In an isosceles triangle with a lateral side equal to 4, the median is drawn to the lateral side.

In an isosceles triangle with a lateral side equal to 4, the median is drawn to the lateral side. find the base of the triangle if the median is 3

Let ABC be a given triangle (AB = BC = 4, AC – base), AE – median, AE = 3.

Let’s draw the median BH (O is the point of intersection of AE and BH). Since ABC is an isosceles triangle, BH will also be the height.

By the property of the medians of the triangle, AE and BH intersect in a ratio of 2 to 1.

Hence, AO refers to EO as 2 to 1. Since AE = 3, then AO = 2, EO = 1.

Let us denote the segment OH by the letter x, then BO will be equal to 2x.

Let us express AH from the triangle AOН (this is a right-angled triangle, since BH is perpendicular to AС) according to the Pythagorean theorem: AH² = AO² – OH² = 4 – x².

Let us express AH from the triangle ABH (this is a right-angled triangle, since BH is perpendicular to AC) according to the Pythagorean theorem: AH² = AB² – BH² = 16 – (3x) ² (BH = BO + OH = 2x + x = 3x).

We equate the obtained expressions: 4 – x² = 16 – 9x².

9x² – x² = 16 – 4; 8x² = 12; x² = 12/8 = 3/2.

Since AH² = 4 – x², then AH² = 4 – 3/2 = 4 – 1.5 = 2.5 = 5/2.

Hence, AH = √ (5/2).

Therefore, AC = 2 * AH = 2 * √ (5/2) = (2 * √5) / √2 = (2 * √5 * √2) / (√2 * √2) = (2 * √5 * √2) / 2 = √ (5 * 2) = √10.

Answer: The base of the triangle is √10.