PSAT Math Quiz: Properties Of Right Triangles

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.

Question 1

1. $2\sqrt{13}$ (correct answer)
2. $6$
3. $\sqrt{13}$
4. $\sqrt{97}$

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

1. 53
2. $10$ (correct answer)
3. 25
4. $15$

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

1. $10$ (correct answer)
2. 100
3. 12
4. 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

1. $\frac{7}{25}$
2. $\frac{24}{25}$ (correct answer)
3. $\frac{7}{24}$
4. $\frac{25}{24}$

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) $7/25$ is $\sin B$ (or $\cos A$). (C) $7/24$ is $\cot A$. (D) $25/24$ is $\sec A$ — reciprocal confusion.

Question 5

1. $25$ (correct answer)
2. $\sqrt{193}$
3. $24$
4. $17$

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

1. $\sqrt{5}$
2. $\sqrt{13}$ (correct answer)
3. $5$
4. $13$

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

1. $\sqrt{65}$ (correct answer)
2. $\sqrt{25}$
3. $5\sqrt{13}$
4. $\sqrt{130}$

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

1. $4$
2. $6$
3. $8$ (correct answer)
4. $96$

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

1. $29$ (correct answer)
2. $41$
3. $\sqrt{41}$
4. $841$

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

1. $15$ (correct answer)
2. 3
3. 25
4. 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

1. 10 (correct answer)
2. 9
3. 7
4. \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

1. $7$
2. 1 (correct answer)
3. 91
4. 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

1. $14$
2. $10$ (correct answer)
3. 20
4. 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

1. $6\sqrt{2}$
2. $12$ (correct answer)
3. $24$
4. $\sqrt{288}$

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

1. $12$
2. $13$ (correct answer)
3. $\frac{30}{13}$
4. $\frac{65}{6}$

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

1. $13$ (correct answer)
2. $\sqrt{119}$
3. $12\sqrt{2}$
4. $\sqrt{181}$

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

1. $24$ (correct answer)
2. $26$
3. $20 + 6\sqrt{2}$
4. $22$

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

1. $2\sqrt{3}$
2. $6$ (correct answer)
3. $2$
4. $3\sqrt{3}$

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

1. $24$ (correct answer)
2. $\frac{96}{5}$
3. $\frac{168}{25}$
4. $20$

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

1. $4$
2. $5$ (correct answer)
3. $6$
4. $7$

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.