The point on the hypotenuse, equidistant from both legs, divides the hypotenuse into segments 30 cm

The point on the hypotenuse, equidistant from both legs, divides the hypotenuse into segments 30 cm and 40 cm long. find the lengths of the legs.

BC = BE + EC = 40 cm + 30 cm = 70 cm.

Point F is equidistant from legs AB and AC, therefore, the lengths of perpendiculars dropped from point F to these legs are equal:

DE = EF = a;

Triangles ABC, BDE and CEF are rectangular and the like.

FC ^ 2 = EC ^ 2 – EF ^ 2;

FC = √ (EC ^ 2 – EF ^ 2) = √ (30 ^ 2 – a ^ 2);

DE / BE = FC / EC;

a / 40 = √ (30 ^ 2 – a ^ 2) / 30;

30a = 40√ (30 ^ 2 – a ^ 2);

900 * a ^ 2 = 1600 * (30 ^ 2 – a ^ 2);

900 * a ^ 2 = 1600 * 30 ^ 2 – 1600 * a ^ 2

2500 * a ^ 2 = 1600 * 30 ^ 2

25 * a ^ 2 = 16 * 30 ^ 2;

5a = 4 * 30 = 120;

a = 24 cm.

DE / BE = AC / BC;

24/40 = AC / 70;

AC = (24 * 70) / 40 = 42 cm;

EF / AB = 30/70;

24 / AB = 30/70;

AB = (24 * 70) / 30 = 56;

Answer: 42 cm and 56 cm.