In ΔABC, angle ∠B is acute, AB = 4, BC = 5. Point M is the midpoint of side AB, point K lies on side BC and BK = 3.

| February 19, 2021 | education

In ΔABC, angle ∠B is acute, AB = 4, BC = 5. Point M is the midpoint of side AB, point K lies on side BC and BK = 3. Find the length of the segment MK if the area of triangle ABC is (7 * √15) / 4 is larger than the area of the MВK triangle.

By condition Sabc = Smbk + (7 * √15) / 4. Angles ABC = MBK = α. MB = AB / 2 = 4/2 = 2. That is, ½ * AB * BC * sinα = ½ * MB * BK * sinα + (7 * √15) / 4. ½ * 4 * 5 * sinα = ½ * 2 * 3 * sinα + (7 * √15) / 4. 40 * sinα = 12 * sinα + 7 * √15. Sinα = √15 / 4. (Sinα) ^ 2 = 15/16. Let’s make trigonometric transformations. 1 – (cosα) ^ 2 = 15/16. (cosα) ^ 2 = 1/16. Since α is an acute angle, cosα> 0, therefore, Cosα = ¼. By the cosine theorem, side MK ^ 2 = MB ^ 2 + BK ^ 2 – 2 * MB * BK * cosα. MK ^ 2 = 2 ^ 2 + 3 ^ 2 – 2 * 2 * 3 * ¼ = 4 + 9 – 3 = 10. MK = √10.



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