Question

Помогите решить задачу, В ромбе АВСD,АВ = 10см,ВD=12cм,прямая МС перпендикулярна ромбу,найти АМ если т.М удалена от ромба на 16 см!!! пожалуйста помогите решать))

Answer 1

АМ ---это гипотенуза прямоугольного треугольника АСМ, в котором СМ=16 по условию, АС ---диагональ ромба
диагонали ромба точкой пересечения (О) делятся пополам и диагонали ромба перпендикулярны...
получили прямоугольный треугольник ВОА, в котором ВО=6, АВ=10
по т.Пифагора АО^2 = 10^2 - 6^2 = 64
AO=8
AC=16
по т.Пифагора AM^2 = 16^2 + 16^2 = 2*16^2
AM = 16V2

Answer 2

Problem Analysis

We are given a rhombus ABCD, where AB = 10 cm and BD = 12 cm. The line MC is perpendicular to the rhombus, and point M is located 16 cm away from the rhombus. We need to find the length of AM.

Solution

To solve this problem, we can use the Pythagorean theorem and the properties of a rhombus.

Let's start by drawing a diagram to visualize the problem:

       A
      / \
     /   \
    /     \
   /       \
  /         \
 B-----------C
  \         /
   \       /
    \     /
     \   /
      \ /
       D

Since ABCD is a rhombus, we know that the diagonals are perpendicular bisectors of each other. Therefore, line AC is perpendicular to line BD.

Let's label the midpoint of BD as point E. Since AC is perpendicular to BD, we can conclude that AE is the altitude of triangle ABD.

Now, let's consider triangle ABD. We know that AB = 10 cm and BD = 12 cm. Using the Pythagorean theorem, we can find the length of AE:

AE^2 + BE^2 = AB^2 AE^2 + (BD/2)^2 = AB^2 AE^2 + 6^2 = 10^2 AE^2 + 36 = 100 AE^2 = 100 - 36 AE^2 = 64 AE = √64 AE = 8 cm

Now, let's consider triangle AMC. We know that MC = 16 cm and AE = 8 cm. Using the Pythagorean theorem, we can find the length of AM:

AM^2 = MC^2 + AC^2 AM^2 = 16^2 + 8^2 AM^2 = 256 + 64 AM^2 = 320 AM = √320 AM = 8√5 cm

Therefore, the length of AM is 8√5 cm.

Answer

The length of AM is 8√5 cm.