Pythagorean Theorem and Special Right Triangles Practice Questions — ACT

A square tabletop has a diagonal length of $8$ feet. What is the area of the tabletop, in square feet?

16

16\sqrt{2}

32

32\sqrt{2}

64

Show answer & explanation

Answer: 32

Answer

32

The correct answer is 32. The diagonal of a square divides it into two congruent 45-45-90 right triangles, where the diagonal is the hypotenuse. The ratio of the diagonal to the side length in a 45-45-90 triangle is

\sqrt{2}

to

1

. Therefore, a square with a diagonal of

8

feet has a side length of

s = \frac{8}{\sqrt{2}}

feet. The area of the square is

s^2 = \left(\frac{8}{\sqrt{2}}\right)^2 = \frac{64}{2} = 32

square feet.

Step-by-Step Solution

1

Relate the diagonal of a square to its side length using special right triangles.

The diagonal of a square splits the square into two 45-45-90 right triangles. The hypotenuse of these triangles is the diagonal,

8

feet, and the legs are the sides of the square,

s

. The relationship is

s\sqrt{2} = 8

.

In a 45-45-90 triangle, the hypotenuse is

\sqrt{2}

times the length of a leg.

2

Solve for the side length

s

of the square tabletop.

s = \frac{8}{\sqrt{2}}

feet.

Divide both sides of the equation by

\sqrt{2}

to isolate the side length

s

.

3

Calculate the area of the square tabletop.

Area

= s^2 = \left(\frac{8}{\sqrt{2}}\right)^2 = \frac{64}{2} = 32

square feet.

The area of a square is calculated by squaring its side length.

Key Concept

Using the properties of 45-45-90 special right triangles to find side lengths and area from the diagonal of a square.

Question 2Question →

A right triangle has two legs of equal length. If the hypotenuse of the triangle is $10\sqrt{2}$ centimeters, what is the length, in centimeters, of one of the legs?

Show answer & explanation

Answer: 10

Answer

The length of one of the legs is 10 centimeters.

An isosceles right triangle possesses acute angles of

45^\circ

and side ratios of

x : x : x\sqrt{2}

, where

x

represents the leg length. Given a hypotenuse of

10\sqrt{2}

centimeters, we equate

x\sqrt{2} = 10\sqrt{2}

. Dividing both sides of the equation by

\sqrt{2}

isolates the leg length, giving

x = 10

centimeters.

Step-by-Step Solution

1

Determine the triangle type from the given properties.

The triangle is a

45^\circ

-

45^\circ

-

90^\circ

special right triangle (isosceles right triangle).

A right triangle with two legs of equal length must have acute angles measuring

45^\circ

each, making it an isosceles right triangle.

2

Set up an equation utilizing the ratios of the side lengths.

Let

x

be the leg length. The hypotenuse length is represented by

x\sqrt{2} = 10\sqrt{2}

centimeters.

The hypotenuse of a

45^\circ

-

45^\circ

-

90^\circ

special right triangle is always

\sqrt{2}

times the length of one of its legs.

3

Solve the equation for the variable

x

.

x = 10

Dividing both sides of the equation by

\sqrt{2}

isolates the variable

x

representing the leg length.

Key Concept

Properties of

45^\circ

-

45^\circ

-

90^\circ

special right triangles.

Alternative Method

Alternatively, you can apply the Pythagorean Theorem:

a^2 + b^2 = c^2

. Since both legs are equal in length, we can set

a = b = x

. This yields the equation

x^2 + x^2 = (10\sqrt{2})^2

. Simplifying both sides gives

2x^2 = 100 \times 2 = 200

. Dividing by 2 yields

x^2 = 100

, and taking the square root of both sides gives

x = 10

centimeters.

Estimated Time:45s

Question 3Question →

A rectangular garden has a straight walking path that connects two opposite corners. The length of the path is $170$ meters, and the width of the garden is $80$ meters. What is the length, in meters, of the garden?

90

120

150

188

250

Show answer & explanation

Answer: 150

Answer

The length of the garden is

150

meters.

The diagonal path, width, and length of the rectangular garden form a right triangle where the path is the hypotenuse. According to the Pythagorean theorem, the square of the length plus the square of the width equals the square of the path:

\text{length}^2 + 80^2 = 170^2

. This simplifies to

\text{length}^2 + 6,400 = 28,900

. Subtracting

6,400

from both sides gives

\text{length}^2 = 22,500

. Taking the square root of

22,500

yields

150

meters.

Step-by-Step Solution

1

Identify the right triangle formed by the length, width, and diagonal path of the garden.

The width (

80

meters) and the unknown length are the legs, while the diagonal path (

170

meters) is the hypotenuse.

The diagonal of a rectangle forms two congruent right triangles with the rectangle's sides.

2

Set up the Pythagorean equation to solve for the unknown leg.

\text{length}^2 + 80^2 = 170^2

The Pythagorean theorem states that

a^2 + b^2 = c^2

, where

c

is the hypotenuse.

3

Calculate the squares of the known lengths.

80^2 = 6,400

and

170^2 = 28,900

Evaluate the exponents to simplify the equation.

4

Isolate the squared unknown variable.

\text{length}^2 = 28,900 - 6,400 = 22,500

Subtract

6,400

from both sides of the equation.

5

Take the square root of both sides to find the length.

\text{length} = \sqrt{22,500} = 150

The square root operation reverses the squaring of the variable.

Key Concept

Applying the Pythagorean theorem to find the length of an unknown leg in a right triangle.

Estimated Time:1m 0s

Question 4Question →

In a right triangle, the length of the side opposite the $60^\circ$ angle is $6\sqrt{3}$ centimeters. What is the length, in centimeters, of the hypotenuse of this triangle?

Show answer & explanation

Answer: 12

Answer

The length of the hypotenuse is

12

centimeters.

In a

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

special right triangle, the sides opposite the

30^\circ

angle, the

60^\circ

angle, and the

90^\circ

(hypotenuse) angle are in the ratio

x : x\sqrt{3} : 2x

. Given that the side opposite the

60^\circ

angle is

6\sqrt{3}

centimeters, we have

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

, which means

x = 6

. The hypotenuse is

2x = 2(6) = 12

centimeters.

Step-by-Step Solution

1

Determine the type of special right triangle.

A

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

right triangle.

Since the triangle is a right triangle and has a

60^\circ

angle, the remaining angle must be

180^\circ - 90^\circ - 60^\circ = 30^\circ

.

2

Set up the relation for the side lengths using the ratio of a

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

triangle.

The side opposite the

60^\circ

angle is

x\sqrt{3}

centimeters, where

x

is the length of the side opposite the

30^\circ

angle.

In any

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

triangle, the side lengths are in the ratio

1 : \sqrt{3} : 2

.

3

Solve for the base variable

x

.

x = 6

We are given that the side opposite the

60^\circ

angle is

6\sqrt{3}

centimeters, so

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

.

4

Calculate the length of the hypotenuse.

The hypotenuse is

2x = 2(6) = 12

centimeters.

The hypotenuse of a

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

triangle is twice the length of the shorter leg, which is

2x

.

Key Concept

Using the side length ratios of a

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

special right triangle to find missing lengths.

Question 5Question →

In right triangle $ABC$ , the measure of angle $B$ is $90^\circ$ and the measure of angle $A$ is $45^\circ$ . If the length of leg $AB$ is $8$ inches, what is the length, in inches, of the hypotenuse $AC$ ?

8

4\sqrt{2}

8\sqrt{2}

8\sqrt{3}

16

Show answer & explanation

Answer: $8\sqrt{2}$

Answer

The length of the hypotenuse is

8\sqrt{2}

inches.

In right triangle

ABC

, the angle measures are

45^\circ

,

45^\circ

, and

90^\circ

. The lengths of the sides of a

45^\circ

-

45^\circ

-

90^\circ

triangle are in the ratio

1 : 1 : \sqrt{2}

. Since the leg is

8

inches, the hypotenuse is

8\sqrt{2}

inches.

Step-by-Step Solution

1

Identify the type of right triangle.

Since angle

B = 90^\circ

and angle

A = 45^\circ

, angle

C

must also be

45^\circ

. This is a

45^\circ

-

45^\circ

-

90^\circ

special right triangle.

The sum of the angles in a triangle is always

180^\circ

.

2

Recall the ratio of the side lengths of a

45^\circ

-

45^\circ

-

90^\circ

triangle.

The ratio of the sides opposite the angles

45^\circ : 45^\circ : 90^\circ

is

1 : 1 : \sqrt{2}

. Thus, the hypotenuse is equal to

\text{leg} \times \sqrt{2}

.

This is a standard geometric property of isosceles right triangles.

3

Calculate the length of the hypotenuse.

Multiply the leg length of

8

inches by

\sqrt{2}

to get

8\sqrt{2}

inches.

The leg adjacent to the

45^\circ

angle is given as

8

inches.

Key Concept

Hypotenuse of a

45^\circ

-

45^\circ

-

90^\circ

special right triangle

Alternative Method

Alternatively, use the Pythagorean theorem:

AB^2 + BC^2 = AC^2

. Since it is an isosceles right triangle,

BC = AB = 8

. Thus,

8^2 + 8^2 = AC^2 \implies 64 + 64 = AC^2 \implies AC = \sqrt{128} = 8\sqrt{2}

.

Estimated Time:45s

Question 6Question →

A regular hexagon $ABCDEF$ has a side length of $8$ centimeters. A point $P$ lies on the side $CD$ such that the ratio of the length of $CP$ to the length of $PD$ is $1:3$ . What is the length, in centimeters, of the segment $AP$ ?

Show answer & explanation

Answer: 14

Answer

The length of the segment

AP

is

14

centimeters.

The correct answer is

14

. Dropping a perpendicular from

P

to the main diagonal

AD

creates a

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

triangle

\triangle PHD

with hypotenuse

PD = 6

. The legs are

DH = 3

and

PH = 3\sqrt{3}

. This leaves

AH = 13

. Applying the Pythagorean Theorem to the right triangle

\triangle AHP

with legs

13

and

3\sqrt{3}

yields

AP = \sqrt{13^2 + (3\sqrt{3})^2} = 14

.

Step-by-Step Solution

1

Determine the length of the main diagonal

AD

of the regular hexagon.

AD = 16

cm

In a regular hexagon with side length

s

, the main diagonal connecting opposite vertices has a length of

2s

. Given

s = 8

, we find

AD = 2 \times 8 = 16

.

2

Calculate the length of the segment

PD

on the side

CD

.

PD = 6

cm

The point

P

divides the side

CD

of length

8

in the ratio

CP:PD = 1:3

. Thus,

PD = \frac{3}{1+3} \times 8 = 6

.

3

Identify the angles and type of triangle formed by dropping a perpendicular from

P

to diagonal

AD

.

\triangle PHD

is a

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

right triangle.

The diagonal

AD

bisects the interior angle

\angle CDE = 120^\circ

of the regular hexagon, making

\angle ADC = 60^\circ

. Since

PH \perp AD

, the triangle

\triangle PHD

has angles

90^\circ

,

60^\circ

, and

30^\circ

.

4

Find the lengths of the legs

DH

and

PH

of the special right triangle

\triangle PHD

.

DH = 3

cm and

PH = 3\sqrt{3}

cm

Using the ratios of a

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

triangle with hypotenuse

PD = 6

, the leg adjacent to the

60^\circ

angle is

DH = 6 \cos(60^\circ) = 3

, and the leg opposite to the

60^\circ

angle is

PH = 6 \sin(60^\circ) = 3\sqrt{3}

.

5

Calculate the length of the segment

AH

.

AH = 13

cm

Since

H

lies on the diagonal

AD

, we subtract the length of

DH

from the total length of the diagonal:

AH = AD - DH = 16 - 3 = 13

.

6

Apply the Pythagorean Theorem to the right triangle

\triangle AHP

to find the length of

AP

.

AP = 14

cm

In the right triangle

\triangle AHP

with legs

AH = 13

and

PH = 3\sqrt{3}

, the hypotenuse is

AP = \sqrt{AH^2 + PH^2} = \sqrt{13^2 + (3\sqrt{3})^2} = \sqrt{169 + 27} = \sqrt{196} = 14

.

Key Concept

Applying special right triangle ratios and the Pythagorean Theorem in multi-step geometric figures.

Question 7Question →

In right triangle $ABC$ , the hypotenuse $AC$ has a length of $15$ centimeters, and leg $AB$ has a length of $9$ centimeters. What is the length, in centimeters, of leg $BC$ ?

Show answer & explanation

Answer: 12

Answer

The length of leg

BC

is

12

centimeters.

The length of leg

BC

is found using the Pythagorean Theorem,

AB^2 + BC^2 = AC^2

. Substituting the given values gives

9^2 + BC^2 = 15^2

, which simplifies to

81 + BC^2 = 225

. Subtracting 81 from both sides yields

BC^2 = 144

. Taking the square root of 144 gives the correct length of 12 centimeters.

Step-by-Step Solution

1

Identify the given dimensions and apply the Pythagorean Theorem.

AB^2 + BC^2 = AC^2

For any right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

2

Substitute the known values

AB = 9

and

AC = 15

into the equation.

9^2 + BC^2 = 15^2

The hypotenuse

AC

is the side opposite the right angle, and

AB

is one of the legs.

3

Simplify the squared terms.

81 + BC^2 = 225

Squaring 9 yields 81, and squaring 15 yields 225.

4

Isolate the unknown term by subtracting 81 from both sides.

BC^2 = 144

Subtracting 81 from both sides isolates

BC^2

on the left side of the equation.

5

Take the square root of both sides to solve for the leg length.

BC = 12

Taking the square root of 144 gives the side length, which must be positive.

Key Concept

Pythagorean Theorem

Question 8Question →

In acute triangle $PQR$ , an altitude $PS$ is drawn from vertex $P$ perpendicular to side $QR$ at point $S$ . The measure of $\angle PQS$ is $60^\circ$ , the length of segment $PQ$ is $24$ , and the length of segment $PR$ is $39$ . What is the length of side $QR$ ?

Show answer & explanation

Answer: 45

Answer

The length of side

QR

is

45

.

The altitude divides the acute triangle into two right triangles. In the first right triangle,

\triangle PQS

, the angles are

30^\circ

,

60^\circ

, and

90^\circ

, with a hypotenuse of

24

. This makes the adjacent leg

QS = 12

and the shared altitude

PS = 12\sqrt{3}

. In the second right triangle,

\triangle PRS

, the hypotenuse is

39

and one leg is

12\sqrt{3}

. Using the Pythagorean Theorem, we find the other leg

SR = \sqrt{39^2 - (12\sqrt{3})^2} = \sqrt{1521 - 432} = \sqrt{1089} = 33

. Summing the two segments gives the total length of side

QR = 12 + 33 = 45

.

Step-by-Step Solution

1

Identify the two right triangles formed by the altitude.

The altitude

PS

divides

\triangle PQR

into two adjacent right triangles:

\triangle PQS

and

\triangle PRS

, which share the side

PS

.

Establishing these right triangles allows us to apply right-triangle trigonometric ratios and the Pythagorean Theorem.

2

Use the properties of the

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle

\triangle PQS

to find

QS

and

PS

.

QS = 12

and

PS = 12\sqrt{3}

.

In a

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle, the leg opposite the

30^\circ

angle is half the hypotenuse, and the leg opposite the

60^\circ

angle is

\sqrt{3}

times the shorter leg. Here, hypotenuse

PQ = 24

, so

QS = 12

and

PS = 12\sqrt{3}

.

3

Apply the Pythagorean Theorem to

\triangle PRS

to find

SR

.

SR = 33

.

In right triangle

\triangle PRS

, the hypotenuse is

PR = 39

. By the Pythagorean Theorem,

PS^2 + SR^2 = PR^2

. Squaring the sides gives

(12\sqrt{3})^2 + SR^2 = 39^2 \implies 432 + SR^2 = 1521

. Solving for

SR

gives

SR^2 = 1089 \implies SR = 33

.

4

Sum the segments

QS

and

SR

to find the total length of

QR

.

QR = 45

.

Because

\triangle PQR

is an acute triangle, the altitude

PS

lands at a point

S

on the segment

QR

, meaning

QR = QS + SR

. Adding the lengths gives

12 + 33 = 45

.

Key Concept

Applying properties of

30^\circ\text{-}60^\circ\text{-}90^\circ

special right triangles and the Pythagorean Theorem in multi-step geometric figures.

Estimated Time:2m 30s

Question 9Question →

In right triangle $ABC$ with $\angle B = 90^\circ$ and $\angle A = 30^\circ$ , the hypotenuse $AC$ has a length of $12$ centimeters. An altitude $BD$ is drawn from vertex $B$ to hypotenuse $AC$ . From point $D$ , a perpendicular segment $DE$ is drawn to side $AB$ , with point $E$ lying on $AB$ . What is the length, in centimeters, of segment $EC$ ?

3\sqrt{7}

\frac{3\sqrt{22}}{2}

\frac{3\sqrt{19}}{2}

\frac{3\sqrt{13}}{2}

\sqrt{39}

Show answer & explanation

Answer: $\frac{3\sqrt{19}}{2}$

Answer

The length of segment EC is \frac{3\sqrt{19}}{2} centimeters.

To find the length of segment EC, we can construct the right triangle EBC with a right angle at B. By using the properties of 30-60-90 right triangles, we determine the side lengths of the triangles in the figure: first finding BC = 6 and AB = 6\sqrt{3} in triangle ABC; then finding AD = 9 in triangle ABD; then finding AE = \frac{9\sqrt{3}}{2} in triangle ADE; and finally finding EB = AB - AE = \frac{3\sqrt{3}}{2}. Applying the Pythagorean theorem to right triangle EBC yields EC = \sqrt{(\frac{3\sqrt{3}}{2})^2 + 6^2} = \frac{3\sqrt{19}}{2}.

Step-by-Step Solution

1

Determine the side lengths of the main right triangle ABC. Since angle A = 30 degrees and angle B = 90 degrees, triangle ABC is a 30-60-90 right triangle. With hypotenuse AC = 12, the leg opposite the 30-degree angle is BC = \frac{12}{2} = 6, and the leg opposite the 60-degree angle is AB = 6\sqrt{3}.

BC = 6 and AB = 6\sqrt{3}

Knowing the side lengths of triangle ABC is necessary to find the dimensions of the smaller inscribed triangles.

2

Find the length of segment AD in right triangle ABD. Altitude BD is perpendicular to AC, making triangle ABD a right triangle with right angle ADB. Since angle A = 30 degrees, triangle ABD is also a 30-60-90 right triangle with hypotenuse AB = 6\sqrt{3}. The side adjacent to the 30-degree angle, AD, is given by AB \times \cos(30^\circ) = 6\sqrt{3} \times \frac{\sqrt{3}}{2} = 9.

AD = 9

Determining AD allows us to analyze the smaller right triangle ADE built on it.

3

Find the lengths of segments AE and EB. In right triangle ADE (where DE is perpendicular to AB), the hypotenuse is AD = 9 and angle A = 30 degrees. The side adjacent to the 30-degree angle is AE = AD \times \cos(30^\circ) = 9 \times \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}. Segment EB is then found by subtracting AE from AB: EB = AB - AE = 6\sqrt{3} - \frac{9\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}.

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

We need the length of segment EB to apply the Pythagorean theorem in the final right triangle EBC.

4

Apply the Pythagorean theorem to right triangle EBC. Since line segment AB is perpendicular to BC, angle EBC is a right angle. In right triangle EBC, the legs are EB = \frac{3\sqrt{3}}{2} and BC = 6. The hypotenuse EC is calculated as EC = \sqrt{EB^2 + BC^2} = \sqrt{(\frac{3\sqrt{3}}{2})^2 + 6^2} = \sqrt{\frac{27}{4} + 36} = \sqrt{\frac{171}{4}} = \frac{3\sqrt{19}}{2}.

EC = \frac{3\sqrt{19}}{2}

Applying the Pythagorean theorem to the legs EB and BC gives the length of the hypotenuse EC.

Key Concept

Applying 30-60-90 right triangle properties and the Pythagorean theorem across multiple connected geometric figures.

Estimated Time:3m 0s

Question 10Question →

A vertical flagpole casts a horizontal shadow on the ground. The distance from the top of the flagpole to the tip of the shadow is $20$ feet. If the length of the shadow is $16$ feet, what is the height, in feet, of the flagpole?

12

16

25.6

36

4

Show answer & explanation

Answer: 12

Answer

12

The flagpole, ground, and line from the top of the pole to the tip of the shadow form a right triangle. The diagonal distance of

20

feet represents the hypotenuse, and the shadow length of

16

feet represents one of the legs. Using the Pythagorean theorem (

a^2 + b^2 = c^2

), we set up the equation

a^2 + 16^2 = 20^2

, which simplifies to

a^2 + 256 = 400

. Subtracting

256

from both sides gives

a^2 = 144

. Taking the square root of both sides gives the height of the flagpole as

12

feet.

Step-by-Step Solution

1

Identify the parts of the right triangle formed by the flagpole, ground, and the line from the top of the flagpole to the shadow's tip.

The hypotenuse (

c

) is

20

feet, and one leg (

b

) is

16

feet.

The flagpole is vertical and the ground is horizontal, forming a right angle. The distance from the top of the pole to the tip of the shadow is the diagonal (hypotenuse).

2

Apply the Pythagorean theorem

a^2 + b^2 = c^2

to find the missing leg (

a

).

a^2 + 16^2 = 20^2

, which simplifies to

a^2 + 256 = 400

.

The Pythagorean theorem relates the lengths of the sides of a right triangle.

3

Solve for the unknown height

a

by subtracting and taking the square root.

a^2 = 144 \implies a = 12

feet.

Isolating

a^2

gives

144

, and taking the square root of

144

gives the height of the flagpole.

Key Concept

Pythagorean Theorem

Question 11Question →

In a right triangle, the measure of one of the acute angles is $30^\circ$ . If the side opposite this $30^\circ$ angle has a length of $6.5$ inches, what is the length, in inches, of the hypotenuse?

Show answer & explanation

Answer: 13

Answer

The length of the hypotenuse is

13

inches.

In a

30^\circ

-

60^\circ

-

90^\circ

special right triangle, the length of the hypotenuse is exactly twice the length of the shorter leg (the side opposite the

30^\circ

angle). Given that the shorter leg has a length of

6.5

inches, the hypotenuse has a length of

2 \times 6.5 = 13

inches.

Step-by-Step Solution

1

Determine the relationship between the given side and the hypotenuse using special right triangle properties.

The triangle is a

30^\circ

-

60^\circ

-

90^\circ

right triangle, meaning the hypotenuse is twice the length of the shorter leg.

By geometric theorem, the sides of a

30^\circ

-

60^\circ

-

90^\circ

triangle are in the ratio

1 : \sqrt{3} : 2

, with the shortest side opposite the

30^\circ

angle and the longest side being the hypotenuse.

2

Multiply the length of the side opposite the

30^\circ

angle by 2.

6.5 \text{ inches} \times 2 = 13 \text{ inches}

Doubling the shorter leg length of

6.5

inches gives the length of the hypotenuse.

Key Concept

In a

30^\circ

-

60^\circ

-

90^\circ

special right triangle, the length of the hypotenuse is always twice the length of the shorter leg (the side opposite the

30^\circ

angle).

Question 12Question →

An equilateral triangle $ABC$ has a side length of $12$ inches. An altitude $AD$ is drawn from vertex $A$ to the side $BC$ . A point $P$ lies on the segment $AD$ such that $\triangle BPC$ is a right triangle with a right angle at $P$ . What is the length, in inches, of the segment $AP$ ?

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

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

6\sqrt{3} - 6

6\sqrt{3} + 6

12\sqrt{3} - 6

Show answer & explanation

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

Answer

The correct answer is

6\sqrt{3} - 6

inches.

The correct answer is

6\sqrt{3} - 6

. Since the side length of the equilateral triangle is

12

, the altitude

AD

splits it into two

30^\circ

-

60^\circ

-

90^\circ

right triangles with base

BD = 6

and altitude

AD = 6\sqrt{3}

. The right triangle

\triangle BPC

has

\angle BPC = 90^\circ

and

PB = PC

, making it an isosceles right triangle. The altitude

PD

splits

\triangle BPC

into two

45^\circ

-

45^\circ

-

90^\circ

right triangles, so

PD = BD = 6

. The length of

AP

is found by subtracting

PD

from

AD

, yielding

6\sqrt{3} - 6

.

Step-by-Step Solution

1

Find the length of the altitude

AD

using the properties of the

30^\circ

-

60^\circ

-

90^\circ

triangle

\triangle ABD

.

The length of

AD

is

6\sqrt{3}

inches.

Since

\triangle ABC

is equilateral with side length

12

inches, the altitude

AD

bisects the base

BC

, making

BD = 6

inches. The altitude splits the equilateral triangle into two

30^\circ

-

60^\circ

-

90^\circ

right triangles. The length of the longer leg is the shorter leg multiplied by

\sqrt{3}

, which gives

AD = 6\sqrt{3}

.

2

Find the length of the segment

PD

using the properties of the

45^\circ

-

45^\circ

-

90^\circ

triangle

\triangle PDB

.

The length of

PD

is

6

inches.

Since

P

lies on the altitude

AD

, which is the axis of symmetry,

\triangle BPC

is an isosceles right triangle with

\angle BPC = 90^\circ

. The altitude

PD

is perpendicular to

BC

and bisects

\angle BPC

, forming two

45^\circ

-

45^\circ

-

90^\circ

right triangles:

\triangle PDB

and

\triangle PDC

. In a

45^\circ

-

45^\circ

-

90^\circ

triangle, the two legs are congruent, so

PD = BD = 6

.

3

Subtract the length of

PD

from the length of

AD

to find the length of segment

AP

.

AP = 6\sqrt{3} - 6

inches.

Since point

P

lies on segment

AD

, the length of

AP

is the difference between the total altitude

AD

and the segment

PD

.

Key Concept

Properties of special right triangles (30-60-90 and 45-45-90) and their multi-step application in geometry.

Estimated Time:2m 0s

Question 13Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $30^\circ$ , and the length of leg $BC$ is $12$ . An altitude $BD$ is drawn perpendicular to the hypotenuse $AC$ . Let $E$ be the midpoint of the altitude $BD$ . A line passing through $E$ is perpendicular to $BD$ and intersects the leg $AB$ at $G$ and the leg $BC$ at $F$ . What is the length of segment $GF$ ?

Show answer & explanation

Answer: 12

Answer

The length of segment

GF

is

12

.

The correct answer is

12

. By analyzing the geometric properties of the 30-60-90 right triangle

ABC

, the altitude

BD

is found to be

6\sqrt{3}

, making the half-segment

BE = 3\sqrt{3}

. The perpendicular line at

E

creates two smaller 30-60-90 right triangles,

\triangle BEF

and

\triangle BEG

. Solving for the legs along the line gives

EF = 3

and

EG = 9

, which sum to

12

.

Step-by-Step Solution

1

Find the length of altitude

BD

in right triangle

ABC

.

BD = 6\sqrt{3}

In right triangle

ABC

, we have

\angle B = 90^\circ

,

\angle A = 30^\circ

, and

\angle C = 60^\circ

. The altitude

BD

forms a smaller 30-60-90 right triangle

BCD

with hypotenuse

BC = 12

. Since

BD

is opposite the

60^\circ

angle

\angle C

, we have

BD = BC \sin(60^\circ) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}

.

2

Calculate the length of segment

BE

.

BE = 3\sqrt{3}

Since

E

is the midpoint of the altitude

BD

, we divide the length of

BD

by 2:

BE = \frac{6\sqrt{3}}{2} = 3\sqrt{3}

.

3

Determine the length of segment

EF

in right triangle

BEF

.

EF = 3

Since the line

GF

is perpendicular to

BD

,

\angle BEF = 90^\circ

. In right triangle

BCD

, we have

\angle DBC = 30^\circ

, which means

\angle EBF = 30^\circ

. This makes

\triangle BEF

a 30-60-90 right triangle where

BE

is adjacent to the

30^\circ

angle and

EF

is opposite to it. Thus,

EF = \frac{BE}{\sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3

.

4

Determine the length of segment

EG

in right triangle

BEG

.

EG = 9

Since

\angle BEG = 90^\circ

and

\angle ABD = 90^\circ - \angle DBC = 60^\circ

, the angle

\angle EBG = 60^\circ

. This makes

\triangle BEG

a 30-60-90 right triangle where

BE

is adjacent to the

60^\circ

angle and

EG

is opposite to it. Thus,

EG = BE \sqrt{3} = 3\sqrt{3} \times \sqrt{3} = 9

.

5

Calculate the total length of segment

GF

.

GF = 12

Since

G

,

E

, and

F

are collinear and

E

lies between

G

and

F

, the length of segment

GF

is the sum of

EG

and

EF

:

GF = 9 + 3 = 12

.

Key Concept

Using properties of 30-60-90 special right triangles to find segment lengths in complex geometric configurations.

Question 14Question →

In right triangle $DEF$ , the measure of $\angle E$ is $90^\circ$ and the measure of $\angle D$ is $60^\circ$ . If the hypotenuse $DF$ has a length of $14$ centimeters, what is the length, in centimeters, of the segment $DE$ ?

7\sqrt{2}

7\sqrt{3}

7

14\sqrt{3}

28

Show answer & explanation

Answer: $7$

Answer

The length of the segment

DE

is

7

centimeters.

The correct option is the one with the value

7

. In right triangle

DEF

, the angles are

90^\circ

,

60^\circ

, and

30^\circ

, making it a special

30^\circ

-

60^\circ

-

90^\circ

right triangle. The side

DE

is opposite the

30^\circ

angle (the shorter leg). By the properties of a

30^\circ

-

60^\circ

-

90^\circ

triangle, the shorter leg is half the length of the hypotenuse. Thus,

DE = 14 / 2 = 7

centimeters.

Step-by-Step Solution

1

Determine the measure of the third angle,

\angle F

.

\angle F = 30^\circ

The sum of angles in a triangle is

180^\circ

. Since

\angle E = 90^\circ

and

\angle D = 60^\circ

, we calculate

\angle F = 180^\circ - 90^\circ - 60^\circ = 30^\circ

.

2

Identify the relationship between the sides of the

30^\circ

-

60^\circ

-

90^\circ

triangle.

DE

is the shorter leg, opposite

\angle F

(

30^\circ

).

The side opposite the

30^\circ

angle is the shorter leg, which is half the length of the hypotenuse.

3

Calculate the length of

DE

.

DE = 7

centimeters

Since the hypotenuse

DF = 14

centimeters, the shorter leg

DE

is

14 / 2 = 7

centimeters.

Key Concept

In a

30^\circ

-

60^\circ

-

90^\circ

right triangle, the lengths of the sides are in the ratio

1 : \sqrt{3} : 2

. The shorter leg (opposite the

30^\circ

angle) is half the length of the hypotenuse.

Estimated Time:1m 0s

Question 15Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $60^\circ$ , and the hypotenuse $AC$ has a length of $20$ centimeters. Point $D$ lies on leg $BC$ such that the length of segment $BD$ is $2\sqrt{3}$ centimeters. A line segment $DE$ is drawn perpendicular to $AC$ such that $E$ lies on $AC$ . What is the length, in centimeters, of segment $AE$ ?

Show answer & explanation

Answer: 8

Answer

8

The correct answer is 8. By solving for the angles and side lengths of the two nested

30^\circ\text{-}60^\circ\text{-}90^\circ

right triangles, we find that the segment

BC

is

10\sqrt{3}

cm, making

DC = 8\sqrt{3}

cm. Using the ratio of sides for the smaller right triangle

DEC

, we find

EC = 12

cm, which leaves

AE = 20 - 12 = 8

cm.

Step-by-Step Solution

1

Determine the third angle of right triangle

ABC

.

\angle C = 30^\circ

The sum of angles in a triangle is

180^\circ

. Since

\angle B = 90^\circ

and

\angle A = 60^\circ

, we have

\angle C = 180^\circ - 90^\circ - 60^\circ = 30^\circ

.

2

Calculate the length of the side

BC

.

BC = 10\sqrt{3}

cm

In the

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle

ABC

, the side

BC

is opposite the

60^\circ

angle, so its length is the hypotenuse

AC

multiplied by

\sin(60^\circ)

or

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

. Thus,

BC = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}

.

3

Find the length of segment

DC

.

DC = 8\sqrt{3}

cm

Since point

D

lies on segment

BC

, the length of

DC

is the total length of

BC

minus the length of

BD

. Since

BD = 2\sqrt{3}

, we have

DC = 10\sqrt{3} - 2\sqrt{3} = 8\sqrt{3}

.

4

Determine the properties of the right triangle

DEC

.

\triangle DEC

is a

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle with hypotenuse

DC = 8\sqrt{3}

cm.

Since segment

DE

is perpendicular to

AC

,

\angle DEC = 90^\circ

. Triangle

DEC

shares the angle

\angle C = 30^\circ

with triangle

ABC

, which makes it a

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle where

DC

is the hypotenuse.

5

Calculate the length of segment

EC

.

EC = 12

cm

In the

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle

DEC

, the leg

EC

is adjacent to the

30^\circ

angle, so its length is the hypotenuse

DC

multiplied by

\cos(30^\circ)

or

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

. Thus,

EC = 8\sqrt{3} \times \frac{\sqrt{3}}{2} = 12

.

6

Calculate the length of segment

AE

.

AE = 8

cm

Since point

E

lies on segment

AC

, we can find

AE

by subtracting

EC

from

AC

. Thus,

AE = AC - EC = 20 - 12 = 8

.

Key Concept

Using the properties of

30^\circ\text{-}60^\circ\text{-}90^\circ

special right triangles to find missing lengths in composite geometric configurations.

Question 16Question →

A right triangle has legs of length $5$ inches and $12$ inches. What is the length, in inches, of the hypotenuse of this triangle?

Show answer & explanation

Answer: 13

Answer

The length of the hypotenuse is 13 inches.

The Pythagorean theorem states that in any right triangle with legs

a

and

b

and hypotenuse

c

,

a^2 + b^2 = c^2

. Substituting the given values:

5^2 + 12^2 = 25 + 144 = 169

. Taking the square root of

169

gives the hypotenuse length of

13

.

Step-by-Step Solution

1

Identify the lengths of the two legs.

a = 5

,

b = 12

These are the given side lengths perpendicular to each other.

2

Apply the Pythagorean theorem

a^2 + b^2 = c^2

.

5^2 + 12^2 = 25 + 144 = 169

To find the square of the hypotenuse.

3

Solve for the hypotenuse

c

by taking the square root.

c = \sqrt{169} = 13

To find the side length of the hypotenuse.

Key Concept

Pythagorean Theorem

Question 17Question →

A $13$ -foot ladder is leaning against a flat vertical wall. The base of the ladder is placed $5$ feet away from the bottom of the wall. How many feet up the wall does the ladder reach?

\sqrt{8}

8

12

\sqrt{194}

18

Show answer & explanation

Answer: $12$

Answer

The height the ladder reaches is

12

feet.

The correct answer is

12

feet. The ladder, wall, and ground form a right triangle where the ladder is the hypotenuse (

13

feet) and the distance along the ground is one leg (

5

feet). By the Pythagorean theorem, the height up the wall,

b

, satisfies

5^2 + b^2 = 13^2

. Solving for

b

gives

b^2 = 169 - 25 = 144

, so

b = 12

.

Step-by-Step Solution

1

Identify the hypotenuse and the given leg from the word problem description.

The ladder length is the hypotenuse (

c = 13

), and the distance from the wall is one of the legs (

a = 5

).

The ladder forms the diagonal side opposite the right angle formed by the vertical wall and the ground.

2

Set up the Pythagorean theorem to find the unknown leg length.

5^2 + b^2 = 13^2

.

The Pythagorean theorem (

a^2 + b^2 = c^2

) relates the side lengths of any right triangle.

3

Solve for the unknown leg length

b

by simplifying terms and taking the square root.

25 + b^2 = 169 \implies b^2 = 144 \implies b = \sqrt{144} = 12

.

Subtracting the square of the known leg from the square of the hypotenuse isolates the squared unknown leg, which can then be solved by finding its square root.

Key Concept

Using the Pythagorean theorem to find an unknown leg of a right triangle when the hypotenuse and one leg are known.

Alternative Method

Recognizing that

5

and

13

are part of the common Pythagorean triple

5

-

12

-

13

allows you to immediately identify the missing leg as

12

without performing calculations.

Estimated Time:45s

Question 18Question →

In the standard $(x,y)$ coordinate plane, a line segment $OQ$ connects the origin $O(0,0)$ to a point $Q$ in the first quadrant. The segment $OQ$ has a length of $10$ units and makes an angle of $60^\circ$ with the positive $x$ -axis. An isosceles right triangle $\triangle OQR$ is constructed such that the right angle is at $Q$ , the leg $QR$ has a length of $10$ units, and point $R$ lies in the first quadrant. What are the coordinates of point $R$ ?

(5\sqrt{3} - 5, 5\sqrt{3} + 5)

(5\sqrt{3} + 5, 5 - 5\sqrt{3})

(5 + 5\sqrt{3}, 5\sqrt{3} - 5)

(20 - 10\sqrt{3}, 10)

(10 + 10\sqrt{3}, 10\sqrt{3} - 10)

Show answer & explanation

Answer: $(5 + 5\sqrt{3}, 5\sqrt{3} - 5)$

Answer

The coordinates

(5 + 5\sqrt{3}, 5\sqrt{3} - 5)

The correct answer is

(5 + 5\sqrt{3}, 5\sqrt{3} - 5)

because constructing two helper

30^\circ\text{-}60^\circ\text{-}90^\circ

right triangles allows us to determine both the coordinates of

Q

as

(5, 5\sqrt{3})

and the horizontal and vertical shifts to

R

as

+5\sqrt{3}

and

-5

respectively.

Step-by-Step Solution

1

Project point

Q

onto the

x

-axis to form a

30^\circ\text{-}60^\circ\text{-}90^\circ

right triangle.

The horizontal leg is

10 \cos(60^\circ) = 5

and the vertical leg is

10 \sin(60^\circ) = 5\sqrt{3}

.

The hypotenuse

OQ

has a length of

10

and makes a

60^\circ

angle with the positive

x

-axis.

2

Determine the coordinates of point

Q

.

Q = (5, 5\sqrt{3})

.

Point

Q

is in the first quadrant, so both coordinates are positive.

3

Determine the orientation of segment

QR

.

Segment

QR

must make a

30^\circ

angle below the horizontal line passing through

Q

(going down and to the right).

Since

\triangle OQR

is a right isosceles triangle with the right angle at

Q

,

QR

is perpendicular to

OQ

and has length

10

. To keep

R

in the first quadrant,

QR

must rotate clockwise from

OQ

by

90^\circ

.

4

Construct a helper

30^\circ\text{-}60^\circ\text{-}90^\circ

right triangle under

QR

to find the changes in

x

and

y

.

The horizontal change is

+10 \cos(30^\circ) = +5\sqrt{3}

and the vertical change is

-10 \sin(30^\circ) = -5

.

The hypotenuse of this triangle is

QR = 10

, and the angle with the horizontal is

30^\circ

.

5

Calculate the coordinates of

R

by applying the changes to the coordinates of

Q

.

R = (5 + 5\sqrt{3}, 5\sqrt{3} - 5)

.

Add the horizontal change to

x_Q

and the vertical change to

y_Q

.

Key Concept

Solving coordinate geometry problems using

30^\circ\text{-}60^\circ\text{-}90^\circ

special right triangles.

Question 19Question →

A rectangular piece of paper $ABCD$ has dimensions $AB = 12$ inches and $BC = 9$ inches. The paper is folded so that vertex $A$ falls directly on vertex $C$ , creating a crease $EF$ where $E$ lies on $AB$ and $F$ lies on $CD$ . What is the length of the crease $EF$ , in inches?

Show answer & explanation

Answer: 11.25

Answer

The length of the crease is 11.25 inches.

By interpreting the fold geometrically, we establish that the crease is the perpendicular bisector of the diagonal. We can solve for the segments along the side using a right triangle formed at the corner, and then construct a second right triangle using an altitude to find the length of the crease hypotenuse.

Step-by-Step Solution

1

Identify the relationship created by the fold.

The crease

EF

is the perpendicular bisector of the diagonal

AC

, meaning

AE = EC = x

.

When a point is folded onto another, the crease line acts as the perpendicular bisector of the segment connecting the two points.

2

Set up an expression for the remaining part of the side

AB

.

Since

AB = 12

and

AE = x

, the length of

EB = 12 - x

.

The point

E

lies on segment

AB

, dividing it into

AE

and

EB

.

3

Use the Pythagorean Theorem in right triangle

EBC

to solve for

x

.

x^2 = (12 - x)^2 + 9^2 \implies x^2 = 144 - 24x + x^2 + 81 \implies 24x = 225 \implies x = 9.375

.

The triangle

EBC

is a right triangle with legs

EB

and

BC

, and hypotenuse

EC

.

4

Form a second right triangle to find the length of the crease

EF

.

Draw

FG \perp AB

with

G

on

AB

. This forms right triangle

EGF

with legs

FG = 9

and

EG = AB - EB - DF = 12 - 2.625 - 2.625 = 6.75

.

Constructing an altitude from

F

to

AB

allows us to create a right triangle that has the crease

EF

as its hypotenuse.

5

Apply the Pythagorean Theorem to right triangle

EGF

to calculate the final length of

EF

.

EF = \sqrt{6.75^2 + 9^2} = \sqrt{45.5625 + 81} = \sqrt{126.5625} = 11.25

inches.

The hypotenuse of right triangle

EGF

represents the length of the crease.

Key Concept

Applying the Pythagorean Theorem to geometric folds and multi-step right triangle relationships

Question 20Question →

In the right trapezoid $ABCD$ below, $AB$ is parallel to $CD$ , and the measures of $\angle A$ and $\angle D$ are both $90^\circ$ . The length of $CD$ is $7$ , the length of $BC$ is $8$ , and the measure of $\angle B$ is $60^\circ$ . What is the length of the diagonal $BD$ ?

Show answer & explanation

Answer: 13

Answer

13

By drawing altitude

CE

perpendicular to

AB

, we form rectangle

AECD

and right triangle

\triangle CEB

. Since

\angle B = 60^\circ

,

\triangle CEB

is a

30^\circ

-

60^\circ

-

90^\circ

triangle with hypotenuse

BC = 8

. The side opposite

30^\circ

is

BE = 8/2 = 4

, and the side opposite

60^\circ

is

CE = 4\sqrt{3}

. Since opposite sides of rectangle

AECD

are equal, we find

AD = CE = 4\sqrt{3}

and

AE = CD = 7

. Thus,

AB = AE + BE = 7 + 4 = 11

. Finally, we apply the Pythagorean Theorem to right triangle

\triangle DAB

:

BD^2 = AD^2 + AB^2 = (4\sqrt{3})^2 + 11^2 = 48 + 121 = 169

, yielding

BD = 13

.

Step-by-Step Solution

1

Decompose the trapezoid by drawing an altitude from

C

perpendicular to

AB

, meeting it at

E

.

This forms a rectangle

AECD

and a right triangle

CEB

.

Decomposing the figure allows us to use right triangle trigonometry and parallel line relationships to determine missing side lengths.

2

Calculate the lengths of the legs of right triangle

CEB

.

BE = 4

and

CE = 4\sqrt{3}

.

Since

\angle B = 60^\circ

,

\triangle CEB

is a

30^\circ

-

60^\circ

-

90^\circ

triangle. The shorter leg

BE

is half the hypotenuse

BC

, and the longer leg

CE

is the shorter leg times

\sqrt{3}

.

3

Determine the lengths of

AD

and

AB

.

AD = 4\sqrt{3}

and

AB = 11

.

In the rectangle

AECD

, opposite sides are equal, so

AD = CE = 4\sqrt{3}

and

AE = CD = 7

. Thus, the base

AB = AE + BE = 7 + 4 = 11

.

4

Use the Pythagorean Theorem in right triangle

DAB

to solve for

BD

.

BD = 13

.

In right triangle

DAB

, the legs are

AD = 4\sqrt{3}

and

AB = 11

. The hypotenuse

BD

satisfies

BD^2 = AD^2 + AB^2 = (4\sqrt{3})^2 + 11^2 = 48 + 121 = 169

, so

BD = \sqrt{169} = 13

.

Key Concept

Solving multi-step geometry problems by decomposing shapes into rectangles and special right triangles (

30^\circ

-

60^\circ

-

90^\circ

), then applying the Pythagorean Theorem.

Pythagorean Theorem and Special Right Triangles

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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Answer

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