Question

Задача. в треугольнике АВС угол А- тупой, ВК и СД- высоты, ВК=12 СМ, АК=9 СМ, СД=10 СМ. найти: площадь треугольника АВС помогите,пожалуйста,при чем надо это найти НЕ через подобные треугольники

Answer 1

$AB= \sqrt{BK^2+AK^2} = 13$
уголBAK=углуCAD (вертикальные)
Поэтому sinBAK=sinCAD; $\frac{12}{13}= \frac{10}{x} \\ x= \frac{130}{12} \\ S= \frac{1}{2}*12*( \frac{130}{12}+9 )- \frac{1}{2}*12*9= 65$
P.S. Благодарность и лучший ответ не помешает.

Answer 2

Finding the Area of Triangle ABC

To find the area of triangle ABC, we can use the given information about the triangle and apply the formula for the area of a triangle.

Given: - Angle A is obtuse - VK and SD are altitudes - VK = 12 cm - AK = 9 cm - SD = 10 cm

Solution

We can use the formula for the area of a triangle using the base and height. The area of a triangle is given by the formula: Area = 0.5 * base * height.

First, we need to find the base and height of the triangle. Since VK and SD are altitudes, we can use them as the base and height of the triangle.

Using the given information: - VK = 12 cm - SD = 10 cm

We can calculate the area of the triangle using the formula: Area = 0.5 * base * height

Substitute the values: Area = 0.5 * VK * SD

Now, we can calculate the area using the given values: Area = 0.5 * 12 cm * 10 cm

Area = 60 square cm

So, the area of triangle ABC is 60 square cm.

I hope this helps! If you have any other questions, feel free to ask.