In a right-angled triangle ACB (C-line) It is known AB = 13, the median AM (on the side of the CB) = √61.

In a right-angled triangle ACB (C-line) It is known AB = 13, the median AM (on the side of the CB) = √61. Find the ABC perimeter and the ABC area.

CM = ВM (AM – conditional median). Introduce the variable x, denote it CM and BM.
Consider a right-angled triangle AСM and write the leg AC according to the Pythagorean theorem:
AC² = AM² – CM² = 61 – x².
Consider a right-angled triangle ABC and write down the AC leg in it:
AC² = AB² – CB² = 13² – (2x) ² = 169 – 4x².
Let’s make the equation:
61 – x² = 169 – 4x².
3x² = 108
x² = 36
x = 6.
AC = √ (169 – 4x²) = √25 = 5;
CB = 2x = 2 * 6 = 12.
P ABC = 13 + 12 + 5 = 30;
S ABC = 1/2 * AC * CB = 1/2 * 5 * 12 = 30.
Answer: The perimeter is 30 linear units, the area is 30 square units.