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

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 30cm and 8cm, and the lateral side of the second triangle is 34cm. Find the perimeter of the second triangle.

If in two isosceles triangles the angles at the apex are equal, then these triangles are similar, since at the base, their corresponding angles will also be equal.

In order to find the perimeter of triangle A1B1C1, you need to find the length of the side of the triangle ABC.

Consider triangle ABH, which is rectangular. The segment AH is equal to half of the side of the AC (since the triangle is isosceles):

AH = AC / 2;

AH = 30/2 = 15 cm.

Now, behind the Pythagorean theorem, we find the lateral side AB:

AB ^ 2 = BH ^ 2 + AH ^ 2;

AB ^ 2 = 15 ^ 2 + 8 ^ 2 = 225 + 64 = 289;

AB = √289 = 17 cm.

Now let’s find the coefficient of similarity (the ratio of the respective sides):

k = A1B1 / AB;

k = 34/17 = 2.

To calculate the unknown side A1C1, it is necessary to multiply the corresponding side by the similarity coefficient:

A1C1 = AC ∙ k;

A1C1 = 30 ∙ 2 = 60 cm.

The sum of all sides of the triangle A1B1C1 is equal to:

P = A1B1 + B1C1 + A1C1;

P = 34 + 34 + 60 = 128 cm.

Answer: the dimension of the A1B1C1 triangle is 128 cm.