Digital SAT Math Practice Question #1169
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Recall the properties of an altitude drawn to the hypotenuse in a right triangle. Specifically, how does the altitude relate to the segments it divides the hypotenuse into?
Consider using both the geometric mean theorem (AD² = BD ⋅ DC) and the Pythagorean theorem in one of the smaller right triangles formed by the altitude (e.g., ΔADB) to set up a system of equations.
⚡ **Backsolving / Plugging in the Answers:** This problem can be efficiently solved by testing the given options. Start with the middle options or the proposed answer. Let's test Option A: BD = 45. 1. **Calculate AD²:** Using the altitude rule AD² = BD ⋅ DC, we get AD² = 45 ⋅ 15 = 675. 2. **Calculate AB²:** Using the Pythagorean theorem in ΔADB, AB² = AD² + BD² = 675 + 45² = 675 + 2025 = 2700. 3. **Check the condition AB = 2 ⋅ AD:** * AD = √675 = √(225 ⋅ 3) = 15√3 * AB = √2700 = √(900 ⋅ 3) = 30√3 * Is AB = 2 ⋅ AD? 30√3 = 2 ⋅ (15√3) => 30√3 = 30√3. Yes, it is! Since Option A satisfies all conditions, it is the correct answer. If it didn't work, you would proceed to test other options. ⚡ **Special Right Triangle Recognition (Advanced):** If you recognize that the condition AB = 2 ⋅ AD implies that triangle ADB is a 30-60-90 right triangle (where AD is the side opposite the 30-degree angle, and AB is the hypotenuse), then: 1. Angle B = 30 degrees. 2. In a 30-60-90 triangle, the sides are in the ratio k : k√3 : 2k. So, if AD = k, then BD = k√3 and AB = 2k. 3. Substitute these into the altitude rule AD² = BD ⋅ DC: k² = (k√3) ⋅ 15 Divide by k (since k ≠ 0): k = 15√3 4. Now find BD: BD = k√3 = (15√3) ⋅ √3 = 15 ⋅ 3 = 45. This method is very fast if you spot the special triangle relationship.
To solve this problem, we need to utilize the properties of right triangles and altitudes drawn to the hypotenuse. 1. **Identify Similar Triangles:** In a right triangle ABC with a right angle at A, when an altitude AD is drawn to the hypot...
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