PSAT Math Quiz: Properties Of Right Triangles

20 questions · exam conditions

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Properties Of Right TrianglesQuestion 1 of 20

In the figure shown, triangle $ABC$ is right-angled at $C$ , and $CD$ is the altitude from $C$ to hypotenuse $AB$ . If $AD = 4$ and $DB = 9$ , what is the length of $AC$ ?

$Question graphic$

2\sqrt{13}

6

\sqrt{13}

\sqrt{97}

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PSAT Math Quiz

PSAT Math Quiz: Properties Of Right Triangles

Practice Properties Of Right Triangles in PSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Properties Of Right Triangles, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In the figure shown, triangle $ABC$ is right-angled at $C$ , and $CD$ is the altitude from $C$ to hypotenuse $AB$ . If $AD = 4$ and $DB = 9$ , what is the length of $AC$ ?

$2\sqrt{13}$ (correct answer)
$6$
$\sqrt{13}$
$\sqrt{97}$

Explanation: By the geometric mean (leg) relationship in a right triangle,

AC^2 = AD \cdot AB = 4 \cdot 13 = 52

, so

AC = \sqrt{52} = 2\sqrt{13}

. (B)

6

comes from computing the altitude

CD = \sqrt{AD \cdot DB} = 6

— that's the wrong segment. (C)

\sqrt{13}

results from

\sqrt{AD + DB}

, a misapplication. (D)

\sqrt{97}

comes from

\sqrt{4^2 + 9^2}

, treating

AD

and

DB

as legs.

Question 2

A right triangle has a $30^$ angle, and the side opposite the $30^$ angle is $5$ meters. What is the length of the hypotenuse?

$53$
$10$ (correct answer)
$25$
$15$

Explanation: This involves a 30°-60°-90° special right triangle with specific side ratios. In this triangle type, if the side opposite 30° = x, then the hypotenuse = 2x, and the side opposite 60° = x√3. Since the side opposite 30° is 5 meters, the hypotenuse = 2 × 5 = 10 meters. A common error is confusing which side corresponds to which angle or mixing up the ratios with those of a 45-45-90 triangle. Always remember: in a 30-60-90 triangle, the hypotenuse is exactly twice the shortest side.

Question 3

In the coordinate plane, what is the distance between the points $(-2, 1)$ and $(4, 9)$ ?

$10$ (correct answer)
$100$
$12$
$20$

Explanation: To find the distance between two points, we use the distance formula, which is derived from the Pythagorean theorem: d = √[(x₂-x₁)² + (y₂-y₁)²]. With points (-2, 1) and (4, 9): d = √[(4-(-2))² + (9-1)²] = √[6² + 8²] = √[36 + 64] = √100 = 10. This forms a right triangle with legs of length 6 and 8, another 3-4-5 triple scaled by 2. A common error is forgetting to square the differences before adding them. The distance formula is essentially the Pythagorean theorem applied to coordinate geometry.

Question 4

In right triangle $ABC$ shown, $\angle C = 90°$ , $AC = 7$ , and $\tan A = \frac{24}{7}$ . What is $\cos B$ ?

$\frac{7}{25}$
$\frac{24}{25}$ (correct answer)
$\frac{7}{24}$
$\frac{25}{24}$

Explanation:

\tan A = BC/AC = 24/7

, so

BC = 24

. Then

AB = \sqrt{7^2 + 24^2} = 25

\cos B = BC/AB = 24/25

. (A)

7/25

\sin B

(or

\cos A

). (C)

7/24

\cot A

. (D)

25/24

\sec A

— reciprocal confusion.

Question 5

In the figure, right triangle $ABC$ has the right angle at $C$ . The altitude from $C$ meets $AB$ at $H$ . If $CH = 12$ and $AH - HB = 7$ , what is the length of the hypotenuse $AB$ ?

$25$ (correct answer)
$\sqrt{193}$
$24$
$17$

Explanation: By the geometric-mean altitude relationship:

CH^2 = AH \cdot HB

, so

AH \cdot HB = 144

. Also

AH - HB = 7

. Solving:

AH

and

HB

are roots of

x^2 - (AH+HB)x + 144 = 0

, but easier:

(AH+HB)^2 = (AH-HB)^2 + 4(AH)(HB) = 49 + 576 = 625

, so

AB = AH + HB = 25

. (B)

\sqrt{193} = \sqrt{49+144}

treats them as legs. (C)

24 = 2\cdot12

is a medain-based guess. (D)

17 = 12 + 5

Question 6

A right triangle has legs 2 and 3. What is the length of the hypotenuse?

$\sqrt{5}$
$\sqrt{13}$ (correct answer)
$5$
$13$

Explanation: The question asks for the length of the hypotenuse in a right triangle with legs of lengths 2 and 3. Apply the Pythagorean theorem: the hypotenuse c is given by c = √(a² + b²), where a=2 and b=3. Substitute the values: c = √(2² + 3²) = √(4 + 9) = √13. The result √13 is already in simplest radical form, as 13 has no perfect square factors other than 1. A frequent mistake is adding the legs instead of their squares, like √(2 + 3) = √5, which underestimates the hypotenuse. For quick verification on tests, remember that hypotenuse is always longer than each leg, so √13 (about 3.6) fits between 3 and the expected larger value.

Question 7

A right triangle has legs of lengths $\sqrt{20}$ and $\sqrt{45}$ . What is the length of the hypotenuse in simplest radical form?

$\sqrt{65}$ (correct answer)
$\sqrt{25}$
$5\sqrt{13}$
$\sqrt{130}$

Explanation: The question asks for the length of the hypotenuse in a right triangle with legs

\sqrt{20}

and

\sqrt{45}

, in simplest radical form. Apply the Pythagorean theorem:

c = \sqrt{(\sqrt{20})^2 + (\sqrt{45})^2} = \sqrt{20 + 45} = \sqrt{65}.

Note that

\sqrt{65}

cannot be simplified further as 65 has no perfect square factors other than 1. Simplifying the legs first to

2\sqrt{5}

and

3\sqrt{5}

shows

c = \sqrt{4 \times 5 + 9 \times 5} = \sqrt{65},

confirming. A common mistake is adding the radicals directly without squaring. For radicals, square them early to combine under one root for simplicity.

Question 8

A right triangle has area $48$ square units and one leg of length $12$ . If the right angle is between the legs, what is the length of the other leg?

$4$
$6$
$8$ (correct answer)
$96$

Explanation: The question asks for the length of the other leg in a right triangle with area 48 square units and one leg of 12, with the right angle between the legs. Use the area formula for a right triangle: area = (1/2) × leg1 × leg2, so 48 = (1/2) × 12 × b. Solve for b: 48 = 6b, so b = 8. This gives legs of 12 and 8. A key error is using the full product instead of half, doubling the area. When area is given, set up the equation directly with (1/2) base × height for quick solution.

Question 9

A firefighter is 20 feet from the base of a building and aims a hose at a window 21 feet above the ground. Assuming a right triangle is formed, how far is the firefighter from the window (straight-line distance), in feet?

$29$ (correct answer)
$41$
$\sqrt{41}$
$841$

Explanation: The question asks for the straight-line distance from a firefighter 20 feet from a building to a window 21 feet up. Use the Pythagorean theorem: distance =

\sqrt{20^2 + 21^2}

\sqrt{400 + 441}

\sqrt{841}

= 29 feet. This matches choice A. Errors include adding distances directly or miscalculating squares. Recognize 20-21-29 as a near-triple to verify.

Question 10

A right triangle has legs of lengths $9$ cm and $12$ cm. What is the length of the hypotenuse, in centimeters?

$15$ (correct answer)
$3$
$25$
$1$

Explanation: This question asks for the hypotenuse of a right triangle given the two legs. We use the Pythagorean theorem: a² + b² = c², where a and b are legs and c is the hypotenuse. Substituting the given values: 9² + 12² = 81 + 144 = 225, so c = √225 = 15. A common error is forgetting to take the square root of the sum, which would give 225 instead of 15. When you see a 9-12 right triangle, recognize it as a multiple of the 3-4-5 Pythagorean triple (multiply each by 3).

Question 11

Refer to the figure. $\triangle ABC$ is a right triangle with $\angle C = 90^{\circ}$ . If $AC = 6\text{ cm}$ and $BC = 8\text{ cm}$ , what is the length of $AB$ in centimeters?

10 (correct answer)
9
7
\sqrt{52}

Explanation: Because

\triangle ABC

is right at C, the Pythagorean theorem applies:

AB^2 = AC^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100

, so

AB = 10

. B: 9 is obtained by mistakenly adding 6 and 8 instead of adding their squares.
C: 7 results from subtracting the legs (8 – 6).
D:

\sqrt{52}

comes from computing

6^2 + 4^2

, mis-reading one leg as 4.

Question 12

A ramp rises 3 feet vertically from the ground to a platform. The ramp is 10 feet long. How far, in feet, is the base of the ramp from the point directly below the platform (the horizontal distance)?

$7$
$1$ (correct answer)
$91$
$100$

Explanation: The ramp forms a right triangle with vertical rise = 3 feet and hypotenuse (ramp length) = 10 feet. We need the horizontal distance using the Pythagorean theorem: horizontal² + 3² = 10². This gives us horizontal² + 9 = 100, so horizontal² = 91, and horizontal = √91 feet. A common error is assuming the horizontal distance equals the ramp length minus the vertical rise (10 - 3 = 7). Always draw the right triangle to identify which measurements correspond to which sides.

Question 13

A rectangular garden is $6$ m wide and $8$ m long. A diagonal path runs from one corner to the opposite corner. What is the length of the diagonal path, in meters?

$14$
$10$ (correct answer)
$20$
$100$

Explanation: The diagonal of a rectangle creates a right triangle, so we find its length using the Pythagorean theorem. With width = 6 m and length = 8 m as the legs: 6² + 8² = 36 + 64 = 100, so diagonal = √100 = 10 m. This is another example of the 3-4-5 Pythagorean triple (multiply by 2 to get 6-8-10). A common mistake is adding the sides directly (6 + 8 = 14) instead of using the Pythagorean theorem. When you see dimensions that are multiples of 3-4-5, the calculation becomes much simpler.

Question 14

A $45^\circ$ - $45^\circ$ - $90^\circ$ triangle has hypotenuse length $12\sqrt{2}$ . What is the length of each leg?

$6\sqrt{2}$
$12$ (correct answer)
$24$
$\sqrt{288}$

Explanation: The question asks for the leg length in a 45°-45°-90° triangle with hypotenuse 12√2. The ratio gives leg = hypotenuse / √2 = 12√2 / √2 = 12. This matches choice B. A mistake is halving to 6 or not simplifying radicals. Rationalize by multiplying numerator and denominator by √2.

Question 15

In the right triangle shown, $\angle B = 90°$ , $\sin A = \frac{5}{13}$ , and the perimeter of the triangle is $30$ . What is the length of the hypotenuse?

$12$
$13$ (correct answer)
$\frac{30}{13}$
$\frac{65}{6}$

Explanation: Since

\sin A = 5/13

, the sides are in ratio

5:12:13

(legs and hypotenuse). Let the sides be

5k, 12k, 13k

. The perimeter is

30k = 30

, so

k = 1

, giving a hypotenuse of

13

. (A)

12

is the longer leg. (C) comes from

30 \div 13

, incorrectly applying the ratio. (D) results from using only

5k + 13k = 30

, ignoring the second leg.

Question 16

In the figure, two right triangles share leg $BD$ . Triangle $ABD$ has a right angle at $D$ with $AD = 9$ , and triangle $BDC$ has a right angle at $D$ with $DC = 5$ . If $AB = 15$ , what is the length of $BC$ ?

$13$ (correct answer)
$\sqrt{119}$
$12\sqrt{2}$
$\sqrt{181}$

Explanation: In triangle

ABD

BD = \sqrt{15^2 - 9^2} = \sqrt{144} = 12

. In triangle

BDC

BC = \sqrt{BD^2 + DC^2} = \sqrt{144 + 25} = \sqrt{169} = 13

. (B)

\sqrt{119}

comes from

\sqrt{144-25}

, subtracting instead of adding. (C)

12\sqrt{2}

treats

BD

and

DC

as equal legs. (D)

\sqrt{181}

comes from

\sqrt{15^2 + 5^2 - 9}

, misapplying the given lengths.

Question 17

In the figure, $\triangle ABC$ has a right angle at $C$ . Squares are drawn externally on each of the three sides. If the area of the square on leg $AC$ is $36$ and the area of the square on the hypotenuse $AB$ is $100$ , what is the perimeter of $\triangle ABC$ ?

$24$ (correct answer)
$26$
$20 + 6\sqrt{2}$
$22$

Explanation:

AC = 6

AB = 10

. By the Pythagorean theorem,

BC = \sqrt{100-36} = 8

. Perimeter

= 6 + 8 + 10 = 24

. (B)

26

mistakenly uses

BC = 10

. (C) treats one side as diagonal. (D)

22

uses

BC = 6

twice.

Question 18

The figure shows a 30-60-90 triangle $ABC$ with the right angle at $C$ and $\angle A = 30°$ . A point $D$ is on $AB$ such that $CD \perp AB$ . If $BC = 4$ , what is the length of $AD$ ?

$2\sqrt{3}$
$6$ (correct answer)
$2$
$3\sqrt{3}$

Explanation: With

BC = 4

opposite

30°

, the hypotenuse

AB = 8

and

AC = 4\sqrt{3}

. By the geometric-mean leg relationship,

AC^2 = AD \cdot AB

, so

AD = \frac{(4\sqrt{3})^2}{8} = \frac{48}{8} = 6

. (A)

2\sqrt{3}

is the altitude

CD

. (C)

2

equals

DB

. (D)

3\sqrt{3}

misuses the 30-60-90 ratios.

Question 19

In the figure, right triangle $ABC$ has a right angle at $C$ , with $AC = 8$ and $\angle A = \theta$ . If $\sin\theta + \cos\theta = \frac{7}{5}$ , what is the area of triangle $ABC$ ?

$24$ (correct answer)
$\frac{96}{5}$
$\frac{168}{25}$
$20$

Explanation: From

\sin\theta + \cos\theta = 7/5

, squaring:

1 + 2\sin\theta\cos\theta = 49/25

, so

\sin\theta\cos\theta = 12/25

. Also

\tan\theta = \sin\theta/\cos\theta

. With legs

AC = 8

(adjacent to

\theta

) and

BC

(opposite),

\tan\theta = BC/8

. Using

\sin\theta\cos\theta = \frac{BC \cdot 8}{AB^2} = 12/25

and the Pythagorean identity, one finds

BC = 6

(check: sides 6,8,10 give

\sin+\cos = 6/10+8/10 = 7/5

✓). Area

= (1/2)(8)(6) = 24

. (B) uses

AB = 12

. (C) misapplies

12/25

. (D) uses wrong legs.

Question 20

A right triangle has legs of lengths $4x$ and $3x$ and hypotenuse $25$ . What is the value of $x$ ?

$4$
$5$ (correct answer)
$6$
$7$

Explanation: We're given a right triangle with legs 4x and 3x and hypotenuse 25, and need to find x. Using the Pythagorean theorem: (4x)² + (3x)² = 25². This gives us 16x² + 9x² = 625, so 25x² = 625, and x² = 25, therefore x = 5. We can verify: legs are 4(5) = 20 and 3(5) = 15, and indeed 20² + 15² = 400 + 225 = 625 = 25². This is a scaled version of the 3-4-5 triple. Watch for problems that use variables with Pythagorean triples.