In an acute-angled triangle ABC, the perpendiculars to the sides AC and BC meet at point D, DC = 5 cm.

| December 1, 2020 | education

In an acute-angled triangle ABC, the perpendiculars to the sides AC and BC meet at point D, DC = 5 cm. Find the distances from point D to the sides of the triangle if the perimeter of triangle ABC is 18 cm, AB: BC: AC = 4: 3: 2.

Since AB: BC: AC = 4: 3: 2 and the perimeter of triangle ABC is 18 cm, we can compose ur-th:
4x + 3x + 2x = 18
9x = 19
x = 2
Then AB = 2 * 4 = 8, BC = 2 * 3 = 6, AC = 2 * 2 = 4.
The mid-perpendicular to the AC side divides its side into two equal parts. equal to 4/2 = 2 cm.Let us consider the tr-ik СDN (DN is the perpendicular to the AC side): By m Pythagoras we find DN: DN = √ (5 ^ 2 – 2 ^ 2) = √21.
From the tr-ka ADN by T of Pythagoras we find AD, AD = 5. In the tr-ik ADM (DM is the middle perpendicular to the side AB 🙂 according to T Pythagoras we find DM: DM = 3
Answer: 3√21



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