This question asks for the area of a right triangle with base 10 cm and height 7 cm. Use the triangle area formula: Area = ½ × base × height. Substituting the values: Area = ½ × 10 × 7 = ½ × 70 = 35 cm². In a right triangle, the two legs serve as the base and height since they are perpendicular to each other. A common error is forgetting the ½ factor or confusing area with perimeter calculations. For right triangle area problems, you only need the two legs - the hypotenuse is not involved in the area calculation.

This question asks for the unknown leg of a right triangle given the hypotenuse (13 cm) and one leg (5 cm). Use the Pythagorean theorem: a² + b² = c², rearranged to solve for the missing leg: b² = c² - a². Substituting: b² = 13² - 5² = 169 - 25 = 144, so b = √144 = 12 cm. This is the famous 5-12-13 Pythagorean triple. A common error is subtracting incorrectly or forgetting to take the square root of the final result. When given the hypotenuse and one leg, always subtract the leg squared from the hypotenuse squared to find the other leg.

This question asks for the hypotenuse of a right triangle with legs of 15 m and 20 m. Use the Pythagorean theorem: a² + b² = c², where the legs are 15 m and 20 m. Substituting: 15² + 20² = 225 + 400 = 625, so c = √625 = 25 m. This is the 3-4-5 triangle scaled by a factor of 5 (since 15 = 3×5, 20 = 4×5, and 25 = 5×5). A common error is making arithmetic mistakes with larger numbers or not recognizing the scaled Pythagorean triple. Look for patterns and multiples of common triples like 3-4-5 to check your work quickly.

This question asks for the hypotenuse length of a right triangle inscribed in a circle with diameter 10 cm. By Thales' theorem, any triangle inscribed in a circle where one side is a diameter must be a right triangle, and the diameter becomes the hypotenuse. Therefore, the hypotenuse length equals the diameter = 10 cm. The triangle is positioned so that the right angle vertex lies on the circle while the hypotenuse spans the full diameter. A common error is trying to calculate using the Pythagorean theorem without recognizing this geometric property. Remember that when a right triangle is inscribed in a circle, the hypotenuse always equals the circle's diameter.

This question asks for the distance between points A(0, 0) and B(3, 4) on a coordinate plane. Use the distance formula, which is derived from the Pythagorean theorem: d = √[(x₂-x₁)² + (y₂-y₁)²]. Substituting the coordinates: d = √[(3-0)² + (4-0)²] = √[3² + 4²] = √[9 + 16] = √25 = 5 units. This creates a right triangle with legs of length 3 and 4, giving us the familiar 3-4-5 triangle. A common error is forgetting to take the square root or mixing up the coordinate differences. The distance formula is essentially the Pythagorean theorem applied to coordinate geometry.

This question asks for the hypotenuse length of a right triangle with legs of 3 cm and 4 cm. Use the Pythagorean theorem: a² + b² = c², where c is the hypotenuse. Substituting the values: 3² + 4² = 9 + 16 = 25, so c = √25 = 5 cm. This is a classic 3-4-5 right triangle, one of the most common Pythagorean triples. A common error is forgetting to take the square root of the final sum or mixing up which sides are legs versus hypotenuse. Recognizing common Pythagorean triples like 3-4-5 can help you solve these problems quickly on the SAT.

This question asks for the longer leg of a 30-60-90 triangle where the shorter leg is 5 cm. In a 30-60-90 triangle, the sides are in the ratio 1:√3:2, where the shorter leg (opposite the 30° angle) is 1, the longer leg (opposite the 60° angle) is √3, and the hypotenuse is 2. Since the shorter leg is 5 cm, the longer leg = 5√3 cm. You could verify using the Pythagorean theorem, but the special ratio is more efficient. A common error is confusing which leg is shorter or mixing up the 30-60-90 ratios with the 45-45-90 ratios. Memorizing both special triangle ratios is essential for quick problem-solving.

This question asks for the length of a ladder that forms the hypotenuse of a right triangle, with height 8 meters and base 6 meters from the wall. Use the Pythagorean theorem: a² + b² = c², where the legs are 8 m and 6 m. Substituting: 8² + 6² = 64 + 36 = 100, so c = √100 = 10 meters. This represents another common Pythagorean triple: 6-8-10 (which is 2 times the 3-4-5 triple). A common error is confusing which measurement represents which side of the triangle or not recognizing this as a right triangle scenario. Real-world problems often disguise basic geometric relationships, so identify the right triangle first.

This question asks for the hypotenuse of a 45-45-90 triangle where one leg is 6 units. In a 45-45-90 triangle, the sides are in the ratio 1:1:√2, where the legs are equal and the hypotenuse is √2 times the length of each leg. Since one leg is 6 units, the hypotenuse = 6√2 units. You could also verify this using the Pythagorean theorem: 6² + 6² = 36 + 36 = 72, so c = √72 = √(36 × 2) = 6√2. A common error is multiplying by 2 instead of √2 or forgetting the special ratio entirely. Memorizing the 45-45-90 triangle ratio saves time compared to using the Pythagorean theorem every time.

This problem involves finding the side length of a square given its diagonal measurement. A square's diagonal creates two congruent 45-45-90 right triangles, where the sides are in the ratio 1:1:√2 (side:side:diagonal). If the diagonal is 10√2 cm and the diagonal equals side × √2, then side × √2 = 10√2, so the side length is 10 cm.