Question

В треугольнике АВС угол С равен 45 градусов, АВ= 10 см, а высота АD делит сторону СВ на отрезки СD=8 см, DВ= 6 см. Найти площадь тругольника и высоту, проведенную к стороне АВ Даю ооочень много балов, только решите без Син и кос ПРАВИЛЬНО!

Answer 1

Решаем без синусов и косинусов

Answer 2

Given Information:

We are given the following information about triangle ABC: - Angle C is 45 degrees. - Side AB is 10 cm. - The height AD divides side CB into segments CD and DB, with CD measuring 8 cm and DB measuring 6 cm.

Finding the Area of the Triangle:

To find the area of triangle ABC, we can use the formula: Area = (1/2) * base * height.

In this case, the base of the triangle is AB, which measures 10 cm. The height of the triangle is AD.

To find the height AD, we can use the fact that the height divides side CB into segments CD and DB. We can use the Pythagorean theorem to find the length of AD.

Using the Pythagorean theorem, we have: AD^2 = CD^2 - DB^2

Substituting the given values, we have: AD^2 = 8^2 - 6^2

Simplifying, we get: AD^2 = 64 - 36 AD^2 = 28

Taking the square root of both sides, we find: AD = √28 AD ≈ 5.29 cm

Now that we have the height AD, we can calculate the area of the triangle: Area = (1/2) * AB * AD Area = (1/2) * 10 * 5.29 Area ≈ 26.45 cm^2

Therefore, the area of triangle ABC is approximately 26.45 cm^2.

Finding the Height Drawn to Side AB:

To find the height drawn to side AB, we can use the formula: Height = 2 * (Area / base).

In this case, the base is AB, which measures 10 cm. The area of the triangle is approximately 26.45 cm^2.

Using the formula, we have: Height = 2 * (26.45 / 10) Height ≈ 5.29 cm

Therefore, the height drawn to side AB is approximately 5.29 cm.

Please note that the calculations are approximate due to rounding.

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