A perpendicular AH of length 1.4 is drawn to the plane of an isosceles triangle ABC

| January 27, 2021 | education

A perpendicular AH of length 1.4 is drawn to the plane of an isosceles triangle ABC with sides AB = BC = 5, AC = 8. Find the distance from point H to side BC of the triangle.

The triangle is isosceles, all sides are known. We find its area according to Heron’s formula, we first find the half-perimeter of the triangle:
p = (AB + BC + AC) / 2 = 9.
S ABC = √ (p * (p – AB) * (p – BC) * (p – AC) = √ (9 * 4 * 4 * 1) = √144 = 12.
From the top A we draw the height of the AK to the side of the BC.
The side of the BC is known by condition, the area was found in the course of the solution, We use the formula for the area through the side and the height and find the height AH:
S ABC = 1/2 * BC * AK → AK = 2 * S ABC / BC = 2 * 12/5 = 4.8.
Consider a right-angled triangle NAC, we find the hypotenuse of NK – the distance from H to BC.
HK = √ (AH² + AK²) = √ (1.96 + 23.04) = √25 = 5.
Answer: the distance from point H to the BC side is 5.



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