In two isosceles triangles, the angles opposite to the bases are equal. The base and the height drawn

In two isosceles triangles, the angles opposite to the bases are equal. The base and the height drawn to it of the first triangle are respectively 30 cm and 8 cm, and the side of the second triangle is 34 cm. Find the perimeter of the second triangle.

Let’s designate triangles ABC and A1B1C1. АС, А1С1 – bases, angles В and В1 – tops of isosceles triangles, ВН, В1Н1 – heights. By the condition ∠ В = ∠ В1, which means that the angles at the base are also equal, and the triangle is similar.
Consider a right-angled triangle BCH:
BH = 8 (cm);
CH = AC / 2 = 30/2 = 15 (cm).
We find the hypotenuse BC, it is the lateral side of the triangle ABC.
BC = √ (BH² + CH²) = √64 + 225 = √289 = 17 (cm).
The side side of A1B1C1 is known by condition, we find the similarity coefficient:
k = B1C1 / BC = 34/17 = 2.
Find the base of the second triangle:
A1C1 = AC * 2 = 30 * 2 = 60 (cm).
Find the perimeter of the second triangle:
P2 = A1B1 + B1C1 + A1C1 = 34 + 34 + 60 = 128 (cm).
Answer: the perimeter of the second triangle is 128 cm.