Triangles: Properties, Perimeter, and Area Alıştırma Soruları — GRE General Test — Sayfa 2

Soru 21Soru →

In acute triangle $ABC$ , point $D$ lies on side $BC$ such that segment $AD$ is perpendicular to $BC$ . The length of side $BC$ is $15$ , and the ratio of the area of triangle $ABD$ to the area of triangle $ADC$ is $2 : 3$ . If the lengths of sides $AB$ and $AC$ are both integers, what is the perimeter of triangle $ABC$ ?

30

45

54

60

75

Cevabı ve açıklamayı göster

Cevap: $60$

Cevap

The perimeter of triangle

ABC

is

60

.

The correct answer is derived by recognizing that the altitude splits the base into segments of lengths 6 and 9 based on the area ratio 2:3. Applying the Pythagorean theorem to both right triangles yields

AC^2 - AB^2 = 45

. Factoring 45 into positive integer pairs shows that the only valid side lengths yielding a real, positive height are

AB = 22

and

AC = 23

, giving a total perimeter of

15 + 22 + 23 = 60

.

Adım Adım Çözüm

1

Determine the lengths of base segments

BD

and

DC

.

BD = 6

and

DC = 9

.

Triangles

ABD

and

ADC

share the common height

AD = h

. The ratio of their areas is equal to the ratio of their bases:

\frac{\text{Area}(ABD)}{\text{Area}(ADC)} = \frac{BD}{DC} = \frac{2}{3}

. Since

BD + DC = 15

, we have

BD = 6

and

DC = 9

.

2

Express the square of height

h^2

using the Pythagorean theorem in right triangles

ABD

and

ADC

.

h^2 = AB^2 - 36 = AC^2 - 81

.

In right triangle

ABD

,

AB^2 = BD^2 + h^2 = 36 + h^2

. In right triangle

ADC

,

AC^2 = DC^2 + h^2 = 81 + h^2

.

3

Set up a difference of squares equation for side lengths

AB

and

AC

.

(AC - AB)(AC + AB) = 45

.

Equating the two expressions for

h^2

gives

AC^2 - 81 = AB^2 - 36 \implies AC^2 - AB^2 = 45

.

4

Find positive integer solutions for

AB

and

AC

.

AB = 22

and

AC = 23

.

Since

AB

and

AC

are positive integers and

AC > AB

, we analyze factor pairs

(AC - AB, AC + AB)

of

45

with same parity (both odd):
- Pair

(1, 45)

:

AC - AB = 1

and

AC + AB = 45 \implies AC = 23, AB = 22

. Here

h^2 = 22^2 - 36 = 448 > 0

, giving a valid non-degenerate triangle.
- Pair

(3, 15)

:

AC - AB = 3

and

AC + AB = 15 \implies AC = 9, AB = 6

. Here

h^2 = 6^2 - 36 = 0

, which means

h = 0

(degenerate line segment, invalid).
- Pair

(5, 9)

:

AC - AB = 5

and

AC + AB = 9 \implies AC = 7, AB = 2

. Here

h^2 = 2^2 - 36 = -32 < 0

(impossible).

5

Calculate the total perimeter of triangle

ABC

.

Perimeter

= 15 + 22 + 23 = 60

.

Summing all three side lengths gives

BC + AB + AC = 15 + 22 + 23 = 60

.

Anahtar Kavram

Triangles: Properties, Perimeter, and Area

Soru 22Soru →

In isosceles triangle $ABC$ , side $AB$ is equal in length to side $AC$ . The perimeter of triangle $ABC$ is $36$ , and the length of the altitude from vertex $A$ to base $BC$ is $12$ . What is the area of triangle $ABC$ ?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

The area of triangle

ABC

is

60

.

Let

x

be the length of the two equal sides

AB

and

AC

, and

b

be the length of base

BC

. The perimeter is

2x + b = 36

, yielding

x = 18 - \frac{b}{2}

. The altitude from

A

to

BC

has length

12

and bisects

BC

into two segments of length

\frac{b}{2}

. Applying the Pythagorean theorem to one of the right triangles gives

x^2 = 12^2 + \left(\frac{b}{2}\right)^2

. Substituting

x = 18 - \frac{b}{2}

gives

\left(18 - \frac{b}{2}\right)^2 = 144 + \frac{b^2}{4} \implies 324 - 18b + \frac{b^2}{4} = 144 + \frac{b^2}{4} \implies 18b = 180 \implies b = 10

. The area is

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

.

Adım Adım Çözüm

1

Set up an equation for the side lengths using the perimeter.

Let

b

be the length of base

BC

, and

x

be the length of sides

AB

and

AC

. Since the perimeter is

36

,

2x + b = 36

, which gives

x = 18 - \frac{b}{2}

.

An isosceles triangle has two sides of equal length, and perimeter is the sum of all three side lengths.

2

Apply the Pythagorean theorem to the right triangle formed by the altitude.

The altitude of length

12

drops perpendicularly to base

BC

, bisecting it into two equal segments of length

\frac{b}{2}

. Thus,

x^2 = 12^2 + \left(\frac{b}{2}\right)^2 = 144 + \frac{b^2}{4}

.

In an isosceles triangle, the altitude to the base bisects the base and creates two congruent right-angled triangles.

3

Solve for the base length

b

.

Substitute

x = 18 - \frac{b}{2}

into the equation:

\left(18 - \frac{b}{2}\right)^2 = 144 + \frac{b^2}{4} \implies 324 - 18b + \frac{b^2}{4} = 144 + \frac{b^2}{4} \implies 18b = 180 \implies b = 10

.

Expanding the squared binomial allows the

\frac{b^2}{4}

terms to cancel out, resulting in a linear equation for

b

.

4

Calculate the area of the triangle.

\text{Area} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 10 \times 12 = 60.

The area of a triangle is evaluated using half the product of its base and corresponding altitude.

Anahtar Kavram

Isosceles triangle properties, altitude-to-base bisector property, Pythagorean theorem, and triangle area calculation.

Soru 23Soru →

A triangle has side lengths of $8$ , $11$ , and $x$ , where $x$ is an integer. If the perimeter of the triangle is a positive integer multiple of $5$ , which of the following could be the value of $x$ ? Select all such values.

Geçerli olan tümünü seçin

1

6

11

16

21

Cevabı ve açıklamayı göster

Cevap: $6$ ; $11$ ; $16$

Cevap

The possible values for

x

are

6

,

11

, and

16

.

According to the Triangle Inequality Theorem, the third side

x

must be strictly greater than

11 - 8 = 3

and strictly less than

11 + 8 = 19

. The perimeter of the triangle is

8 + 11 + x = 19 + x

. For

19 + x

to be a positive multiple of

5

,

19 + x

can be

25

,

30

, or

35

within the allowed range for

x

, giving

x = 6

,

x = 11

, and

x = 16

.

Adım Adım Çözüm

1

Apply the Triangle Inequality Theorem to determine the valid range for the unknown side

x

.

The length of

x

must satisfy

11 - 8 < x < 11 + 8

, which simplifies to

3 < x < 19

.

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

2

Set up an expression for the perimeter

P

of the triangle.

P = 8 + 11 + x = 19 + x

.

Perimeter is the total sum of all three side lengths.

3

Determine which values of

x

within the range

3 < x < 19

make

P = 19 + x

a multiple of

5

.

If

x = 6

,

P = 25

(multiple of

5

). If

x = 11

,

P = 30

(multiple of

5

). If

x = 16

,

P = 35

(multiple of

5

).

Adding

19

to

6

,

11

, and

16

gives multiples of

5

within the strict inequality bounds.

4

Test boundary values outside the inequality bounds.

x = 1

gives

P = 20

, but

1 \le 3

(invalid).

x = 21

gives

P = 40

, but

21 \ge 19

(invalid).

Values outside

3 < x < 19

cannot form a valid triangle.

Anahtar Kavram

Triangle Inequality Theorem and Perimeter Constraints

Soru 24Soru →

In triangle $ABC$ , the ratio of the side lengths $AB : BC : AC$ is $3 : 4 : 5$ , and the total area of triangle $ABC$ is $24$ square units. A line segment $DE$ is drawn parallel to side $BC$ , where point $D$ lies on side $AB$ and point $E$ lies on side $AC$ . If the perimeter of triangle $ADE$ is exactly half the perimeter of triangle $ABC$ , what is the area of trapezoid $DBCE$ in square units?

6

12

16

18

20

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The area of trapezoid

DBCE

is

18

square units.

Because line segment

DE

is parallel to side

BC

, triangle

ADE

is similar to triangle

ABC

. Given that the perimeter of triangle

ADE

is half the perimeter of triangle

ABC

, the ratio of their side lengths (the linear scale factor) is

1/2

. The area ratio of similar triangles is the square of the linear scale factor, which is

(1/2)^2 = 1/4

. Thus, the area of triangle

ADE

is

1/4 \times 24 = 6

square units. Subtracting this from the total area gives the area of trapezoid

DBCE

:

24 - 6 = 18

square units.

Adım Adım Çözüm

1

Determine the linear scale factor between triangle

ADE

and triangle

ABC

.

Linear scale factor

k = \frac{\text{Perimeter}(ADE)}{\text{Perimeter}(ABC)} = \frac{1}{2}

.

Since

DE \parallel BC

, triangle

ADE

is similar to triangle

ABC

, so the ratio of their perimeters equals the ratio of corresponding side lengths.

2

Calculate the area of triangle

ADE

using the area scale factor

k^2

.

\text{Area}(ADE) = k^2 \times \text{Area}(ABC) = \left(\frac{1}{2}\right)^2 \times 24 = \frac{1}{4} \times 24 = 6

square units.

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

3

Subtract the area of triangle

ADE

from the area of triangle

ABC

to find the area of trapezoid

DBCE

.

\text{Area}(DBCE) = \text{Area}(ABC) - \text{Area}(ADE) = 24 - 6 = 18

square units.

Trapezoid

DBCE

is formed by removing triangle

ADE

from triangle

ABC

.

Anahtar Kavram

Properties of Similar Triangles and Area Scaling

Tahmini Süre:1m 30s

Soru 25Soru →

In the $xy$ -plane, triangle $PQR$ has vertices at $P(0, 0)$ , $Q(8, 0)$ , and $R(2, 6)$ . Point $S$ lies on segment $PQ$ such that segment $RS$ divides triangle $PQR$ into two regions of equal area. Point $T$ lies on segment $QR$ such that segment $ST$ is parallel to segment $PR$ . What is the area of triangle $QST$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

The area of triangle

PQR

is calculated using base

PQ = 8

and height

h = 6

, yielding

\frac{1}{2} \times 8 \times 6 = 24

. Since segment

RS

divides triangle

PQR

into two regions of equal area that share the height from vertex

R

, point

S

must be the midpoint of

PQ

, making

QS = 4

. Because segment

ST

is parallel to segment

PR

, triangle

QST

is similar to triangle

QPR

with a side length ratio of

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

. The ratio of the areas of similar triangles is the square of the side ratio,

\left(\frac{1}{2}\right)^2 = \frac{1}{4}

. Multiplying the total area

24

by

\frac{1}{4}

gives the area of triangle

QST

as

6

.

Adım Adım Çözüm

1

Calculate the area of the main triangle

PQR

.

\text{Area}(PQR) = \frac{1}{2} \times 8 \times 6 = 24

.

The base

PQ

lies along the x-axis with length

8 - 0 = 8

, and the perpendicular height from vertex

R(2,6)

to the base is

6

.

2

Determine the length of segment

QS

.

QS = 4

.

Line segment

RS

splits

\triangle PQR

into two smaller triangles,

\triangle PSR

and

\triangle QSR

, which share the same altitude from vertex

R

. For their areas to be equal, their base lengths

PS

and

SQ

must be equal. Therefore,

S

is the midpoint of

PQ

, giving

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

.

3

Establish the similarity relationship and scale factor between

\triangle QST

and

\triangle QPR

.

\triangle QST \sim \triangle QPR

with scale factor

k = \frac{1}{2}

.

Because segment

ST

is parallel to segment

PR

, corresponding angles are equal (

\angle QST = \angle QPR

and

\angle QTS = \angle QRP

). Thus,

\triangle QST

is similar to

\triangle QPR

. The ratio of corresponding side lengths is

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

.

4

Compute the area of triangle

QST

.

\text{Area}(QST) = 6

.

The ratio of the areas of similar triangles is equal to the square of their linear scale factor:

\text{Area}(QST) = \left(\frac{1}{2}\right)^2 \times \text{Area}(PQR) = \frac{1}{4} \times 24 = 6

.

Anahtar Kavram

Area of triangles, midpoint area partitioning, and area ratio scaling in similar triangles

Tahmini Süre:2m 0s

Soru 26Soru →

An altitude of an acute triangle divides its base into two adjacent segments of lengths $5$ and $9$ . If the area of the triangle is $84$ square units, what is the perimeter of the triangle?

28

36

42

48

52

Cevabı ve açıklamayı göster

Cevap: $42$

Cevap

The perimeter of the triangle is

42

.

The base of the triangle is the sum of the two adjacent segments,

5 + 9 = 14

. Setting up the area formula gives

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

, which yields an altitude height of

h = 12

. The altitude creates two right triangles: one with legs

5

and

12

(hypotenuse

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

) and another with legs

9

and

12

(hypotenuse

\sqrt{9^2 + 12^2} = 15

). Adding all three side lengths (

14 + 13 + 15

) yields

42

.

Adım Adım Çözüm

1

Find the total base length and calculate the height (altitude) of the triangle.

Base

= 5 + 9 = 14

. Height

h = \frac{2 \times \text{Area}}{\text{base}} = \frac{2 \times 84}{14} = 12

.

The area of a triangle is given by

A = \frac{1}{2} b h

.

2

Calculate the lengths of the two non-base sides using the Pythagorean theorem on the two right triangles formed by the altitude.

Left side

= \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13

. Right side

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

.

An altitude drawn to the base creates two right triangles sharing the altitude as a common leg.

3

Sum all three boundary sides to find the perimeter.

Perimeter

= 14 + 13 + 15 = 42

.

Perimeter is the total length around the outside of the triangle.

Anahtar Kavram

Area and perimeter of triangles split by an altitude using the Pythagorean theorem.

Tahmini Süre:1m 30s

Soru 27Soru →

In right triangle $ABC$ , the measure of angle $C$ is $90^\circ$ . Point $D$ lies on segment $AC$ such that $AD = 11$ and $BD = 13$ . If $DC = 5$ , what is the perimeter of triangle $ABD$ ?

Cevabı ve açıklamayı göster

Cevap: 44

Cevap

The perimeter of triangle

ABD

is

44

.

To find the perimeter of triangle

ABD

, we need the lengths of its three sides:

AD

,

BD

, and

AB

. We are given

AD = 11

and

BD = 13

. To find

AB

, we first analyze right triangle

BCD

with right angle at

C

, hypotenuse

BD = 13

, and leg

DC = 5

. Using the Pythagorean theorem, leg

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

. Next, the full leg

AC = AD + DC = 11 + 5 = 16

. In right triangle

ABC

, legs are

AC = 16

and

BC = 12

, making hypotenuse

AB = \sqrt{16^2 + 12^2} = 20

. Finally, the perimeter of triangle

ABD

is

11 + 13 + 20 = 44

.

Adım Adım Çözüm

1

Find the length of side

BC

using right triangle

BCD

BC = 12

In right triangle

BCD

, angle

C = 90^\circ

, hypotenuse

BD = 13

, and leg

DC = 5

. By the Pythagorean theorem,

BC = \sqrt{13^2 - 5^2} = \sqrt{144} = 12

.

2

Determine the total length of side

AC

AC = 16

Since point

D

lies on segment

AC

,

AC = AD + DC = 11 + 5 = 16

.

3

Calculate the hypotenuse

AB

of the main right triangle

ABC

AB = 20

In right triangle

ABC

, legs are

AC = 16

and

BC = 12

. By the Pythagorean theorem,

AB = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

.

4

Calculate the perimeter of triangle

ABD

Perimeter

= 44

The sides of triangle

ABD

are

AD = 11

,

BD = 13

, and

AB = 20

. Adding these gives

11 + 13 + 20 = 44

.

Anahtar Kavram

Pythagorean Theorem and Multi-Step Triangle Properties

Soru 28Soru →

An isosceles triangle has two sides of length $10$ units each and a third side of integer length $x$ units. If the area of the triangle is strictly greater than $24$ square units and less than or equal to $48$ square units, which of the following could be the value of $x$ ? Select all such values.

Geçerli olan tümünü seçin

4

6

14

16

20

Cevabı ve açıklamayı göster

Cevap: 6; 16

Cevap

The values 6 and 16 are the valid side lengths.

The values 6 and 16 produce valid areas of approximately 28.62 and exactly 48 square units respectively, both of which satisfy the given condition that the area must be strictly greater than 24 and less than or equal to 48.

Adım Adım Çözüm

1

Express the height and area of the isosceles triangle in terms of

x

.

Height

h = \sqrt{10^2 - (x/2)^2} = \sqrt{100 - \frac{x^2}{4}}

, so Area

A = \frac{1}{2} x \sqrt{100 - \frac{x^2}{4}} = \frac{1}{4} x \sqrt{400 - x^2}

.

The altitude to the base of an isosceles triangle bisects the base into two equal segments of length

\frac{x}{2}

.

2

Set up the inequality for the area constraints

24 < A \le 48

.

24 < \frac{1}{4} x \sqrt{400 - x^2} \le 48 \implies 96 < x \sqrt{400 - x^2} \le 192

.

Multiplying all parts by

4

isolates the radical expression.

3

Square all terms to analyze the function

f(x^2) = x^2(400 - x^2)

.

9216 < x^2(400 - x^2) \le 36864

.

Squaring positive quantities preserves the inequality direction.

4

Evaluate the area function for each given option choice.

For

x=4

:

A \approx 19.6

(too small). For

x=6

:

A \approx 28.6

(valid). For

x=14

:

A \approx 50.0

(too large). For

x=16

:

A = 48

(valid). For

x=20

: degenerate triangle with

A = 0

(invalid).

Direct evaluation identifies which integer choices satisfy

24 < A \le 48

.

Anahtar Kavram

Properties of isosceles triangles, Pythagorean theorem for altitude, area bounds, and triangle inequality.

Soru 29Soru →

In triangle $ABC$ , the length of side $AB$ is $16$ units and the length of side $AC$ is $13$ units. If the area of triangle $ABC$ is $96$ square units and the altitude from vertex $C$ to side $AB$ intersects the line segment $AB$ at point $D$ , what is the length of segment $AD$ ?

5

11

12

7

1

Cevabı ve açıklamayı göster

Cevap: $5$

Cevap

The length of segment

AD

is

5

units.

The correct answer is

5

. First, find altitude

CD

using the triangle area equation:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 96 = \frac{1}{2} \times 16 \times CD \implies CD = 12

. Next, because

CD \perp AB

, triangle

ADC

is a right triangle with hypotenuse

AC = 13

and leg

CD = 12

. Applying the Pythagorean theorem yields

AD = \sqrt{13^2 - 12^2} = \sqrt{25} = 5

.

Adım Adım Çözüm

1

Calculate the length of altitude

CD

using the triangle area formula.

CD = 12

units.

The area of triangle

ABC

is given by

\text{Area} = \frac{1}{2} \times AB \times CD

. Substituting the given values:

96 = \frac{1}{2} \times 16 \times CD \implies 96 = 8 \times CD \implies CD = 12

.

2

Apply the Pythagorean theorem in right triangle

ADC

to find

AD

.

AD = 5

units.

Altitude

CD

is perpendicular to

AB

, forming right triangle

ADC

with hypotenuse

AC = 13

and leg

CD = 12

. By the Pythagorean theorem,

AD^2 + CD^2 = AC^2 \implies AD^2 + 12^2 = 13^2 \implies AD^2 + 144 = 169 \implies AD^2 = 25 \implies AD = 5

.

Anahtar Kavram

Triangles: Properties, Perimeter, and Area

Tahmini Süre:1m 30s

Soru 30Soru →

A triangle has side lengths of $x$ , $x + 4$ , and $14$ units, where $x$ is an integer. Which of the following could be the perimeter of the triangle? Select all that apply.

Geçerli olan tümünü seçin

26

28

30

34

40

Cevabı ve açıklamayı göster

Cevap: $30$ ; $34$ ; $40$

Cevap

The possible perimeters of the triangle are 30, 34, and 40.

The Triangle Inequality Theorem dictates that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the remaining side. For side lengths

x

,

x + 4

, and

14

, setting

x + (x + 4) > 14

yields

2x > 10

, so

x > 5

. Since

x

is specified as an integer,

x

can be any integer greater than or equal to 6. Substituting valid values of

x

into the perimeter formula

P = 2x + 18

gives possible perimeters of 30 (when

x = 6

), 34 (when

x = 8

), and 40 (when

x = 11

). Thus, the values 30, 34, and 40 are all valid perimeters.

Adım Adım Çözüm

1

Apply the Triangle Inequality Theorem

x > 5

According to the Triangle Inequality Theorem, the sum of any two side lengths must be strictly greater than the third side length. Therefore,

x + (x + 4) > 14

, which simplifies to

2x + 4 > 14 \implies 2x > 10 \implies x > 5

. The other two inequality conditions (

(x + 4) + 14 > x

and

x + 14 > x + 4

) are satisfied for all positive values of

x

.

2

Express the perimeter in terms of

x

P = 2x + 18

The perimeter

P

is the sum of the three side lengths:

P = x + (x + 4) + 14 = 2x + 18

.

3

Evaluate the given choices against the valid bounds of

x

Valid integer values of

x \ge 6

correspond to perimeters

P \ge 30

that are even numbers.

Since

x

is an integer and

x > 5

, the minimum integer value for

x

is 6, which yields a minimum perimeter of

P = 2(6) + 18 = 30

. Testing each option:
- For 26:

2x + 18 = 26 \implies x = 4

(Invalid,

x \le 5

)
- For 28:

2x + 18 = 28 \implies x = 5

(Invalid,

x \le 5

)
- For 30:

2x + 18 = 30 \implies x = 6

(Valid)
- For 34:

2x + 18 = 34 \implies x = 8

(Valid)
- For 40:

2x + 18 = 40 \implies x = 11

(Valid)

Anahtar Kavram

Triangle Inequality Theorem and Perimeter Calculation

Soru 31Soru →

In triangle $PQR$ , point $S$ lies on segment $QR$ such that segment $PS$ is perpendicular to $QR$ . The length of altitude $PS$ is $8$ units. If the area of triangle $PQR$ is $56$ square units and the ratio of the area of triangle $PQS$ to the area of triangle $PSR$ is $3 : 4$ , what is the length of side $PQ$ ?

10

4\sqrt{13}

\sqrt{174.25}

4\sqrt{5}

14

Cevabı ve açıklamayı göster

Cevap: $10$

Cevap

10

The area of triangle

PQR

is given as 56 square units and altitude

PS = 8

. Using

\text{Area} = \frac{1}{2} \times QR \times 8

, we find

QR = 14

. Because triangles

PQS

and

PSR

share altitude

PS

, the ratio of their areas equals the ratio of their bases

QS : SR = 3 : 4

. Dividing

QR = 14

into 7 equal parts yields

QS = 6

. In right-angled triangle

PQS

, legs are

PS = 8

and

QS = 6

, giving hypotenuse

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

.

Adım Adım Çözüm

1

Calculate the total length of base

QR

using the area of triangle

PQR

.

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 56 = \frac{1}{2} \times QR \times 8 \implies 56 = 4 \times QR \implies QR = 14

.

The area formula for any triangle relates base, height, and total area.

2

Determine the length of segment

QS

using the given area ratio.

Since triangles

PQS

and

PSR

share the same altitude

PS

, their areas are proportional to their base lengths

QS

and

SR

. Therefore,

QS : SR = 3 : 4

. The total parts are

3 + 4 = 7

. Thus,

QS = 14 \times \frac{3}{7} = 6

.

Triangles sharing a common altitude have areas proportional to their respective bases.

3

Apply the Pythagorean theorem in right triangle

PQS

to find hypotenuse

PQ

.

PQ^2 = PS^2 + QS^2 = 8^2 + 6^2 = 64 + 36 = 100 \implies PQ = 10

.

Segment

PS

is perpendicular to

QR

, forming right-angled triangle

PQS

with legs of length 8 and 6.

Anahtar Kavram

Area of triangles, common altitude area ratio, and the Pythagorean theorem.

Soru 32Soru →

In $\triangle ABC$ , angle $B$ is a right angle, and line segment $BD$ is an altitude drawn to side $AC$ with point $D$ lying on $AC$ . If $AD = 4$ units and $DC = 16$ units, what is the area, in square units, of $\triangle ABC$ ?

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The area of triangle ABC is 80 square units.

In right triangle

ABC

with right angle at

B

, altitude

BD

drawn to hypotenuse

AC

divides the hypotenuse into segments

AD

and

DC

. By the geometric mean theorem,

BD^2 = AD \times DC = 4 \times 16 = 64

, which gives

BD = 8

units. The length of hypotenuse

AC

is

AD + DC = 4 + 16 = 20

units. The area of triangle

ABC

is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8 = 80

square units.

Adım Adım Çözüm

1

Calculate the height (altitude)

BD

of the triangle.

BD = 8

units.

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that

BD^2 = AD \times DC

. Substituting the given values yields

BD^2 = 4 \times 16 = 64

, so

BD = \sqrt{64} = 8

.

2

Calculate the total length of hypotenuse

AC

.

AC = 20

units.

Since point

D

lies on segment

AC

, the total length is the sum of its parts:

AC = AD + DC = 4 + 16 = 20

.

3

Calculate the area of

\triangle ABC

.

Area = 80 square units.

Using the triangle area formula

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

, with base

AC = 20

and altitude

BD = 8

, we get

\text{Area} = \frac{1}{2} \times 20 \times 8 = 80

.

Anahtar Kavram

Geometric mean theorem for right triangle altitude and area of a triangle

Soru 33Soru →

In $\triangle ABC$ , the length of side $AB$ is $13$ units, the length of side $BC$ is $21$ units, and the area of $\triangle ABC$ is $126$ square units. If $\angle ABC$ is an acute angle, what is the length of side $AC$ ?

15

18

20

22

25

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

20

The area of triangle

ABC

is

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

, which yields an altitude

AH = 12

perpendicular to side

BC

. In right triangle

ABH

, the base segment

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

. Because angle

ABC

is acute,

H

falls between

B

and

C

, making

HC = 21 - 5 = 16

. Finally, in right triangle

AHC

,

AC = \sqrt{12^2 + 16^2} = \sqrt{400} = 20

. Therefore, 20 is the correct answer.

Adım Adım Çözüm

1

Calculate the altitude

h

from vertex

A

to base

BC

.

h = 12

units.

The area formula for a triangle is

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

. Substituting

BC = 21

and

\text{Area} = 126

yields

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

.

2

Find the length of segment

BH

where

H

is the foot of the altitude on

BC

.

BH = 5

units.

In right triangle

\triangle ABH

,

AB = 13

and

AH = 12

. By the Pythagorean theorem,

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

.

3

Determine the length of segment

HC

.

HC = 16

units.

Since

\angle ABC

is an acute angle, point

H

lies on segment

BC

. Therefore,

HC = BC - BH = 21 - 5 = 16

.

4

Calculate the length of side

AC

.

AC = 20

units.

In right triangle

\triangle AHC

,

AH = 12

and

HC = 16

. By the Pythagorean theorem,

AC = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20

.

Anahtar Kavram

Triangles: Altitude, Area, and Pythagorean Theorem

Tahmini Süre:1m 30s

Soru 34Soru →

In the coordinate plane, triangle $JKL$ has vertices $J(0,0)$ , $K(14,0)$ , and $L(x, 12)$ , where $x > 0$ . If the perimeter of triangle $JKL$ is $42$ units and the length of side $JL$ is less than the length of side $KL$ , what is the value of $x$ ?

4

5

7

9

12

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5

The correct answer is 5. Using the distance formula, the base length

JK = 14

. Expressing the side lengths as

JL = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + 144}

, setting the perimeter

JK + JL + KL = 42

leads to the quadratic equation

x^2 - 14x + 45 = 0

. This yields

x = 5

or

x = 9

. Evaluating the sides for

x = 5

gives

JL = 13

and

KL = 15

, which satisfies the problem condition

JL < KL

.

Adım Adım Çözüm

1

Calculate the length of base

JK

using the distance formula.

The distance between

J(0,0)

and

K(14,0)

is

14 - 0 = 14

units.

Base

JK

lies along the horizontal

x

-axis.

2

Express side lengths

JL

and

KL

in terms of

x

.

JL = \sqrt{(x-0)^2 + (12-0)^2} = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + (12-0)^2} = \sqrt{(14-x)^2 + 144}

.

Apply the distance formula between coordinates

L(x,12)

and vertices

J

and

K

.

3

Set up and solve the perimeter equation.

14 + \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 42 \implies \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 28

. Squaring both sides systematically yields

x^2 - 14x + 45 = 0

, giving roots

x = 5

and

x = 9

.

The total perimeter is given as 42 units.

4

Apply the constraint

JL < KL

to choose the valid root.

For

x = 5

,

JL = \sqrt{25 + 144} = 13

and

KL = \sqrt{81 + 144} = 15

, satisfying

JL < KL

.

For

x = 9

,

JL = 15

and

KL = 13

, which violates

JL < KL

.

Anahtar Kavram

Coordinate Geometry and Triangle Side Length Constraints

Tahmini Süre:1m 30s

Triangles: Properties, Perimeter, and Area

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