# The two sides of the triangle are 6 and 8 cm. The medians drawn to these sides are mutually

The two sides of the triangle are 6 and 8 cm. The medians drawn to these sides are mutually perpendicular. Find the third side of the triangle.

Given an arbitrary △ ABC: AC = 6 cm, BC = 8 cm, AK and BM are medians intersecting at point O.
1. The length of the median is calculated by the formula:
m = (√ (2b² + 2c² – a²)) / 2,
where a is the side of the triangle to which the median is drawn, b and c are the other sides of the triangle.
Let us express the length of the median AK:
AK = (√ (2AB² + 2AC² – BC²)) / 2 = (√ (2AB² + 2 * 6² – 8²)) / 2 = (√ (2AB² + 2 * 36 – 64)) / 2 = (√ (2AB² + 72 – 64)) / 2 = (√ (2AB² + 8)) / 2.
Let us express the length of the median BM:
BM = (√ (2AB² + 2BC² – AC²)) / 2 = (√ (2AB² + 2 * 8² – 6²)) / 2 = (√ (2AB² + 2 * 64 – 36)) / 2 = (√ (2AB² + 128 – 36)) / 2 = (√ (2AB² + 92)) / 2.
2. The medians are divided by the point of intersection in a ratio of 2: 1 starting from the top.
2.1. OA / OK = 2/1;
OA + OK = AK.
We get the system of equations:
OA / OK = 2/1;
OA + OK = (√ (2AB² + 8)) / 2.
In the second equation, we express OK:
OK = (√ (2AB² + 8)) / 2 – OA.
Substitute the expression into the first equation:
OA / ((√ (2AB² + 8)) / 2 – OA) = 2/1;
2 ((√ (2AB² + 8)) / 2 – OA) = OA (proportional);
√ (2AB² + 8) – 2OA = OA;
3OA = √ (2AB² + 8);
OA = (√ (2AB² + 8)) / 3.
2.2. OB / OM = 2/1;
OB + OM = BM.
We get the system of equations:
OB / OM = 2/1;
OB + OM = (√ (2AB² + 92)) / 2.
In the second equation, we express OM:
OM = (√ (2AB² + 92)) / 2 – OB.
Substitute the expression into the first equation:
OB / ((√ (2AB² + 92)) / 2 – OB) = 2/1;
2 (√ (2AB² + 92) / 2 – OB) = OB (proportional);
√ (2AB² + 92) – 2OB = OB;
3OB = √ (2AB² + 92);
OB = (√ (2AB² + 92)) / 3.
3. Consider △ AOB: ∠AOB = 90 ° (since the medians are perpendicular), OA = (√ (2AB² + 8)) / 3 and OB = (√ (2AB² + 92)) / 3 – legs, AB – hypotenuse ( since lie opposite a right angle).
By the Pythagorean theorem:
AB = √ (OA² + OB²);
AB = √ (((√ (2AB² + 8)) / 3) ² + ((√ (2AB² + 92)) / 3) ²);
AB = √ ((2AB² + 8) / 9 + (2AB² + 92) / 9);
AB = √ ((2AB² + 8 + 2AB² + 92) / 9);
AB = √ ((4AB² + 100) / 9) (square both sides of the equation);
AB² = (√ ((4AB² + 100) / 9)) ²;
(4AB² + 100) / 9 = AB²;
9AB² = 4AB² + 100 (in proportion);
9AB² – 4AB² = 100;
5AB² = 100;
AB² = 100/5;
AB² = 20;
AB = √20;
AB = 2√5 cm.