ОА perpendicular ОВ perpendicular to ОА, ОА = ОВ = 6cm, OC = 8cm Р = AB + BC + CA-?

Let us find the hypotenuse of the BC of the right-angled triangle of the COB according to the Pythagorean theorem (the square of the hypotenuse is equal to the sum of the squares of the legs):

ВС² = OС² + ОВ²;

ВС² = 8² + 6²;

ВС² = 64 + 36;

ВС² = 100;

BC = √100;

BC = 10 cm.

Let us find by the Pythagorean theorem the hypotenuse AB of the right-angled triangle AOB:

AB² = ОВ² + ОА²;

AB² = 6² + 6²;

AB² = 36 + 36;

AB² = 72;

AB = √72;

AB = 8√8 cm.

AC = OC + OA = 8 + 6 = 14 cm.

The perimeter of the triangle ABC is:

P = AB + AC + BC;

P = 8√8 + 14 + 10;

P = 24 + 8√8 cm

Answer: 24 + 8√8 cm