Question

Дан ромб ABCD. BM перпендикулярна AD, BM = 8, Pdoc = 10, Найти P ромба.

Answer 1

Ромб ABCD, его высота ВМ=8, диагонали перпендикулярны и в точке пересечения делятся пополам.
Пусть сторона ромба равна а, половины диагоналей равны Х и Y.
Площадь ромба - произведение стороны на высоту, высота=8.Тогда Sр=8а.
Площадь треугольника DОС, образованного половинами диагоналей и стороной, равна 1/4 площади ромба, то есть 2а.
Тогда имеем: Х+Y+а=10 (периметр треугольника DОС) или
X+Y=10-a. В треугольнике DOC: X²+Y²=a² (по Пифагору). Sdoc=8а/4=2а.
Но Sdoc = (1/2)Х*Y, отсюда Х*Y=4а.
Итак, имеем:
 (1) X+Y=10-a
 (2) X²+Y²=a²
 (3) X*Y=4a.
Возведем (1) в квадрат, тогда (X+Y)²=(10-a)² или Х²+2ХY+Y²=100-20a+a².
Вставим сюда (2) и (3):  а²+8а=100-20a+a² или 28а=100, отсюда а=25/7.
Тогда периметр ромба равен 4*25/7=100/7 = 14и2/7.

Answer 2

Given Information:

We are given a rhombus ABCD, where BM is perpendicular to AD, BM = 8, and Pdoc = 10. We need to find the perimeter (P) of the rhombus.

Solution:

To find the perimeter of the rhombus, we need to know the lengths of its sides. Let's analyze the given information and use it to find the missing lengths.

From the given information, we know that BM is perpendicular to AD and BM = 8. This means that BM is the height of the rhombus.

We also know that Pdoc = 10, which is the length of the diagonal of the rhombus.

To find the lengths of the sides of the rhombus, we can use the properties of a rhombus. In a rhombus, the diagonals are perpendicular bisectors of each other. This means that the diagonal Pdoc divides the rhombus into two congruent right-angled triangles.

Let's denote the length of the sides of the rhombus as a. Using the Pythagorean theorem in one of the right-angled triangles, we can find the value of a.

Step 1: Apply the Pythagorean theorem in one of the right-angled triangles formed by the diagonal Pdoc and the side a of the rhombus.

Using the Pythagorean theorem, we have: a^2 = (BM/2)^2 + (Pdoc/2)^2

Substituting the given values, we get: a^2 = (8/2)^2 + (10/2)^2 a^2 = 4^2 + 5^2 a^2 = 16 + 25 a^2 = 41

Taking the square root of both sides, we get: a = √41

Step 2: Calculate the perimeter (P) of the rhombus.

The perimeter of a rhombus is given by the formula: P = 4a

Substituting the value of a, we get: P = 4 * √41

Therefore, the perimeter of the given rhombus is 4√41.

Answer:

The perimeter (P) of the given rhombus ABCD is 4√41.