Question

9. В трапеции ABCD диагонали пересекаются в точке О. Точка пересечения диагоналей трапеции делит диагональ АС на отрезки 6см и 4см. Найдите основания трапеции AD и BC, если их разность равна 25см. Выполните чертеж по условию задачи Срочно пжпжпжпжпжп

Answer 1

50 см, 75 см

Треугольник ВОС подобен треугольнику DOA: <BOC=<DOA - как вертикальные, < ВСО = < DAO - как внутренние накрест лежащие при параллельных прямых ВС и AD и секущей АС. Первый признак подобия.

Пусть ВС=х см, тогда AD= (x+25) см

Из подобия треугольников следует подобие соответствующих сторон :

$\dfrac{BC}{AD} = \dfrac{OC}{OA} \\ \\ \frac{x}{x + 25} = \frac{4}{6} \\ \\ 6x = 4(x + 25) \\ 6x = 4x + 100 \\ 2x = 100 \\ x = 50$

BC=50см, AD= 25+50=75см

Answer 2

Problem Analysis

We are given a trapezoid ABCD with diagonals intersecting at point O. The point of intersection of the diagonals divides the diagonal AC into segments of 6 cm and 4 cm. We need to find the lengths of the bases AD and BC, given that their difference is 25 cm.

Solution

Let's denote the length of AD as x and the length of BC as y.

According to the given information, the difference between the lengths of the bases AD and BC is 25 cm. Therefore, we can write the equation:

x - y = 25 We also know that the point of intersection of the diagonals divides the diagonal AC into segments of 6 cm and 4 cm. This means that AO = 6 cm and OC = 4 cm.

Using the property of similar triangles, we can write the following proportion:

AD/OA = BC/OC Substituting the given values, we get:

x/6 = y/4

Cross-multiplying, we have:

4x = 6y

Simplifying, we get:

2x = 3y Now, we have a system of two equations with two variables:

Equation 1: x - y = 25 Equation 2: 2x = 3y We can solve this system of equations to find the values of x and y.

Solving the System of Equations

To solve the system of equations, we can use the substitution method or the elimination method. Let's use the elimination method:

Multiplying Equation 1 by 2, we get:

2x - 2y = 50

Now, we can subtract Equation 2 from this new equation:

(2x - 2y) - (2x) = 50 - (3y)

Simplifying, we have:

-2y = 50 - 3y

Adding 3y to both sides, we get:

y = 50 - 3y + 3y

Simplifying further, we have:

4y = 50

Dividing both sides by 4, we find:

y = 12.5

Substituting this value of y back into Equation 1, we can solve for x:

x - 12.5 = 25

Adding 12.5 to both sides, we get:

x = 37.5

Therefore, the lengths of the bases AD and BC are 37.5 cm and 12.5 cm, respectively.

Drawing the Trapezoid

Here is a rough sketch of the trapezoid ABCD:

      A ________ B
       |        |
       |        |
       |        |
       |________|
      D          C

Please note that this is a rough sketch and not drawn to scale.

I hope this helps! Let me know if you have any further questions.