Find the median of a right-angled triangle, drawn to the hepotenuse

Find the median of a right-angled triangle, drawn to the hepotenuse, if the perimeter of the triangle is 60 cm and the difference between the legs is 14 cm.

Since the difference between the legs a and b of this triangle is 14, then a – b = 14. Hence, b = a – 14. The hypotenuse is equal to the difference between the perimeter and the sum of the lengths of the legs: c = 60 – a – b = 60 – a – a + 14 = 74 – 2a.

The square of the hypotenuse is equal to the sum of the squares of the legs, which means:

a ^ 62 + b ^ 2 = c ^ 2;

a ^ 2 + (a – 14) ^ 2 = (74 – 2a) ^ 2;

a ^ 2 + a ^ 2 + 196 – 28a = 5476 + 4a ^ 2 – 296a;

2a ^ 2 – 268a + 5280 = 0;

a ^ 2 – 134a + 2640 = 0.

D = 1342 – 4 * 1 * 2640 = 17956 – 10560 = 7396 = 862;

a1 = (134 + 86) / 2 = 110 – this solution does not satisfy the condition of the problem, since the perimeter is 60 cm, therefore the leg cannot be 110 cm.

a2 = (134 – 86) / 2 = 48/2 = 24 cm.

c = 74 – 2a = 74 – 2 * 24 = 74 – 48 = 26 cm.

The median of a right-angled triangle, drawn to the hypotenuse, is equal to half of it, therefore:

m = c / 2 = 26/2 = 13 cm.