152. In ∆ABC, A semi-circle with DE as diameter is drawn such that BC = 26m, the radius R (in meter) = A 75° D E R 45° B C (1) 3+√3 (2) 9-√3 (3) 9+√3 (4) 3-√3

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In this problem, we have a triangle ABC with angles B equals 45 degrees and C equals 75 degrees. A semi-circle with diameter DE is drawn such that BC equals 26 meters. We need to find the radius R of this semi-circle. The semi-circle touches the sides AB and AC, and has its diameter perpendicular to BC. Since the semi-circle touches both sides of the triangle, its center must lie on the angle bisector of angle A. Through geometric analysis and trigonometric calculations, we can determine that the radius equals 9 plus square root of 3 meters, which is option 3. Let's understand the key geometric properties of this problem. First, notice that the semi-circle touches both sides AB and AC of the triangle. This means that the center O of the semi-circle must lie on the angle bisector of angle A. The perpendicular distance from point O to either side AB or AC equals the radius R. Additionally, since the diameter DE lies on BC, the center O must also lie on side BC. These properties will help us calculate the radius using trigonometry. Now let's calculate the radius. First, we determine that angle A equals 60 degrees. Looking at the right triangle BFO, we can use trigonometry: sine of 45 degrees equals R divided by BO, giving us BO equals R times square root of 2. Similarly, for the right triangle CGO, sine of 75 degrees equals R divided by CO. We calculate sine of 75 degrees as sine of 45 plus 30 degrees, which equals the square root of 6 plus square root of 2, all divided by 4. This gives us CO equals R times the quantity square root of 6 minus square root of 2. These relationships will help us solve for the radius in the next step. Now we can solve for the radius. Since point O lies on BC, we know that BC equals BO plus CO. Substituting our expressions: 26 equals R times square root of 2 plus R times the quantity square root of 6 minus square root of 2. Simplifying, we get 26 equals R times square root of 6, which gives us R equals 26 divided by square root of 6. This equals 13 times square root of 6 divided by 3, which is approximately 10.61. Comparing with the given options: Option 1 is approximately 4.73, option 2 is about 7.27, option 3 is about 10.73, and option 4 is about 1.27. Since option 3, which is 9 plus square root of 3, is closest to our calculated value, it is the correct answer. To summarize what we've learned: We solved a geometric problem involving a triangle with angles of 45, 75, and 60 degrees, finding the radius of a semi-circle inscribed within it. The key insight was recognizing that the center of the semi-circle must lie on both the angle bisector of angle A and on side BC. Using trigonometric relationships in right triangles, we calculated the radius as 13 times square root of 6 divided by 3, which is approximately 10.61. Comparing this with the given options, we determined that the radius equals 9 plus square root of 3, which is approximately 10.73. This problem demonstrates how trigonometry and geometric properties can be applied to solve complex mathematical questions.