Soru 70Soru →

In a geometric plane, line $L_1$ is parallel to line $L_2$ . Points $A$ and $C$ lie on line $L_1$ , and points $B$ and $D$ lie on line $L_2$ . Line segments $AB$ and $CD$ intersect at point $X$ located between lines $L_1$ and $L_2$ . If measure of $\angle XAC = 42^\circ$ and measure of $\angle XDB = 35^\circ$ , what is the measure, in degrees, of $\angle AXC$ ?

Cevabı ve açıklamayı göster

Cevap: 103

Cevap

The measure of

\angle AXC

is

103^\circ

.

Line

L_1

is parallel to line

L_2

, meaning segment

AC

is parallel to segment

BD

. Transversal line

CD

intersects both parallel lines, creating alternate interior angles

\angle XCA

and

\angle XDB

. Hence,

\angle XCA = \angle XDB = 35^\circ

. Inside triangle

ACX

, the three interior angles must sum to

180^\circ

. Substituting the values yields

\angle AXC = 180^\circ - (42^\circ + 35^\circ) = 103^\circ

.

Adım Adım Çözüm

1

Identify parallel lines and the transversal line

Line segment

CD

acts as a transversal line intersecting parallel lines

L_1

and

L_2

.

Points

A

and

C

lie on line

L_1

while points

B

and

D

lie on line

L_2

with

L_1 \parallel L_2

.

2

Apply the alternate interior angles theorem

\angle XCA = \angle XDB = 35^\circ

When a transversal intersects two parallel lines, alternate interior angles are equal.

3

Calculate the target angle using the sum of interior angles in a triangle

\angle AXC = 180^\circ - (42^\circ + 35^\circ) = 103^\circ

The sum of interior angles in triangle

ACX

is

180^\circ

.

Anahtar Kavram

Properties of parallel lines intersected by a transversal and the triangle angle sum theorem.

Soru 71Soru →

In the $xy$ -plane, point $P$ has coordinates $(3, -4)$ . Point $P$ is reflected across the $y$ -axis to form point $Q$ . Point $Q$ is then translated upward by $2$ units to form point $R$ . Which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

The coordinates of point

Q

are

(-3, -4)

.

Point

R

lies in Quadrant III.

The distance between point

P

and point

Q

is

8

units.

The

y

-coordinate of point

R

is

2

.

Cevabı ve açıklamayı göster

Cevap: The coordinates of point $Q$ are $(-3, -4)$ .; Point $R$ lies in Quadrant III.

Cevap

The correct statements are that the coordinates of point

Q

are

(-3, -4)

and that point

R

lies in Quadrant III.

Reflecting

(3, -4)

across the

y

-axis yields

(-3, -4)

for point

Q

. Translating

(-3, -4)

upward by

2

units gives

(-3, -2)

for point

R

. Because both coordinates of

R

are negative, it resides in Quadrant III.

Adım Adım Çözüm

1

Perform the reflection across the

y

-axis

Point

Q = (-3, -4)

Reflecting across the

y

-axis negates the

x

-coordinate while keeping the

y

-coordinate the same.

2

Perform the vertical translation upward by

2

units

Point

R = (-3, -4 + 2) = (-3, -2)

Translating upward adds

2

to the

y

-coordinate of point

Q

.

3

Determine the quadrant of point

R

Quadrant III

Points with

x < 0

and

y < 0

are located in Quadrant III.

4

Calculate the distance between point

P(3, -4)

and point

Q(-3, -4)

Distance =

|3 - (-3)| = 6

units

Points

P

and

Q

lie on the horizontal line

y = -4

, so the distance is the difference in their

x

-coordinates.

Anahtar Kavram

Coordinate transformations including axis reflections and translations in the Cartesian plane

Soru 72Soru →

In circle $O$ , points $P$ and $Q$ lie on the circumference such that the ratio of the minor arc length $PQ$ to the radius $r$ of the circle is $\frac{5\pi}{6}$ . If the area of sector $POQ$ is $30\pi$ , what is the total perimeter of sector $POQ$ ?

12\sqrt{2} + 5\sqrt{2}\pi

12 + 5\pi

5\sqrt{2}\pi

12 + 10\pi

12\sqrt{2} + 12\sqrt{2}\pi

Cevabı ve açıklamayı göster

Cevap: $12\sqrt{2} + 5\sqrt{2}\pi$

Cevap

12\sqrt{2} + 5\sqrt{2}\pi

The central angle

\theta

in radians is equal to the ratio of minor arc length to radius, so

\theta = \frac{5\pi}{6}

. Using the area of a sector formula

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

, we substitute

A = 30\pi

to get

\frac{1}{2}r^2\left(\frac{5\pi}{6}\right) = 30\pi

, which simplifies to

r^2 = 72

, or

r = 6\sqrt{2}

. The arc length is then

s = r\theta = (6\sqrt{2})\left(\frac{5\pi}{6}\right) = 5\sqrt{2}\pi

. The total perimeter of the sector includes both bounding radii and the arc length:

2r + s = 12\sqrt{2} + 5\sqrt{2}\pi

.

Adım Adım Çözüm

1

Relate the central angle in radians to the given ratio of arc length to radius.

The central angle

\theta

in radians is given by

\theta = \frac{\text{arc length}}{r} = \frac{5\pi}{6}

.

By definition of radian measure, arc length

s = r\theta

, so

\frac{s}{r} = \theta

.

2

Use the sector area formula to solve for the radius

r

.

Setting

\frac{1}{2}r^2\left(\frac{5\pi}{6}\right) = 30\pi

yields

\frac{5\pi}{12}r^2 = 30\pi \implies r^2 = 72 \implies r = 6\sqrt{2}

.

The area of a sector with central angle

\theta

(in radians) is

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

.

3

Calculate the minor arc length

s

.

Arc length

s = r\theta = (6\sqrt{2})\left(\frac{5\pi}{6}\right) = 5\sqrt{2}\pi

.

Multiplying the radius by the central angle in radians gives the length of the subtended arc.

4

Calculate the total perimeter of sector

POQ

.

Perimeter

= 2r + s = 2(6\sqrt{2}) + 5\sqrt{2}\pi = 12\sqrt{2} + 5\sqrt{2}\pi

.

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

Anahtar Kavram

Relationship between radian measure, sector area, arc length, and sector perimeter

Tahmini Süre:2m 0s

Soru 73Soru →

A triangle has integer side lengths $a$ , $b$ , and $c$ such that its perimeter is $18$ . Which of the following could be the area of the triangle? Select all such values.

Geçerli olan tümünü seçin

12

6\sqrt{6}

9\sqrt{3}

4\sqrt{15}

8\sqrt{5}

Cevabı ve açıklamayı göster

Cevap: $12$ ; $6\sqrt{6}$ ; $9\sqrt{3}$

Cevap

The possible areas of the triangle are

12

,

6\sqrt{6}

, and

9\sqrt{3}

.

To form a valid triangle with integer sides and a perimeter of 18, each side length must be a positive integer strictly less than 9 by the triangle inequality theorem. Evaluating Heron's formula

A = \sqrt{s(s-a)(s-b)(s-c)}

with

s=9

for all valid side combinations

(a,b,c)

produces the set of areas

\{3\sqrt{7}, 6\sqrt{3}, 3\sqrt{15}, 12, 6\sqrt{5}, 6\sqrt{6}, 9\sqrt{3}\}

. Among the options given, the values 12 (from side lengths 8, 5, 5),

6\sqrt{6}

(from side lengths 7, 6, 5), and

9\sqrt{3}

(from side lengths 6, 6, 6) are correct.

Adım Adım Çözüm

1

Determine the structural bounds on side lengths using the Triangle Inequality Theorem.

For any side length

x

,

x < \frac{\text{Perimeter}}{2} = 9

. Thus,

a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}

.

The sum of any two side lengths must be strictly greater than the third side length.

2

List all unique integer side length combinations

(a, b, c)

where

a \ge b \ge c

and

a + b + c = 18

.

The valid triples are

(8, 8, 2)

,

(8, 7, 3)

,

(8, 6, 4)

,

(8, 5, 5)

,

(7, 7, 4)

,

(7, 6, 5)

, and

(6, 6, 6)

.

Exhaustive enumeration of integer partitions of 18 satisfying

a \le 8

.

3

Calculate the semi-perimeter

s

of the triangle.

s = \frac{18}{2} = 9

.

Required parameter for Heron's formula

A = \sqrt{s(s-a)(s-b)(s-c)}

.

4

Compute the area for candidate triples using Heron's formula.

For

(8, 5, 5)

:

A = \sqrt{9(1)(4)(4)} = 12

.
For

(7, 6, 5)

:

A = \sqrt{9(2)(3)(4)} = 6\sqrt{6}

.
For

(6, 6, 6)

:

A = \sqrt{9(3)(3)(3)} = 9\sqrt{3}

.

Direct evaluation of geometric area for the valid triangles.

Anahtar Kavram

Triangle Inequality Theorem and Area calculation via Heron's Formula

Soru 74Soru →

In the $xy$ -plane, line $k$ passes through the point $(1, -2)$ and is perpendicular to line $m$ , which is defined by the equation $3x - 4y = 12$ . Line $k$ intersects line $n$ , defined by the equation $y = 2x + 1$ , at point $P$ . What is the distance between point $P$ and the point $(3.5, 3)$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5

Rewriting the equation of line

m

,

3x - 4y = 12

, in slope-intercept form gives

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

, so line

m

has a slope of

\frac{3}{4}

. Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal,

-\frac{4}{3}

. Using the point

(1, -2)

, line

k

has the equation

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

. Setting this equal to the equation of line

n

(

y = 2x + 1

) yields

-\frac{4}{3}x - \frac{2}{3} = 2x + 1

, which solves to

x = -0.5

and

y = 0

, giving the intersection point

P(-0.5, 0)

. Finally, calculating the distance between

P(-0.5, 0)

and

(3.5, 3)

using the distance formula gives

\sqrt{(3.5 - (-0.5))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

.

Adım Adım Çözüm

1

Determine the slope of line

m

The slope of line

m

is

\frac{3}{4}

Rewriting

3x - 4y = 12

in slope-intercept form gives

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

, so the slope is

\frac{3}{4}

.

2

Determine the slope of line

k

The slope of line

k

is

-\frac{4}{3}

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Write the equation of line

k

The equation of line

k

is

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

Using point-slope form with

(1, -2)

gives

y - (-2) = -\frac{4}{3}(x - 1)

, which simplifies to

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

.

4

Find the coordinates of intersection point

P

Point

P

has coordinates

(-0.5, 0)

Setting

-\frac{4}{3}x - \frac{2}{3} = 2x + 1

yields

-\frac{10}{3}x = \frac{5}{3} \implies x = -0.5

. Substituting

x = -0.5

into

y = 2x + 1

yields

y = 0

.

5

Calculate the distance between

P(-0.5, 0)

and

(3.5, 3)

The distance is

5

Applying the distance formula yields

d = \sqrt{(3.5 - (-0.5))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

.

Anahtar Kavram

Perpendicular line slopes, finding intersection of two lines, and applying the distance formula.

Soru 75Soru →

In right triangle $PQR$ , the measure of angle $P$ is $30^\circ$ , the measure of angle $Q$ is $60^\circ$ , and the length of the hypotenuse $PR$ is $12$ . Which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

The length of the shortest side is

6

.

The length of the side opposite the

60^\circ

angle is

6\sqrt{3}

.

The length of the side opposite the

60^\circ

angle is

6\sqrt{2}

.

The area of triangle

PQR

is

36\sqrt{3}

.

The perimeter of triangle

PQR

is

18\sqrt{3}

.

Cevabı ve açıklamayı göster

Cevap: The length of the shortest side is $6$ .; The length of the side opposite the $60^\circ$ angle is $6\sqrt{3}$ .

Cevap

The statements confirming that the shortest side has length

6

and that the side opposite the

60^\circ

angle has length

6\sqrt{3}

are both correct.

In any

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

right triangle, the sides opposite the

30^\circ

,

60^\circ

, and

90^\circ

angles are in the ratio

1 : \sqrt{3} : 2

. With a hypotenuse of

12

, the side opposite

30^\circ

is

\frac{12}{2} = 6

, and the side opposite

60^\circ

is

6\sqrt{3}

. Thus, both statements specifying these values are correct.

Adım Adım Çözüm

1

Identify the side length ratio for a

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

right triangle.

The side lengths follow the standard ratio

1 : \sqrt{3} : 2

, corresponding to the sides opposite the

30^\circ

,

60^\circ

, and

90^\circ

angles respectively.

Special right triangle properties establish fixed proportional relationships between sides based on interior angles.

2

Calculate the length of the shortest side (opposite the

30^\circ

angle).

Shortest side =

\frac{\text{Hypotenuse}}{2} = \frac{12}{2} = 6

.

Since the hypotenuse corresponds to

2x = 12

,

x = 6

.

3

Calculate the length of the longer leg (opposite the

60^\circ

angle).

Longer leg =

x\sqrt{3} = 6\sqrt{3}

.

The side opposite

60^\circ

is

\sqrt{3}

times the side opposite

30^\circ

.

Anahtar Kavram

Side length ratios of a

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

special right triangle (

1 : \sqrt{3} : 2

).

Soru 76Soru →

In circle $O$ , sector $AOB$ has an area of $45\pi$ square units and arc $AB$ has a length of $6\pi$ units. What is the perimeter of sector $AOB$ ?

15 + 6\pi

30 + 6\pi

45 + 6\pi

60 + 6\pi

30\pi

Cevabı ve açıklamayı göster

Cevap: $30 + 6\pi$

Cevap

30 + 6\pi

The area of a sector is given by

\frac{1}{2} r L

, where

r

is the radius and

L

is the arc length. Substituting

L = 6\pi

and Area

= 45\pi

yields

45\pi = \frac{1}{2} r (6\pi)

, which simplifies to

45\pi = 3\pi r

, so

r = 15

. The perimeter of the sector is

L + 2r = 6\pi + 2(15) = 30 + 6\pi

. Thus, the option equal to

30 + 6\pi

is correct.

Adım Adım Çözüm

1

Express sector area and arc length in terms of radius

r

and central angle

\theta

in degrees.

Sector Area =

\frac{\theta}{360^\circ} \times \pi r^2 = 45\pi

and Arc Length =

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

.

Relating both formulas allows solving for the radius directly using the ratio of area to arc length.

2

Divide the sector area formula by the arc length formula to find radius

r

.

\frac{\text{Sector Area}}{\text{Arc Length}} = \frac{\frac{\theta}{360^\circ} \pi r^2}{\frac{\theta}{360^\circ} 2\pi r} = \frac{r}{2} = \frac{45\pi}{6\pi} = 7.5 \implies r = 15.

The central angle fraction

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

and

\pi

cancel out, leaving

\frac{r}{2} = 7.5

.

3

Calculate the total perimeter of sector

AOB

.

\text{Perimeter} = \text{Arc Length} + 2r = 6\pi + 2(15) = 30 + 6\pi.

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

OA

and

OB

).

Anahtar Kavram

The relationship between sector area, arc length, radius, and sector perimeter

Soru 77Soru →

In $\triangle ABC$ , the lengths of sides $AB$ , $BC$ , and $AC$ are $13$ , $14$ , and $15$ , respectively. A line segment $DE$ is drawn parallel to side $BC$ , with point $D$ lying on side $AB$ and point $E$ lying on side $AC$ . If the perimeter of $\triangle ADE$ is equal to the perimeter of quadrilateral $DBCE$ , what is the area of $\triangle ADE$ ?

Cevabı ve açıklamayı göster

Cevap: 47.25

Cevap

47.25

The area of the original triangle

\triangle ABC

is computed as

84

using Heron's formula. By defining the linear scale factor

k

between

\triangle ADE

and

\triangle ABC

, the perimeters of

\triangle ADE

and quadrilateral

DBCE

are expressed as

42k

and

42 - 14k

, respectively. Setting these equal yields

k = 0.75

. The area of

\triangle ADE

is then

k^2 \times 84 = 0.5625 \times 84 = 47.25

.

Adım Adım Çözüm

1

Calculate the perimeter and area of the main triangle

\triangle ABC

.

The perimeter of

\triangle ABC

is

13 + 14 + 15 = 42

. Using Heron's formula with semi-perimeter

s = 21

,

\text{Area}(\triangle ABC) = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = 84

.

Finding the area and perimeter of the full triangle sets the required baseline for proportional scaling.

2

Set up expressions for the perimeters of

\triangle ADE

and quadrilateral

DBCE

using a scale factor

k

.

Because

DE \parallel BC

,

\triangle ADE \sim \triangle ABC

with scale factor

k = \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}

. Thus,

\text{Perimeter}(\triangle ADE) = 13k + 15k + 14k = 42k

. The segments

DB = 13(1-k)

and

EC = 15(1-k)

, so

\text{Perimeter}(DBCE) = 13(1-k) + 14 + 15(1-k) + 14k = 42 - 14k

.

Parallel lines create similar triangles, which allows all perimeter segment lengths to be represented in terms of one variable

k

.

3

Solve for the scale factor

k

by equating the two perimeters.

42k = 42 - 14k \implies 56k = 42 \implies k = \frac{42}{56} = \frac{3}{4} = 0.75

.

Equating the perimeters satisfies the condition specified in the question stem.

4

Calculate the area of

\triangle ADE

using the square of the linear scale factor.

\text{Area}(\triangle ADE) = k^2 \times \text{Area}(\triangle ABC) = \left(\frac{3}{4}\right)^2 \times 84 = \frac{9}{16} \times 84 = \frac{189}{4} = 47.25

.

The area ratio of similar geometric figures is proportional to the square of their linear scale factor.

Anahtar Kavram

Properties of Similar Triangles, Area via Heron's Formula, and Perimeter Scaling

Soru 78Soru →

In $\triangle ABC$ , the length of side $AB$ is $10$ and the length of side $BC$ is $17$ . Point $D$ lies on the line containing segment $AC$ such that segment $BD$ is perpendicular to line $AC$ . If the length of altitude $BD$ and the length of side $AC$ are both integers, and the area of $\triangle ABC$ is strictly greater than $36$ , what is the perimeter of $\triangle ABC$ ?

36

42

48

52

56

Cevabı ve açıklamayı göster

Cevap: $48$

Cevap

The perimeter of

\triangle ABC

is

48

.

Applying the Pythagorean theorem to both right triangles formed by altitude

BD = h

gives

100 - m^2 = h^2

and

289 - n^2 = h^2

, where

m = AD

and

n = CD

. Subtracting these equations yields

n^2 - m^2 = 189

, which factors as

(n - m)(n + m) = 189

. Testing integer factor pairs of

189

while enforcing

m \le 10

isolates a single non-degenerate solution:

m = 6

,

n = 15

, and

h = 8

. When point

D

lies between

A

and

C

,

AC = 6 + 15 = 21

. This gives an area of

\frac{1}{2} \times 21 \times 8 = 84

(which is strictly greater than

36

) and a total perimeter of

10 + 17 + 21 = 48

.

Adım Adım Çözüm

1

Set up Pythagorean relationships for the right triangles formed by altitude

BD

.

Let

BD = h

,

AD = m

, and

CD = n

. In right

\triangle ABD

,

m^2 + h^2 = 10^2 = 100

. In right

\triangle CBD

,

n^2 + h^2 = 17^2 = 289

.

Altitude

BD

divides the figure into two right triangles sharing leg

h

.

2

Subtract the two equations to eliminate

h^2

and factor the difference of squares.

n^2 - m^2 = 289 - 100 = 189 \implies (n - m)(n + m) = 189.

Since

h

and

AC

are integers,

m

and

n

must also be integers for

AC = n \pm m

to be an integer.

3

Analyze integer factor pairs

(u, v)

of

189

where

u = n - m

and

v = n + m

.

Factor pairs

(u, v)

with

u \cdot v = 189

:
-

(1, 189) \implies m = 94

(invalid,

m \le 10

)
-

(3, 63) \implies m = 30

(invalid,

m \le 10

)
-

(7, 27) \implies m = 10, h = 0

(degenerate triangle)
-

(9, 21) \implies n = 15, m = 6, h = \sqrt{100 - 36} = 8

.

The leg

m = AD

cannot exceed the hypotenuse

AB = 10

.

4

Evaluate side

AC

, area, and perimeter for valid geometric configurations.

Case 1:

D

lies on segment

AC \implies AC = n + m = 15 + 6 = 21

.
Area =

\frac{1}{2} \times 21 \times 8 = 84 > 36

.
Perimeter =

10 + 17 + 21 = 48

.
Case 2:

D

lies outside segment

AC \implies AC = n - m = 15 - 6 = 9

.
Area =

\frac{1}{2} \times 9 \times 8 = 36

, which does not satisfy area

> 36

.

The problem specifies that the area must be strictly greater than

36

.

Anahtar Kavram

Properties of triangles, Pythagorean theorem system solver, and geometric area constraints

Soru 79Soru →

In triangle $ABC$ , point $D$ lies on side $AB$ such that $AD : DB = 3 : 1$ , and point $E$ lies on side $AC$ such that segment $DE$ is parallel to side $BC$ . If the area of triangle $ABC$ is $64$ , what is the area of triangle $ADE$ ?

16

27

36

48

52

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

36

Because segment

DE

is parallel to

BC

, triangle

ADE

is similar to triangle

ABC

. The ratio of side

AD

to side

AB

is

3 : (3 + 1) = 3 : 4

. For similar triangles, the ratio of their areas equals the square of the ratio of their corresponding side lengths, which is

(3/4)^2 = 9/16

. Multiplying the area of triangle

ABC

(

64

) by

9/16

yields an area of

36

for triangle

ADE

.

Adım Adım Çözüm

1

Determine the scale factor between similar triangles

ADE

and

ABC

.

Since

DE \parallel BC

, triangle

ADE

is similar to triangle

ABC

. The side length ratio is

AD / AB = AD / (AD + DB) = 3 / (3 + 1) = 3/4

.

Parallel lines create corresponding equal angles, making triangles

ADE

and

ABC

similar.

2

Calculate the ratio of the areas of the similar triangles.

The area ratio is the square of the side scale factor:

(3/4)^2 = 9/16

.

The ratio of the areas of two similar figures is equal to the square of their scale factor.

3

Compute the area of triangle

ADE

.

\text{Area}(\triangle ADE) = \frac{9}{16} \times 64 = 36.

Multiply the total area of triangle

ABC

by the area ratio

9/16

.

Anahtar Kavram

Area Ratio of Similar Triangles

Soru 80Soru →

In the $xy$ -plane, line $l_1$ intersects the positive $x$ -axis at $(a, 0)$ and the positive $y$ -axis at $(0, b)$ , where $a > b > 0$ . Line $l_2$ is perpendicular to line $l_1$ and passes through the point $(a, b)$ . If the perpendicular distance from the origin $(0, 0)$ to line $l_1$ is equal to the perpendicular distance from the origin to line $l_2$ , what is the value of $\frac{a}{b}$ ?

\frac{\sqrt{5} - 1}{2}

1

\frac{\sqrt{5} + 1}{2}

\sqrt{2}

2

Cevabı ve açıklamayı göster

Cevap: $\frac{\sqrt{5} + 1}{2}$

Cevap

The ratio

\frac{a}{b}

is

\frac{\sqrt{5} + 1}{2}

.

The option stating

\frac{\sqrt{5} + 1}{2}

is correct because equating the distances of both lines from the origin leads to the equation

a^2 - ab - b^2 = 0

. Expressed in terms of the ratio

r = \frac{a}{b}

, this becomes

r^2 - r - 1 = 0

. Because

a > b > 0

,

r

must be greater than

1

, yielding

r = \frac{\sqrt{5} + 1}{2}

.

Adım Adım Çözüm

1

Find the equation and distance from origin for line

l_1

.

The equation of

l_1

is

\frac{x}{a} + \frac{y}{b} = 1

, or

bx + ay - ab = 0

. The perpendicular distance from

(0,0)

to

l_1

is

d_1 = \frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}}

.

Standard intercept form of a line and point-to-line distance formula.

2

Find the slope and equation of line

l_2

.

The slope of

l_1

is

-\frac{b}{a}

, so the slope of perpendicular line

l_2

is

\frac{a}{b}

. Since

l_2

passes through

(a, b)

, its equation is

y - b = \frac{a}{b}(x - a)

, which simplifies to

ax - by - (a^2 - b^2) = 0

.

Perpendicular lines have slopes that are negative reciprocals.

3

Calculate the distance from the origin to line

l_2

and set

d_1 = d_2

.

The distance

d_2 = \frac{|-(a^2 - b^2)|}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}}

(since

a > b > 0

). Equating

d_1 = d_2

gives

\frac{ab}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}}

, so

ab = a^2 - b^2

.

The distances from the origin to both lines are specified to be equal.

4

Solve for the ratio

r = \frac{a}{b}

.

Divide

a^2 - ab - b^2 = 0

by

b^2

to obtain

r^2 - r - 1 = 0

. Applying the quadratic formula gives

r = \frac{1 \pm \sqrt{5}}{2}

. Since

a > b > 0

,

r > 1

, so

r = \frac{\sqrt{5} + 1}{2}

.

Determines the unique positive value greater than 1 that satisfies the algebraic relationship.

Anahtar Kavram

Perpendicular line slopes, point-to-line distance formula, and algebraic substitution for ratio determination.

Geometry

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