Triangles: Properties, Perimeter, and Area Practice Questions — GRE General Test

Question 1Question →

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

Show answer & explanation

Answer: $10\sqrt{13}$

Answer

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}

.

Step-by-Step Solution

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

.

Key Concept

Triangle Area, Altitude Projection, and Extended Pythagorean Theorem

Question 2Question →

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.

Select all that apply

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

.

Show answer & explanation

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

Answer

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.

Step-by-Step Solution

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.

Key Concept

Triangle area formula

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

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

Question 3Question →

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?

Show answer & explanation

Answer: 6

Answer

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

.

Step-by-Step Solution

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.

Key Concept

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

Question 4Question →

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

Select all that apply

4

6

10

14

16

Show answer & explanation

Answer: $6$ ; $10$

Answer

The possible lengths for the third side are

6

centimeters and

10

centimeters.

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

x

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

9 - 5 = 4

) and strictly less than their sum (

9 + 5 = 14

). This establishes the valid range as

4 < x < 14

. The options

6

centimeters and

10

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

Step-by-Step Solution

1

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

The length of the third side,

x

, must satisfy

|9 - 5| < x < 9 + 5

, which simplifies to

4 < x < 14

.

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

2

Check each candidate option against the inequality

4 < x < 14

.

The values

6

and

10

satisfy

4 < x < 14

, whereas

4

,

14

, and

16

do not.

Only numbers strictly within the open interval

(4, 14)

can form a valid triangle with side lengths

5

and

9

.

Key Concept

Triangle Inequality Theorem

Question 5Question →

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

Select all that apply

2\sqrt{37}

16

20

22

\sqrt{274}

Show answer & explanation

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

Answer

The possible lengths of side

AC

are

2\sqrt{37}

and

20

.

Using the triangle area formula

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

, we find

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

. Because

\sin B

is positive in both Quadrant I and Quadrant II, angle

B

can be either acute or obtuse. If angle

B

is acute,

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

, giving

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

. If angle

B

is obtuse,

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

, giving

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

. Both values represent valid triangle configurations.

Step-by-Step Solution

1

Determine the sine of angle

B

using the area formula.

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

.

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

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

.

2

Find the possible values for

\cos B

.

Since

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

,

\cos B

can be either

\frac{3}{5}

(if angle

B

is acute) or

-\frac{3}{5}

(if angle

B

is obtuse).

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

ABC

.

3

Calculate the length of side

AC

when angle

B

is acute.

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

.

Apply the Law of Cosines:

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

with

\cos B = 0.6

.

4

Calculate the length of side

AC

when angle

B

is obtuse.

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

.

Apply the Law of Cosines with

\cos B = -0.6

.

Key Concept

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

Question 6Question →

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

18

39

78

117

234

Show answer & explanation

Answer: $39$

Answer

The area of triangle

PQR

is 39.

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

QS

satisfies

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

, so

QS = 6

. The hypotenuse

PR = PS + SR = 4 + 9 = 13

. Substituting base

13

and height

6

into the triangle area formula

\frac{1}{2}bh

gives

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

.

Step-by-Step Solution

1

Find the length of hypotenuse

PR

PR = PS + SR = 4 + 9 = 13

Point

S

lies on segment

PR

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

2

Calculate altitude

QS

using the geometric mean theorem for right triangles

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

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

3

Calculate the area of triangle

PQR

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

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

Key Concept

Altitude to the hypotenuse in right triangles and triangle area calculation

Alternative Method

Alternatively, use similar triangles

\triangle PQS \sim \triangle QRS

. The ratio of corresponding sides gives

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

, and

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

. Since

\triangle PQR

is a right triangle at

Q

, its area is

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

.

Estimated Time:1m 30s

Question 7Question →

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

Select all that apply

The area of

\triangle ABD

is

36

square units.

The area of

\triangle ABE

is

18

square units.

The area of

\triangle BEC

is

30

square units.

The area of

\triangle ADC

is

30

square units.

The area of

\triangle ECD

is

15

square units.

Show answer & explanation

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

Answer

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

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

Step-by-Step Solution

1

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

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

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

2

Calculate the area of triangle ABE.

Area of triangle ABE = 18 square units.

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

3

Calculate the area of triangle ECD.

Area of triangle ECD = 12 square units.

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

4

Calculate the area of triangle BEC.

Area of triangle BEC = 30 square units.

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

Key Concept

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

Estimated Time:2m 0s

Question 8Question →

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

Show answer & explanation

Answer: 28

Answer

The perimeter of the trapezoid is 28.

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

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

. The area of the original triangle is

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

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

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

, which means the linear scale factor is

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

. Thus, the top triangle has legs of length

5

and a base of length

6

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

10 - 5 = 5

units, and its bottom base is

12

units. Summing these four side lengths gives

5 + 5 + 6 + 12 = 28

.

Step-by-Step Solution

1

Calculate the height and area of the original isosceles triangle

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

6

. The altitude length is

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

. The area of the original triangle is

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

square units.

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

2

Determine the linear scale factor of the smaller top triangle

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

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

. Taking the square root yields a linear scale factor of

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

.

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

3

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

The sides of the smaller triangle are

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

,

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

, and base

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

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

10 - 5 = 5

.

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

4

Calculate the perimeter of the trapezoid

Perimeter =

5 + 5 + 6 + 12 = 28

.

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

Key Concept

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

Question 9Question →

In triangle $ABC$ , the measure of angle $A$ is $40^\circ$ and the measure of angle $B$ is $70^\circ$ . If the perimeter of triangle $ABC$ is $22$ centimeters and the length of side $AB$ is $8$ centimeters, what is the length, in centimeters, of side $BC$ ?

4

6

7

8

14

Show answer & explanation

Answer: $6$

Answer

6 centimeters

First, find the measure of angle

C

:

180^\circ - 40^\circ - 70^\circ = 70^\circ

. Because angle

B

and angle

C

both measure

70^\circ

, triangle

ABC

is isosceles with sides

AC

and

AB

of equal length (

AC = AB = 8

cm). Since the perimeter is the sum of all three sides (

AB + AC + BC = 22

), substituting the known side lengths gives

8 + 8 + BC = 22

, which yields

BC = 6

cm.

Step-by-Step Solution

1

Calculate the measure of the third angle, angle

C

.

Angle

C = 180^\circ - (40^\circ + 70^\circ) = 70^\circ

.

The sum of interior angles in any triangle is

180^\circ

.

2

Determine the relationship between the side lengths using the angle measures.

Side

AC = AB = 8

centimeters.

Since angle

B = 70^\circ

and angle

C = 70^\circ

, triangle

ABC

is isosceles. Sides opposite equal angles are equal in length, so the side opposite angle

C

(which is

AB

) equals the side opposite angle

B

(which is

AC

).

3

Use the perimeter formula to find the length of side

BC

.

Length of side

BC = 22 - (8 + 8) = 6

centimeters.

Perimeter is the total boundary distance (

AB + AC + BC = 22

). Substituting known values gives

8 + 8 + BC = 22

, so

BC = 6

.

Key Concept

Isosceles Triangle Side-Angle Properties and Perimeter

Estimated Time:45s

Question 10Question →

An isosceles triangle has a base of length $10$ inches and a perimeter of $36$ inches. What is the area of the triangle, in square inches?

30

60

65

120

130

Show answer & explanation

Answer: $60$

Answer

60

square inches

Subtracting the base length of

10

inches from the total perimeter of

36

inches leaves

26

inches for the remaining two congruent sides, giving

13

inches each. The perpendicular height divides the isosceles triangle into two right triangles with base

5

inches and hypotenuse

13

inches. By the Pythagorean theorem, the height is

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

inches. Substituting base

10

and height

12

into the area formula

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

gives

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

square inches.

Step-by-Step Solution

1

Determine the length of the two congruent sides of the isosceles triangle.

Each side has a length of

13

inches.

Since the perimeter is

36

inches and the base is

10

inches, the sum of the two equal sides is

36 - 10 = 26

inches. Dividing by

2

yields

13

inches per side.

2

Calculate the perpendicular height of the triangle using the Pythagorean theorem.

The height of the triangle is

12

inches.

Dropping an altitude from the top vertex to the base bisects the base into two segments of

5

inches. This creates two right triangles with a base of

5

inches and hypotenuse of

13

inches. Thus,

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

inches.

3

Calculate the area of the isosceles triangle.

The area is

60

square inches.

Using the triangle area formula

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

, we compute

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

.

Key Concept

Properties of Isosceles Triangles and Area Formula

Question 11Question →

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.

Select all that apply

26

23

44

62

31

Show answer & explanation

Answer: 26; 44

Answer

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.

Step-by-Step Solution

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.

Key Concept

Triangle Inequality Theorem & Algebraic Bounds on Side Lengths

Estimated Time:2m 0s

Question 12Question →

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?

Show answer & explanation

Answer: 54

Answer

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.

Step-by-Step Solution

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

.

Key Concept

Perimeter and area of right triangles using standard side ratios

Question 13Question →

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

Show answer & explanation

Answer: 126

Answer

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

.

Step-by-Step Solution

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.

Key Concept

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

Question 14Question →

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

Show answer & explanation

Answer: 150

Answer

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

.

Step-by-Step Solution

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.

Key Concept

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

Estimated Time:2m 0s

Question 15Question →

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.

Select all that apply

12

6\sqrt{6}

9\sqrt{3}

4\sqrt{15}

8\sqrt{5}

Show answer & explanation

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

Answer

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.

Step-by-Step Solution

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.

Key Concept

Triangle Inequality Theorem and Area calculation via Heron's Formula

Question 16Question →

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 $\triangle ADE$ ?

Show answer & explanation

Answer: 47.25

Answer

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

.

Step-by-Step Solution

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.

Key Concept

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

Question 17Question →

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

Show answer & explanation

Answer: $48$

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

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

Question 18Question →

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

Show answer & explanation

Answer: 36

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Area Ratio of Similar Triangles

Question 19Question →

A non-degenerate triangle has integer side lengths $a$ , $b$ , and $c$ such that $a \le b \le c$ . If the perimeter of the triangle is $30$ and its area is an integer, what is the length of the longest side $c$ ?

Show answer & explanation

Answer: 13

Answer

13

Using Heron's formula with semi-perimeter

s = 15

, the area of the triangle is

\sqrt{15(15-a)(15-b)(15-c)}

. By defining

x = 15-a

,

y = 15-b

, and

z = 15-c

, we have

x+y+z = 15

with

x \ge y \ge z \ge 1

. For the area to be an integer,

15xyz

must be a perfect square. The AM-GM inequality bounds

xyz \le 125

, leaving

xyz = 60

as the only valid case (

k=2

). The system

x+y+z=15

and

xyz=60

has a unique positive integer solution

(10, 3, 2)

, giving side lengths

(5, 12, 13)

. Thus, the longest side length

c

is

13

.

Step-by-Step Solution

1

Find the semi-perimeter of the triangle.

s = \frac{a+b+c}{2} = \frac{30}{2} = 15

Heron's formula requires the semi-perimeter

s

to compute the area of a triangle.

2

Set up Heron's formula for the area.

\text{Area} = \sqrt{15(15-a)(15-b)(15-c)}

Heron's formula relates the side lengths of a triangle directly to its area.

3

Define variables for the differences from the semi-perimeter.

Let

x = 15-a

,

y = 15-b

, and

z = 15-c

. Then

x+y+z = 15

and

x \ge y \ge z \ge 1

.

Because

a \le b \le c

, their complements with respect to

15

satisfy

x \ge y \ge z

.

4

Determine the condition for the area to be an integer.

\text{Area} = \sqrt{15xyz}

must be an integer, so

15xyz = m^2

for some integer

m

, implying

xyz = 15k^2

for an integer

k \ge 1

.

For

\sqrt{15xyz}

to be an integer,

15xyz

must be a perfect square containing

15

as a factor.

5

Bound

xyz

using the AM-GM inequality.

xyz \le \left(\frac{x+y+z}{3}\right)^3 = 5^3 = 125 \implies 15k^2 \le 125 \implies k=1 \text{ or } k=2

The AM-GM inequality provides a maximum upper bound for the product of three positive numbers with a fixed sum.

6

Evaluate the possible values of

k

.

If

k=1

,

xyz = 15

, which has no integer solutions for

x+y+z=15

. If

k=2

,

xyz = 60

, which yields

(x, y, z) = (10, 3, 2)

.

Solving the system

x+y+z=15

and

xyz=60

under

x \ge y \ge z \ge 1

gives

z=2

,

y=3

,

x=10

uniquely.

7

Calculate the side lengths and identify the longest side

c

.

a = 5

,

b = 12

,

c = 13

, so the longest side length is

13

.

Since

c = 15 - z = 15 - 2 = 13

, the longest side of the right triangle

(5, 12, 13)

is

13

.

Key Concept

Heron's formula, integer partitioning, and the AM-GM inequality for triangle area optimization

Question 20Question →

The side lengths of triangle $ABC$ , in units, are given by $AB = x$ , $BC = x + 2$ , and $CA = x + 4$ , where $x$ is a positive integer. If the perimeter of triangle $ABC$ is strictly less than $36$ , how many possible integer values of $x$ exist such that triangle $ABC$ is an acute triangle?

2

3

4

5

7

Show answer & explanation

Answer: 3

Answer

There are 3 possible integer values of

x

.

To form an acute triangle with side lengths

x

,

x+2

, and

x+4

, three conditions must be met: the perimeter bound (

3x+6 < 36 \implies x < 10

), the non-degeneracy condition (

x + x + 2 > x + 4 \implies x > 2

), and the acute angle condition (

(x+4)^2 < x^2 + (x+2)^2 \implies x^2 - 4x - 12 > 0 \implies x > 6

). Combining these bounds restricts

x

to integer values in the range

6 < x < 10

, giving exactly three valid integer values:

7, 8,

and

9

.

Step-by-Step Solution

1

Apply the perimeter constraint to find an upper bound for

x

.

The perimeter is

P = x + (x + 2) + (x + 4) = 3x + 6

. Given

3x + 6 < 36

, subtracting

6

yields

3x < 30

, so

x < 10

.

The total length around the triangle must be strictly less than

36

.

2

Apply the triangle inequality theorem to find a lower bound for non-degenerate triangles.

AB + BC > CA \implies x + (x + 2) > x + 4 \implies 2x + 2 > x + 4 \implies x > 2

.

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

3

Formulate and solve the acute triangle condition.

For an acute triangle with longest side

CA = x + 4

, we require

CA^2 < AB^2 + BC^2

.

(x + 4)^2 < x^2 + (x + 2)^2

x^2 + 8x + 16 < x^2 + x^2 + 4x + 4

x^2 + 8x + 16 < 2x^2 + 4x + 4

0 < x^2 - 4x - 12

(x - 6)(x + 2) > 0

Since

x

is a positive integer,

x + 2 > 0

, which implies

x - 6 > 0

, so

x > 6

.

In any acute triangle, the square of the longest side is strictly less than the sum of the squares of the remaining two sides.

4

Combine the bounds to determine the number of valid integer values for

x

.

Combining

x > 6

and

x < 10

gives integer values

x \in \{7, 8, 9\}

. There are

3

such values.

Only integer values in the open interval

(6, 10)

satisfy all conditions simultaneously.

Key Concept

Acute Triangle Criteria and Triangle Inequality

Triangles: Properties, Perimeter, and Area

Answer

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