Find the sides of a right-angled triangle in which: a) the hypotenuse is 10 cm, the difference between the legs is 2 cm;

Find the sides of a right-angled triangle in which: a) the hypotenuse is 10 cm, the difference between the legs is 2 cm; b) the hypotenuse is 26 cm, and the ratio of the legs is 5:12.

According to the Pythagorean theorem, in a right-angled triangle, the hypotenuse in a square is equal to the sum of the squares of its legs.
a) Since the hypotenuse AB is 10 cm, and the leg BC is 2 centimeters larger than the leg AC, we will express them as follows:
x is the length of the AC leg;
x + 2 – length of the BC leg;
x^2 + (x + 2) 2 = 10^2;
x^2 + x^2 + 2x + 2x + 4 = 100;
2x^2 + 4x + 4 = 100;
2x^2 + 4x – 96 = 0;
x^2 + 2x – 48 = 0;
D = b2 – 4ac;
D = 2^2 – 2 1 (-48) = 4 – (-192) = 4 + 192 = 196 = 14^2.
Since in this case there can be no negative number, then:
x = (-b + √D) / 2a;
x = (-2 + 14) / 2 = 12/2 = 6;
AC = 6 cm;
BC = 6 + 2 = 8 cm.
Answer: the legs are 6 cm and 8 cm.

b) Since the hypotenuse is 26 cm, and the legs are at 5:12, then:
5x – length of the AC leg;
12x – BC leg length;
5x^2 + 12x^2 = 26^2;
25×2^ + 144x^2 = 676;
169x^2 = 676;
x^2 = 676/169 = 4;
x = √4 = 2;
AC = 5 2 = 10 cm;
AB = 12 2 = 24 cm.
Answer: the legs are 10 cm and 24 cm.