# In triangle ABC, the height AD is 4 seconds less than the side BC. The AC side is 5 cm.

**In triangle ABC, the height AD is 4 seconds less than the side BC. The AC side is 5 cm. Find the perimeter of the ABC triangle if its area is 16 cm ^ 2.**

Let the length of the height AD = X cm, then, by condition, BC = X + 4 cm.

The area of the triangle will be equal to: Sавс = ВС * АD / 2.

16 = (X + 4) * X / 2

32 = X ^ 2 + 4 * X.

X ^ 2 + 4 * X – 32 = 0.

Let’s solve the quadratic equation.

X1 = -8. Doesn’t match because <0.

X2 = 4 cm.

AD = 4 cm, then BC = 4 + 4 = 8 cm.

From the right-angled triangle ACD, according to the Pythagorean theorem, we determine the length of the leg CD.

CD ^ 2 = AC ^ 2 – AD ^ 2 = 25 – 16 = 9.

CD = 3 cm.

Then the segment ВD = ВС – СD = 8 – 3 = 5 cm.

From the right-angled triangle ABD, by the Pythagorean theorem, we determine the length of the hypotenuse AB.

AB ^ 2 = AD ^ 2 + BD ^ 2 = 16 + 25 – 41.

AB = √41 cm.

Let’s define the perimeter of the triangle ABC.

Ravs = AC + BC + AB = 5 + 8 + √41 = 13 + √41 cm.

Answer: the perimeter of the triangle is 13 + √41 cm.