Properties of Right Triangles - PSAT Math

1. 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)?
   Answer: $1$ 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.
2. A triangular sign is a right triangle with legs 9 inches and 14 inches. What is the area of the sign, in square inches?
   Answer: $63$ The question asks for the area of a right triangular sign with legs 9 inches and 14 inches. Area = (1/2) × 9 × 14 = (1/2) × 126 = 63 square inches. This matches choice B. A key error is omitting the 1/2, giving 126, or using hypotenuse. Use legs as base and height for right triangles.
3. 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?
   Answer: $8$ 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.
4. A $45^\circ$-$45^\circ$-$90^\circ$ triangle has hypotenuse length $12\sqrt{2}$. What is the length of each leg?
   Answer: $12$ 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.
5. A right triangle has legs of lengths $9$ cm and $12$ cm. What is the length of the hypotenuse, in centimeters?
   Answer: $15$ 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).
6. A ramp rises 3 feet vertically over a horizontal distance of 10 feet, forming a right triangle. What is the length of the ramp, in feet?
   Answer: $\sqrt{109}$ The question asks for the length of a ramp that rises 3 feet vertically over 10 feet horizontally. Apply the Pythagorean theorem to find the hypotenuse of this right triangle. The length l = $\sqrt{3^2 + 10^2}$ = $\sqrt{9 + 100}$ = $\sqrt{109}$ feet. This matches choice A. Errors often occur by adding instead of squaring, giving 13, or swapping legs and hypotenuse. Visualize the ramp as the hypotenuse to confirm the calculation setup.
7. A grayscale diagram shows a right triangle with hypotenuse $c=17$ and one leg $a=8$. The right angle is marked between the two legs. What is the length of the other leg?
   Answer: $15$ The question asks for the length of the missing leg in a right triangle with hypotenuse 17 and one leg 8. Use the Pythagorean theorem rearranged to find the leg: b = $\sqrt{c^2 - a^2}$. Here, b = $\sqrt{17^2 - 8^2}$ = $\sqrt{289 - 64}$ = $\sqrt{225}$ = 15. This matches choice B. A common mistake is adding squares instead of subtracting, leading to $\sqrt{353}$, or confusing hypotenuse with a leg. Check if the sides satisfy 8-15-17 as a Pythagorean triple for verification.
8. A right triangle has a 30^ angle, and the side opposite the 30^ angle is $5$ meters. What is the length of the hypotenuse?
   Answer: $10$ 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.
9. A grayscale diagram shows a right triangle with legs labeled $6$ and $8$. A second step asks for the area of the triangle. What is the area, in square units?
   Answer: $24$ The question asks for the area of a right triangle with legs $6$ and $8$. Use the area formula for a right triangle: area = $\frac{1}{2} \times \text{leg1} \times \text{leg2} = \frac{1}{2} \times 6 \times 8 = \frac{1}{2} \times 48 = 24$ square units. The legs are the base and height. This is straightforward multiplication halved. Key errors include forgetting to halve or using hypotenuse in the formula. When legs are given, apply the $\frac{1}{2}$ base $\times$ height directly without needing the hypotenuse.
10. A $30^\circ$-$60^\circ$-$90^\circ$ triangle has a longer leg of length $12\sqrt{3}$. What is the length of the hypotenuse?
    Answer: $24$ The question asks for the length of the hypotenuse in a 30°-60°-90° triangle where the longer leg is 12√3. Use the special ratios for a 30°-60°-90° triangle, where the sides are in the ratio x : x√3 : 2x, with the longer leg opposite the 60° angle being x√3. Here, x√3 = 12√3, so divide both sides by √3 to find x = 12. The hypotenuse is then 2x = 2 × 12 = 24. A key error is confusing the longer leg with the hypotenuse, leading to multiplying by √3 instead of finding 2x. For special triangles, memorize the ratios to quickly identify the hypotenuse as twice the shorter leg.
11. A rectangular screen is 15 inches wide and 20 inches tall. What is the length of the diagonal of the screen, in inches?
    Answer: $25$ The question asks for the length of the diagonal of a rectangular screen that is 15 inches wide and 20 inches tall. Treat the diagonal as the hypotenuse of a right triangle with legs 15 and 20, using Pythagorean theorem: d = √(15² + 20²) = √(225 + 400) = √625 = 25 inches. This is a scaled 3-4-5 triangle. The width and height form the legs. Errors involve adding dimensions without squaring or confusing with area. For rectangles, remember the diagonal formula √(l² + w²) for direct application.
12. A grayscale diagram shows a right triangle with legs labeled $5\text{ in}$ and $12\text{ in}$. What is the area of the triangle, in square inches?
    Answer: $30$ The question asks for the area of a right triangle with legs 5 inches and 12 inches. The area of a right triangle is (1/2) × leg1 × leg2, using the legs as base and height. Thus, area = (1/2) × 5 × 12 = (1/2) × 60 = 30 square inches. This matches choice B. A common error is forgetting the 1/2 factor, yielding 60, or using the hypotenuse incorrectly. Always identify the legs for area calculations in right triangles to avoid mistakes.
13. 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 $BC$?
    Answer: $13$ 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.
14. A right triangle has legs of lengths $\sqrt{20}$ and $\sqrt{45}$. What is the length of the hypotenuse in simplest radical form?
    Answer: $\sqrt{65}$ 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.
15. In the coordinate plane, what is the distance between the points $(-2, 1)$ and $(4, 9)$?
    Answer: $10$ 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.
16. A 20-foot ladder leans against a vertical wall. The base of the ladder is 12 feet from the wall on level ground. How high up the wall does the ladder reach, in feet?
    Answer: $16$ We need to find how high the ladder reaches on the wall, which forms a right triangle with the ladder as hypotenuse. Using the Pythagorean theorem with hypotenuse = 20 feet and base = 12 feet: 12² + h² = 20². This gives us 144 + h² = 400, so h² = 256, and h = √256 = 16 feet. A common mistake is treating the 20-foot ladder length as a leg instead of the hypotenuse. Always identify which side is the hypotenuse (the longest side, opposite the right angle) before applying the formula.
17. A right triangle has legs of lengths $4x$ and $3x$ and hypotenuse $25$. What is the value of $x$?
    Answer: $5$ 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.
18. A grayscale diagram shows a square with side length $10\text{ cm}$. A diagonal is drawn, forming a right triangle. What is the length of the diagonal?
    Answer: $10\sqrt{2}\text{ cm}$ The question asks for the length of the diagonal in a square with side 10 cm. The diagonal forms a $45^\circ$-$45^\circ$-$90^\circ$ triangle, where diagonal = side $\times \sqrt{2}$. Thus, diagonal = $10 \times \sqrt{2} = 10\sqrt{2}$ cm. This matches choice A. Errors include multiplying by 2 to get 20 or incorrect radicals like $\sqrt{2000}$. Use the special ratio for quick computation in squares.
19. 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?
    Answer: 10 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.
20. 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 $AD$?
    Answer: $6$ 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.

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